{"id":3605,"date":"2015-05-06T03:51:00","date_gmt":"2015-05-06T03:51:00","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=3605"},"modified":"2020-12-31T23:06:42","modified_gmt":"2020-12-31T23:06:42","slug":"multiple-equilibria-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/chapter\/multiple-equilibria-2\/","title":{"raw":"Coupled Equilibria","rendered":"Coupled Equilibria"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe examples of systems involving two (or more) coupled chemical equilibria<\/li>\r\n \t<li>Calculate reactant and product concentrations for coupled equilibrium systems<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idm703568\">As discussed in preceding chapters on equilibrium,\u00a0<em>coupled equilibria<\/em>\u00a0involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.<\/p>\r\n<p id=\"fs-idp11174336\">An environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean\u2019s coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates).<\/p>\r\n\u00a0Carbon dioxide in the air dissolves in sea water, forming carbonic acid (H<sub>2<\/sub>CO<sub>3<\/sub>). The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions [latex]\\left({\\text{HCO}}_{3}{}^{-}\\right)[\/latex], which can further ionize into more hydrogen ions and carbonate ions [latex]\\left({\\text{CO}}_{3}{}^{2-}\\right):[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}{\\text{CO}}_{2}\\left(g\\right)&amp;\\rightleftharpoons&amp;{\\text{CO}}_{2}\\left(aq\\right)\\\\{\\text{CO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}&amp;\\rightleftharpoons&amp;{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\\\{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)&amp;\\rightleftharpoons&amp;{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)\\\\{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)&amp;\\rightleftharpoons&amp;{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\r\nThe excess H<sup>+<\/sup> ions make seawater more acidic. Increased ocean acidification can then have negative impacts on reef-building coral, as they cannot absorb the calcium carbonate they need to grow and maintain their skeletons (Figure\u00a01). This in turn disrupts the local biosystem that depends upon the health of the reefs for its survival. If enough local reefs are similarly affected, the disruptions to sea life can be felt globally. The world\u2019s oceans are presently in the midst of a period of intense acidification, believed to have begun in the mid-nineteenth century, and which is now accelerating at a rate faster than any change to oceanic pH in the last 20 million years.\r\n\r\n[caption id=\"attachment_5431\" align=\"aligncenter\" width=\"975\"]<img class=\"size-full wp-image-5431\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042406\/CNX_Chem_15_03_CoralReef.jpg\" alt=\"This figure contains two photographs of coral reefs. In a, a colorful reef that includes hues of purple and pink corals is shown in blue green water with fish swimming in the background. In b, grey-green mossy looking coral is shown in a blue aquatic environment. This photo does not have the colorful appearance or fish that were shown in figure a.\" width=\"975\" height=\"386\" \/> Figure\u00a01. Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonite skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by \u201cprilfish\u201d\/Flickr)[\/caption]\r\n\r\n<div class=\"textbox\">\r\n\r\nLearn more about ocean acidification and how it affects other marine creatures.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=5559071&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=kxPwbhFeZSw&amp;video_target=tpm-plugin-0cebssbe-kxPwbhFeZSw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/OceanAcidification_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Ocean Acidification\" here (opens in new window)<\/a>.\r\n\r\n<\/div>\r\n<div class=\"textbox\">This site has detailed information about <a href=\"http:\/\/www.teachoceanscience.net\/teaching_resources\/education_modules\/coral_reefs_and_climate_change\/how_does_climate_change_affect_coral_reefs\/\" target=\"_blank\" rel=\"noopener\">how ocean acidification specifically affects coral reefs<\/a>.<\/div>\r\nThe dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (Figure 2), a sparingly soluble ionic compound whose dissolution equilibrium is\r\n<p style=\"text-align: center;\">[latex]{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\left(s\\right)\\longrightarrow 5{\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+3{\\text{PO}}_{4}^{3-}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_5434\" align=\"aligncenter\" width=\"650\"]<img class=\"wp-image-5434 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042527\/CNX_Chem_15_03_Apatite.jpg\" alt=\"This figure includes an image of two large light blue apatite crystals in a mineral conglomerate that includes white, grey, and tan crystals. The blue apatite crystals have a dull, dusty, or powdered appearance.\" width=\"650\" height=\"383\" \/> Figure\u00a02. Crystal of the mineral hydroxylapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, is shown here. Pure apatite is white, but like many other minerals, this sample is colored because of the presence of impurities.[\/caption]\r\n\r\nThis compound dissolved to yield two different basic ions: triprotic phosphate ions\r\n<p style=\"text-align: center;\">[latex]{\\text{PO}}_{4}{}^{3-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{2-}+{\\text{H}}_{2}\\text{O}[\/latex]\r\n[latex]{\\text{PO}}_{4}{}^{2-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}+{\\text{H}}_{2}\\text{O}[\/latex]\r\n[latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{3}{\\text{PO}}_{4}+{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\r\nand monoprotic hydroxide ions:\r\n<p style=\"text-align: center;\">[latex]{\\text{OH}}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow 2{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\r\nOf the two basic productions, the hydroxide is, of course, by far the stronger base (it\u2019s the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added NaF or SnF<sub>2<\/sub>\u00a0that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:\r\n<p style=\"text-align: center;\">[latex]\\text{NaF}+{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\rightleftharpoons {\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{F}+{\\text{Na}}^{+}+{\\text{OH}}^{-}[\/latex]<\/p>\r\nThe weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.\r\n<div class=\"textbox shaded\">\r\n<h3>Role of Fluoride in Preventing Tooth Decay<\/h3>\r\nAs we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure\u00a03).\r\n\r\n[caption id=\"attachment_5435\" align=\"aligncenter\" width=\"650\"]<img class=\"wp-image-5435 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042617\/CNX_Chem_15_03_Toothpaste.jpg\" alt=\"A tube of toothpaste\" width=\"650\" height=\"269\" \/> Figure 3. Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).[\/caption]\r\n\r\nUnfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm\u00a0(4 mg\/L)\u00a0of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.\r\n\r\n<\/div>\r\n<p id=\"fs-idp19645632\">The solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion\u00a0<span class=\"os-math-in-para\"><span id=\"MathJax-Element-554-Frame\" class=\"MathJax\" style=\"overflow: initial; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; letter-spacing: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px;\" role=\"presentation\"><span id=\"MathJax-Span-18138\" class=\"math\"><span id=\"MathJax-Span-18139\" class=\"mrow\"><span id=\"MathJax-Span-18140\" class=\"semantics\"><span id=\"MathJax-Span-18141\" class=\"mrow\"><span id=\"MathJax-Span-18142\" class=\"mrow\"><span id=\"MathJax-Span-18143\" class=\"mtext\">Al<\/span><span id=\"MathJax-Span-18144\" class=\"msub\"><span id=\"MathJax-Span-18145\" class=\"mrow\"><span id=\"MathJax-Span-18146\" class=\"mo\">(<\/span><span id=\"MathJax-Span-18147\" class=\"mtext\">OH<\/span><span id=\"MathJax-Span-18148\" class=\"mo\">)<\/span><\/span><span id=\"MathJax-Span-18149\" class=\"mn\">4<\/span><\/span><span id=\"MathJax-Span-18150\" class=\"msup\"><span id=\"MathJax-Span-18151\" class=\"mrow\"><\/span><span id=\"MathJax-Span-18152\" class=\"mtext\">\u2212<\/span><\/span><span id=\"MathJax-Span-18153\" class=\"mo\">.<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"MJX_Assistive_MathML\" role=\"presentation\">Al(OH)4\u2212.<\/span><\/span><\/span><\/p>\r\n<span id=\"fs-idp137004736\" class=\"scaled-down\"><img id=\"7\" src=\"https:\/\/openstax.org\/resources\/ec0583eb0fb6238971af3b9efbce4e52b7ee02c8\" alt=\"An H atom is bonded to an O atom. The O atom has 2 dots above it and 2 dots below it. The O atom is bonded to an A l atom, which has three additional O atoms bonded to it as well. Each of these additional O atoms has 4 dots arranged around it, and is bonded to an H atom. This entire molecule is contained in brackets, to the right of which is a superscripted negative sign.\" \/><\/span>\r\n<p id=\"fs-idm55619648\">The equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of Al(OH)<sub>3<\/sub>.<\/p>\r\n\r\n<center>[latex]\\text{Al(OH)}_3 (s) \\rightleftharpoons \\text{Al}^{3+} (aq) + 3\\text{OH}^- (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_{sp} = 2 \\times 10^{-32}[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]\\text{Al}^{3+} (aq) + 4\\text{OH}^- (aq) \\rightleftharpoons \\text{Al(OH)}_4^{-} (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_{f} = 1.1 \\times 10^{33}[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]\\text{Net: Al(OH)}_3 (s) + \\text{OH}^- (aq) \\rightleftharpoons \\text{Al(OH)}_4^{-} (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K = K_{sp} K_f = 22[\/latex]<\/center>\r\n<div class=\"textbox examples\">\r\n<h3>Example 1: Increased solubility in acidic solutions<\/h3>\r\nCompute and compare the molar solublities for aluminum hydroxide, Al(OH)<sub>3<\/sub>, dissolved in (a) pure water and (b) a buffer containing 0.100\u00a0<em>M<\/em>\u00a0acetic acid and 0.100\u00a0<em>M<\/em>\u00a0sodium acetate.\r\n<h4>Solution<\/h4>\r\n[reveal-answer q=\"576838\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"576838\"]\r\n(a) The molar solubility of aluminum hydroxide in water is computed considering the dissolution equilibrium only as demonstrated in several previous examples:\r\n\r\n<center>[latex]\\text{Al(OH)}_3 (s) \\rightleftharpoons \\text{Al}^{3+} (aq) + 3\\text{OH}^- (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_{sp} = 2 \\times 10^{-32}[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]\\text{molar solubility in water} = [\\text{Al}^{3+}] = (2\\times 10^{-32} \/ 27)^{1\/4} = 5\\times 10^{-9} M[\/latex]<\/center>\r\n<p id=\"fs-idm324374096\">(b) The concentration of hydroxide ion of the buffered solution is conveniently calculated by the Henderson-Hasselbalch equation:<\/p>\r\n\r\n<center>[latex]\\text{pH} = \\text{pKa} +\\text{log}[\\text{CH}_3\\text{COO}^-] \/ [\\text{CH}_3\\text{COOH}][\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]\\text{pH} = 4.74 +\\text{log} (0.100\/0.100) = 4.74[\/latex]<\/center>\r\n<p id=\"fs-idm378269776\">At this pH, the concentration of hydroxide ion is<\/p>\r\n\r\n<center>[latex]\\text{pOH} = 14.00-4.74=9.26[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex][\\text{OH}^-] = 10^{-9.26} = 5.5\\times 10^{-10}[\/latex]<\/center>\r\n<p id=\"fs-idm373349040\">The solubility of Al(OH)<sub>3<\/sub>\u00a0in this buffer is then calculated from its solubility product expressions:<\/p>\r\n\r\n<center>[latex]K_{sp} = [\\text{Al}^{3+}][\\text{OH}^{-}]^3[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]\\text{molar solubility in buffer} = [\\text{Al}^{3+}] = K_{sp}\/[\\text{OH}^{-}]^3 = (2\\times 10^{-32}) \/ (5\\times 10^{-10})^3 = 1.2\\times 10^{-4} M[\/latex]<\/center>\r\n<p id=\"fs-idm329760656\">Compared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low).<\/p>\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nWhat is the solubility of aluminum hydroxide in a buffer comprised of 0.100\u00a0<em>M<\/em>\u00a0formic acid and 0.100\u00a0<em>M<\/em>\u00a0sodium formate?\r\n<div id=\"fs-idp14780320\" class=\"ui-has-child-title\"><header>\r\n<h4 class=\"os-title\"><span id=\"8\" class=\"os-title-label\">ANSWER:<\/span><\/h4>\r\n<\/header><section>[reveal-answer q=\"576839\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"576839\"]\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idp15140704\">0.1\u00a0<em>M<\/em><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0Multiple Equilibria<\/h3>\r\nUnexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>, called hypo) to form the complex ion [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex] (<em>K<\/em><sub>f<\/sub> = 4.7 [latex]\\times [\/latex] 10<sup>13<\/sup>).\r\n\r\n<img class=\"aligncenter size-full wp-image-5437\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042721\/CNX_Chem_15_03_AgBr_img.jpg\" alt=\"A chemical reaction is shown using structural formulas. On the left, A g superscript plus is followed by a plus sign, the number 2, and a structure in brackets. The structure is composed of a central S atom which has O atoms single bonded above, right, and below. A second S atom is single bonded to the left. Each of these bonded atoms has 6 dots around it. Outside the brackets is a superscript 2 negative. Following a bidirectional arrow is a structure in brackets with a central A g atom. To the left and right, S atoms are single bonded to the A g atom. Each of these S atoms has four dots around it, and an S atom connected with a single bond moving out from the central A g atom, forming the ends of the structure. Each of these atoms has three O atoms attached with single bonds above, below, and at the end of the structure. Each O atom has six dots around it. Outside the brackets is a superscript 3 negative.\" width=\"1000\" height=\"176\" \/>\r\n\r\n&nbsp;\r\n\r\nWhat mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}?[\/latex]\r\n\r\n[reveal-answer q=\"693371\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"693371\"]\r\n\r\nTwo equilibria are involved when AgBr dissolves in a solution containing the [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] ion:\r\n<ul>\r\n \t<li>Reaction 1 (dissolution): [latex]\\text{AgBr}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Br}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{sp}}=5.0\\times {10}^{\\text{-13}}[\/latex]<\/li>\r\n \t<li>Reaction 2 (complexation): [latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}\\left(aq\\right)\\qquad{K}_{\\text{f}}=4.7\\times {10}^{13}[\/latex]<\/li>\r\n<\/ul>\r\nFirst, calculate the concentration of bromide that will result when the 1.00 g of AgBr is completely dissolved via the cited complexation reaction:\r\n<p style=\"text-align: center;\">[latex]1.00\\text{ g AgBr }\/(187.77\\text{ g\/mol})(1\\text{ mol Br}^{-}\/1\\text{ mol AgBr})=0.00532\\text{ mol Br}^{-}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]0.00532\\text{ mol Br}^{-}\/1.00\\text{ L}=0.00532 M\\text{ Br}^{-}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Next, use this bromide molarity and the solubility product for silver bromide to calculate the silver ion molarity in the solution:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Ag}}^{+}\\right]=K_{\\text{sp}}\/ \\left[\\text{Br}^{-}\\right]=5.0\\times{10}^{-13}\/0.00532=9.4\\times {10}^{-11}M[\/latex]<\/p>\r\nBased on the stoichiometry of the complex ion formation, the concentration of complex ion produced is\r\n<p style=\"text-align: center;\">[latex]0.00532-9.4\\times{10}^{-11}=0.00521M[\/latex]<\/p>\r\nUse the silver ion and complex ion concentrations and the formation constant for the complex ion to compute the concentration of thiosulfate ion.\r\n<p style=\"padding-left: 60px;\">[latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]^{2}=\\left[\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{\\text{3-}}\\right]\/\\left[\\text{Ag}^{+}\\right]K_{\\text{f}}=0.00521\/(9.6\\times{10}^{-11})(4.7\\times{10}^{13})=1.15\\times{10}^{-6}[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">[latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]=1.1\\times{10}^{-3}M[\/latex]<\/p>\r\nFinally, use this molar concentration to derive the required mass of sodium thiosulfate:\r\n<p style=\"text-align: center;\">[latex](1.1\\times {10}^{\\text{-3}}\\text{ mol}{\\text{ S}}_{2}{\\text{O}}_{3}{}^{2-}\/\\text{L})\\times(1\\text{ mol}{\\text{ Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}\/1\\text{ mol}{\\text{ S}}_{2}{\\text{O}}_{3}{}^{2-})\\times(158.1\\text{ g}{\\text{ Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}\/\\text{mol})=1.7\\text{ g}[\/latex]<\/p>\r\nThus, 1.00 L of a solution prepared from 1.7 g Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> dissolves 1.0 g of AgBr.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nAgCl(s), silver chloride, is well known to have a very low solubility:\r\n<p style=\"text-align: center;\">[latex]\\text{AgCl}(\\text{s})\\rightleftharpoons\\text{Ag}^{+}(\\text{aq}) + \\text{Cl}^{-}(\\text{aq})\\qquad{K}_{\\text{sp}}=1.77\\times{10}^{-10}[\/latex]<\/p>\r\nAdding ammonia significantly increases the solubility of AgCl because a complex ion is formed:\r\n<p style=\"text-align: center;\">[latex]\\text{Ag}^{+}(\\text{aq})+2\\text{NH}_{3}(\\text{aq})\\rightleftharpoons\\text{Ag(NH}_{3})_{2}{}^{+}(\\text{aq})\\qquad{K}_{\\text{f}}=1.7\\times{10}^{7}[\/latex]<\/p>\r\nWhat mass of NH<sub>3<\/sub> is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of Ag(NH<sub>3<\/sub>)<sub>2<\/sub><sup>+<\/sup>?\r\n\r\n[reveal-answer q=\"88870\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"88870\"]\r\n\r\n1.00 L of a solution prepared with 4.81 g NH<sub>3<\/sub> dissolves 2.0 g of AgCl.[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\n<div class=\"PageContent-ny9bj0-0 iapMdy\">\r\n<div id=\"main-content\" class=\"MainContent__HideOutline-sc-6yy1if-0 bdVAq\">\r\n<div id=\"composite-page-59\" class=\"os-eoc os-summary-container\"><section id=\"fs-idm305648\" class=\"summary\">\r\n<p id=\"fs-idm304720\">Systems involving two or more chemical equilibria that share one or more reactant or product are called coupled equilibria. Common examples of coupled equilibria include the increased solubility of some compounds in acidic solutions (coupled dissolution and neutralization equilibria) and in solutions containing ligands (coupled dissolution and complex formation). The equilibrium tools from other chapters may be applied to describe and perform calculations on these systems.<\/p>\r\n\r\n<\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"PrevNextBar__BarWrapper-sc-13m2i12-3 fEZPiF\"><\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li id=\"fs-idm301296\">A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?<\/li>\r\n \t<li>Calculate the equilibrium concentration of Ni<sup>2+<\/sup> in a 1.0-<em>M<\/em> solution [Ni(NH<sub>3<\/sub>)<sub>6<\/sub>](NO<sub>3<\/sub>)<sub>2<\/sub>.<\/li>\r\n \t<li>Calculate the equilibrium concentration of Zn<sup>2+<\/sup> in a 0.30-<em>M<\/em> solution of [latex]\\text{Zn}{\\left(\\text{CN}\\right)}_{4}{}^{2-}[\/latex].<\/li>\r\n \t<li>Calculate the equilibrium concentration of Cu<sup>2+<\/sup> in a solution initially with 0.050 <em>M<\/em> Cu<sup>2+<\/sup> and 1.00 <em>M<\/em> NH<sub>3<\/sub>.<\/li>\r\n \t<li>Calculate the equilibrium concentration of Zn<sup>2+<\/sup> in a solution initially with 0.150 <em>M<\/em> Zn<sup>2+<\/sup> and 2.50 <em>M<\/em> CN<sup>\u2013<\/sup>.<\/li>\r\n \t<li>Calculate the Fe<sup>3+<\/sup> equilibrium concentration when 0.0888 mole of K<sub>3<\/sub>[Fe(CN)<sub>6<\/sub>] is added to a solution with 0.0.00010 <em>M<\/em> CN<sup>\u2013<\/sup>.<\/li>\r\n \t<li>Calculate the Co<sup>2+<\/sup> equilibrium concentration when 0.100 mole of [Co(NH<sub>3<\/sub>)<sub>6<\/sub>](NO<sub>3<\/sub>)<sub>2<\/sub> is added to a solution with 0.025 <em>M<\/em> NH<sub>3<\/sub>. Assume the volume is 1.00 L.<\/li>\r\n \t<li>The equilibrium constant for the reaction [latex]{\\text{Hg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right)\\rightleftharpoons {\\text{HgCl}}_{2}\\left(aq\\right)[\/latex] is 1.6 [latex]\\times [\/latex] 10<sup>13<\/sup>. Is HgCl<sub>2<\/sub> a strong electrolyte or a weak electrolyte? What are the concentrations of Hg<sup>2+<\/sup> and Cl<sup>\u2013<\/sup> in a 0.015-<em>M<\/em> solution of HgCl<sub>2<\/sub>?<\/li>\r\n \t<li>Calculate the molar solubility of Sn(OH)<sub>2<\/sub> in a buffer solution containing equal concentrations of NH<sub>3<\/sub> and [latex]{\\text{NH}}_{4}{}^{+}[\/latex].<\/li>\r\n \t<li>Calculate the molar solubility of Al(OH)<sub>3<\/sub> in a buffer solution with 0.100 <em>M<\/em> NH<sub>3<\/sub> and 0.400 <em>M<\/em> [latex]{\\text{NH}}_{4}{}^{+}[\/latex].<\/li>\r\n \t<li>What is the molar solubility of CaF<sub>2<\/sub> in a 0.100-<em>M<\/em> solution of HF? <em>K<\/em><sub>a<\/sub> for HF = 7.2 [latex]\\times [\/latex] 10<sup>\u20134<\/sup>.<\/li>\r\n \t<li>What is the molar solubility of BaSO<sub>4<\/sub> in a 0.250-<em>M<\/em> solution of NaHSO<sub>4<\/sub>? <em>K<\/em><sub>a<\/sub> for [latex]{\\text{HSO}}_{4}{}^{-}[\/latex] = 1.2 [latex]\\times [\/latex] 10<sup>\u20132<\/sup>.<\/li>\r\n \t<li>What is the molar solubility of Tl(OH)<sub>3<\/sub> in a 0.10-<em>M<\/em> solution of NH<sub>3<\/sub>?<\/li>\r\n \t<li>What is the molar solubility of Pb(OH)<sub>2<\/sub> in a 0.138-<em>M<\/em> solution of CH<sub>3<\/sub>NH<sub>2<\/sub>?<\/li>\r\n \t<li>A solution of 0.075 <em>M<\/em> CoBr<sub>2<\/sub> is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). What is the minimum pH at which CoS begins to precipitate?\r\n[latex]\\text{CoS}\\left(s\\right)\\rightleftharpoons {\\text{Co}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{sp}}=4.5\\times {10}^{-27}[\/latex][latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times {10}^{\\text{-26}}[\/latex]<\/li>\r\n \t<li>A 0.125-<em>M<\/em> solution of Mn(NO<sub>3<\/sub>)<sub>2<\/sub> is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). At what pH does MnS begin to precipitate?[latex]\\text{MnS}\\left(s\\right)\\rightleftharpoons {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{sp}}=4.3\\times {10}^{-22}[\/latex][latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times {10}^{-26}[\/latex]<\/li>\r\n \t<li>Calculate the molar solubility of BaF<sub>2<\/sub> in a buffer solution containing 0.20 <em>M<\/em> HF and 0.20 <em>M<\/em> NaF.<\/li>\r\n \t<li>Calculate the molar solubility of CdCO<sub>3<\/sub> in a buffer solution containing 0.115 <em>M<\/em> Na<sub>2<\/sub>CO<sub>3<\/sub> and 0.120 <em>M<\/em> NaHCO<sub>3<\/sub><\/li>\r\n \t<li>To a 0.10-<em>M<\/em> solution of Pb(NO<sub>3<\/sub>)<sub>2<\/sub> is added enough HF(<em>g<\/em>) to make [HF] = 0.10 <em>M<\/em>.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Does PbF<sub>2<\/sub> precipitate from this solution? Show the calculations that support your conclusion.<\/li>\r\n \t<li>What is the minimum pH at which PbF<sub>2<\/sub> precipitates?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Calculate the concentration of Cd<sup>2+<\/sup> resulting from the dissolution of CdCO<sub>3<\/sub> in a solution that is 0.250 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H, 0.375 <em>M<\/em> in NaCH<sub>3<\/sub>CO<sub>2<\/sub>, and 0.010 <em>M<\/em> in H<sub>2<\/sub>CO<sub>3<\/sub>.<\/li>\r\n \t<li>Both AgCl and AgI dissolve in NH<sub>3<\/sub>.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>What mass of AgI dissolves in 1.0 L of 1.0 <em>M<\/em> NH<sub>3<\/sub>?<\/li>\r\n \t<li>What mass of AgCl dissolves in 1.0 L of 1.0 <em>M<\/em> NH<sub>3<\/sub>?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Calculate the volume of 1.50 <em>M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H required to dissolve a precipitate composed of 350 mg each of CaCO<sub>3<\/sub>, SrCO<sub>3<\/sub>, and BaCO<sub>3<\/sub>.<\/li>\r\n \t<li>Even though Ca(OH)<sub>2<\/sub> is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)<sub>2<\/sub>?<\/li>\r\n \t<li>What mass of NaCN must be added to 1 L of 0.010 <em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> in order to produce the first trace of Mg(OH)<sub>2<\/sub>?<\/li>\r\n \t<li>Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)<\/li>\r\n \t<li>The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Solve the following problem: [latex]{\\text{MgF}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{F}}^{-}\\left(aq\\right)[\/latex]In a saturated solution of MgF<sub>2<\/sub> at 18 \u00b0C, the concentration of Mg<sup>2+<\/sup> is 1.21 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em>. The equilibrium is represented by the equation above.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Write the expression for the solubility-product constant, <em>K<\/em><sub>sp<\/sub>, and calculate its value at 18 \u00b0C.<\/li>\r\n \t<li>Calculate the equilibrium concentration of Mg<sup>2+<\/sup> in 1.000 L of saturated MgF<sub>2<\/sub> solution at 18 \u00b0C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.<\/li>\r\n \t<li>Predict whether a precipitate of MgF<sub>2<\/sub> will form when 100.0 mL of a 3.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>-<em>M<\/em> solution of Mg(NO<sub>3<\/sub>)<sub>2<\/sub> is mixed with 200.0 mL of a 2.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>-<em>M<\/em> solution of NaF at 18 \u00b0C. Show the calculations to support your prediction.<\/li>\r\n \t<li>At 27 \u00b0C the concentration of Mg<sup>2+<\/sup> in a saturated solution of MgF<sub>2<\/sub> is 1.17 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em>. Is the dissolving of MgF<sub>2<\/sub> in water an endothermic or an exothermic process? Give an explanation to support your conclusion.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Which of the following compounds, when dissolved in a 0.01-<em>M<\/em> solution of HClO<sub>4<\/sub>, has a solubility greater than in pure water: AgBr, BaF<sub>2<\/sub>, Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>, ZnS, PbI<sub>2<\/sub>? Explain your answer.<\/li>\r\n \t<li>What is the effect on the amount of solid Mg(OH)<sub>2<\/sub> that dissolves and the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> when each of the following are added to a mixture of solid Mg(OH)<sub>2<\/sub> and water at equilibrium?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>MgCl<sub>2<\/sub><\/li>\r\n \t<li>KOH<\/li>\r\n \t<li>HClO<sub>4<\/sub><\/li>\r\n \t<li>NaNO<sub>3<\/sub><\/li>\r\n \t<li>Mg(OH)<sub>2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What is the effect on the amount of CaHPO<sub>4<\/sub> that dissolves and the concentrations of Ca<sup>2+<\/sup> and [latex]{\\text{HPO}}_{4}{}^{-}[\/latex] when each of the following are added to a mixture of solid CaHPO<sub>4<\/sub> and water at equilibrium?\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>CaCl<sub>2<\/sub><\/li>\r\n \t<li>HCl<\/li>\r\n \t<li>KClO<sub>4<\/sub><\/li>\r\n \t<li>NaOH<\/li>\r\n \t<li>CaHPO<sub>4<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Identify all chemical species present in an aqueous solution of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> and list these species in decreasing order of their concentrations. (Hint: Remember that the [latex]{\\text{PO}}_{4}{}^{\\text{3-}}[\/latex] ion is a weak base.)<\/li>\r\n \t<li>A volume of 50 mL of 1.8 <em>M<\/em> NH<sub>3<\/sub> is mixed with an equal volume of a solution containing 0.95 g of MgCl<sub>2<\/sub>. What mass of NH<sub>4<\/sub>Cl must be added to the resulting solution to prevent the precipitation of Mg(OH)<sub>2<\/sub>?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"348957\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"348957\"]\r\n\r\n2. [latex]{\\text{Ni}}^{\\text{2+}}\\left(aq\\right)+6{\\text{NH}}_{3}\\left(aq\\right)\\rightleftharpoons {\\left[\\text{Ni}{\\left({\\text{NH}}_{3}\\right)}_{6}\\right]}^{\\text{2+}}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{f}}=1.8\\times {10}^{8}[\/latex]\r\n\r\nLet <em>x<\/em> be the change in concentration as Ni<sup>2+<\/sup> dissociates. Because the initial Ni<sup>2+<\/sup> concentration is 0, the concentration at any times is <em>x<\/em>:\r\n<p style=\"text-align: center;\">[latex]1.8\\times {10}^{8}=\\frac{{\\left[\\text{Ni}{\\left({\\text{NH}}_{3}\\right)}_{6}\\right]}^{\\text{2+}}}{\\left[{\\text{Ni}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{6}}=\\frac{\\left(1.0-x\\right)}{x{\\left(6x\\right)}^{6}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">1.8 [latex]\\times [\/latex] 10<sup>8<\/sup>(46656<em>x<\/em><sup>7<\/sup>) = 1.0 \u2013 <em>x<\/em><\/p>\r\n<p style=\"text-align: center;\">8.40 [latex]\\times [\/latex] 10<sup>12<\/sup>(<em>x<\/em><sup>2<\/sup>) = 1.0 \u2013 <em>x<\/em><\/p>\r\nSince <em>x<\/em> is small in comparison with 1.0, drop <em>x<\/em>:\r\n<p style=\"text-align: center;\">8.40 [latex]\\times [\/latex] 10<sup>12<\/sup>(<em>x<\/em><sup>7<\/sup>) = 1.0<\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em><sup>7<\/sup> = 1.19 [latex]\\times [\/latex] 10<sup>\u201313<\/sup><\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 0.014 <em>M<\/em><\/p>\r\n4.\u00a0Assume that all Cu<sup>2+<\/sup> forms the complex whose concentration is 0.050 <em>M<\/em> and the remaining NH<sub>3<\/sub> has a concentration of 1.00 <em>M<\/em> \u2013 4(0.050 <em>M<\/em>) = 0.80 <em>M<\/em>. The complex dissociates:\r\n<p style=\"text-align: center;\">[latex]{\\left[\\text{Cu}{\\left({\\text{NH}}_{3}\\right)}_{4}\\right]}^{\\text{2+}}\\rightleftharpoons \\left[{\\text{Cu}}^{\\text{2+}}\\right]+4\\left[{\\text{NH}}_{3}\\right][\/latex]<\/p>\r\nLet <em>x<\/em> be the change in concentration of Cu<sup>2+<\/sup> that dissociates:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[Cu(NH<sub>3<\/sub>)<sub>4<\/sub><sup>2+<\/sup>]<\/th>\r\n<th>[Cu<sup>2+<\/sup>]<\/th>\r\n<th>[NH<sub>3<\/sub>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.050<\/td>\r\n<td>0<\/td>\r\n<td>0.80<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.050\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>4<em>x<\/em> + 0.80<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Cu}{\\left({\\text{NH}}_{3}\\right)}_{4}{}^{\\text{2+}}\\right]}{\\left[{\\text{Cu}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{4}}=1.2\\times {10}^{12}=\\frac{0.050-x}{x{\\left(4x+0.80\\right)}^{4}}[\/latex]<\/p>\r\nAssume that 4<em>x<\/em> is small when compared with 0.80 and that <em>x<\/em> is small when compared with 0.050:\r\n<p style=\"text-align: center;\">(0.80)<sup>4<\/sup> [latex]\\times [\/latex] 1.2 [latex]\\times [\/latex] 10<sup>12<\/sup><em>x<\/em> = 0.050<\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 1.0 [latex]\\times [\/latex] 10<sup>\u201313<\/sup><em>M<\/em><\/p>\r\n6.\u00a0Set up a table listing initial and equilibrium concentrations for the reaction:\r\n<p style=\"text-align: center;\">[latex]{\\text{Fe}}^{\\text{3+}}+6{\\text{CN}}^{-}\\rightleftharpoons {\\left(\\text{Fe}{\\left(\\text{CN}\\right)}_{6}\\right]}^{3-}{K}_{\\text{f}}=1\\times {10}^{44}[\/latex]<\/p>\r\nLet <em>x<\/em> be the concentration of Fe<sup>3+<\/sup> that dissociates when 0.0888 mol dissolves in 1.00 L of 0.00010 <em>M<\/em> CN<sup>\u2013<\/sup>. Assume no volume change upon dissolution:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[Fe(CN)<sub>6<\/sub><sup>3\u2212<\/sup>]<\/th>\r\n<th>[Fe<sup>3+<\/sup>]<\/th>\r\n<th>[CN<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.0888<\/td>\r\n<td>0<\/td>\r\n<td>0.00010<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.0888\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>0.00010\u00a0\u2212 6<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Fe}{\\left(\\text{CN}\\right)}_{6}{}^{3-}\\right]}{\\left[{\\text{Fe}}^{\\text{3+}}\\right]{\\left[{\\text{CN}}^{-}\\right]}^{6}}=\\frac{0.0888-x}{x{\\left(0.000010 - 6x\\right)}^{6}}=1\\times {10}^{44}[\/latex]<\/p>\r\nAssume that <em>x<\/em> is small when compared with the terms from which it is subtracted:\r\n<p style=\"text-align: center;\">0.0888 = (0.00010)<sup>6<\/sup>(<em>x<\/em>)(1 [latex]\\times [\/latex] 10<sup>44<\/sup>)<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\frac{0.0888}{1\\times {10}^{26}}=9\\times {10}^{-22}M[\/latex]<\/p>\r\n8.\u00a0Let <em>x<\/em> be the change in the number of moles of Hg<sup>2+<\/sup> that form per liter:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[HgCl<sub>2<\/sub>]<\/th>\r\n<th>[Hg<sup>2+<\/sup>]<\/th>\r\n<th>[Cl<sup>\u2212<\/sup>]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Initial concentration (<em>M<\/em>)<\/th>\r\n<td>0.015<\/td>\r\n<td>0<\/td>\r\n<td>0<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Equilibrium (<em>M<\/em>)<\/th>\r\n<td>0.015\u00a0\u2212\u00a0<em>x<\/em><\/td>\r\n<td><em>x<\/em><\/td>\r\n<td>2<em>x<\/em><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{\\left[{\\text{HgCl}}_{2}\\right]}{\\left[{\\text{Hg}}^{\\text{2+}}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}}=K=1.6\\times {10}^{13}\\\\ \\frac{0.015-x}{\\left(x\\right){\\left(2x\\right)}^{2}}\\approx \\frac{0.015}{4{x}^{3}}=1.6\\times {10}^{13}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em><sup>3<\/sup> = 2.3 [latex]\\times [\/latex] 10<sup>\u201316<\/sup><\/p>\r\n<p style=\"text-align: center;\"><em>x<\/em> = 6.2 [latex]\\times [\/latex] 10<sup>\u20136<\/sup><em>M<\/em> = [Hg<sup>2+<\/sup>]<\/p>\r\n<p style=\"text-align: center;\">2x = 1.2 [latex]\\times [\/latex] 10<sup>\u20135<\/sup>] <em>M<\/em> = [Cl<sup>\u2013<\/sup>]<\/p>\r\nThe substance is a weak electrolyte because very little of the initial 0.015 <em>M<\/em> HgCl<sub>2<\/sub> dissolved.\r\n\r\n10.\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(0.400\\right)\\left[{\\text{OH}}^{-}\\right]}{\\left(0.100\\right)}=1.8\\times {10}^{-5}[\/latex]\r\n\r\n[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{\\left(0.100\\right)\\left(1.8\\times {10}^{-5}\\right)}{0.0400}\\text{=}4.5\\times {10}^{-5}[\/latex]\r\n\r\n<em>K<\/em><sub>sp<\/sub> = [Al<sup>3+<\/sup>][OH<sup>\u2013<\/sup>]<sup>3<\/sup> = [Al<sup>3+<\/sup>](4.5 [latex]\\times [\/latex] 10<sup>\u20135<\/sup>)<sup>3<\/sup> = 1.9 [latex]\\times [\/latex] 10<sup>\u201333<\/sup>\r\n\r\n[Al<sup>3+<\/sup>] = 2.1 [latex]\\times [\/latex] 10<sup>\u201320<\/sup> (molar solubility)\r\n\r\n12.\u00a0Find the amount of [latex]{\\text{SO}}_{4}{}^{2-}[\/latex] present from <em>K<\/em><sub>a<\/sub> for the equilibrium:\r\n<p style=\"text-align: center;\">[latex]{\\text{HSO}}_{4}{}^{-}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{SO}}_{4}{}^{2-}[\/latex]<\/p>\r\nLet <em>x<\/em> be the change in [latex]\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex] :\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{SO}}_{4}{}^{2-}\\right]}{\\left[{\\text{HSO}}_{4}{}^{-}\\right]}=\\frac{{x}^{2}}{0.250-x}=1.2\\times {10}^{-2}[\/latex]<\/p>\r\nBecause <em>K<\/em><sub>a<\/sub> is too large to disregard <em>x<\/em> in the expression 0.250 \u2013 <em>x<\/em>, we must solve the quadratic equation:\r\n<p style=\"text-align: center;\"><em>x<\/em><sup>2<\/sup> + 1.2 [latex]\\times [\/latex] 10<sup>\u20132<\/sup><em>x<\/em> \u2013 0.250(1.2 [latex]\\times [\/latex] 10<sup>\u20132<\/sup>) = 0<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\frac{-1.2\\times {10}^{-2}\\pm \\sqrt{{\\left(1.2\\times {10}^{-2}\\right)}^{2}+4\\left(3.0\\times {10}^{\\text{-3}}\\right)}}{2}=\\frac{-1.2\\times {10}^{-2}\\pm 0.11}{2}=0.049M[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Ba<sup>2+<\/sup>] [latex]\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex] = [Ba<sup>2+<\/sup>](0.049) = 1.08 [latex]\\times [\/latex] 10<sup>\u201310<\/sup><\/p>\r\n<p style=\"text-align: center;\">[Ba<sup>2+<\/sup>] = 2.2 [latex]\\times [\/latex] 10<sup>\u20139<\/sup> (molar solubility)<\/p>\r\n14.\u00a0[latex]{\\text{CH}}_{3}{\\text{NH}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{CH}}_{3}{\\text{NH}}_{3}{}^{+}+{\\text{OH}}^{-}[\/latex]\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{CH}}_{3}{\\text{NH}}_{2}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{0.138-x}=4.4\\times {10}^{-4}[\/latex]<\/p>\r\nSolve the quadratic equation using the quadratic formula:\r\n<p style=\"text-align: center;\"><em>x<\/em><sup>2<\/sup> + 4.4 [latex]\\times [\/latex] 10<sup>\u20134<\/sup><em>x<\/em> \u2013 0.138(4.4 [latex]\\times [\/latex] 10<sup>\u20134<\/sup>) = 0<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\frac{-4.4\\times {10}^{-4}\\pm \\sqrt{{\\left(4.4\\times {10}^{-4}\\right)}^{2}+4\\left(6.07\\times {10}^{\\text{-5}}\\right)}}{2}=\\frac{-4.4\\times {10}^{\\text{-4}}\\pm \\sqrt{2.43\\times {10}^{-4}}}{2}=\\frac{0.0152}{2}=7.6\\times {10}^{-3}M[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = [Pb<sup>2+<\/sup>](7.6 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 2.8 [latex]\\times [\/latex] 10<sup>\u201316<\/sup><\/p>\r\n<p style=\"text-align: center;\">[Pb<sup>2+<\/sup>] = 4.8 [latex]\\times [\/latex] 10<sup>\u201312<\/sup> (molar solubility)<\/p>\r\n16.\u00a0Two equilibria are in competition for the ions and must be considered simultaneously. Precipitation of MnS will occur when the concentration of S<sup>2\u2013<\/sup> in conjunction with 0.125 <em>M<\/em> Mn<sup>2+<\/sup> exceeds the <em>K<\/em><sub>sp<\/sub> of MnS. The [S<sup>2\u2013<\/sup>] must come from the ionization of H<sub>2<\/sub>S as defined by the equilibrium:\r\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left[{\\text{S}}^{2-}\\right]}{\\left[{\\text{H}}_{2}\\text{S}\\right]}={K}_{1}{K}_{2}\\left({\\text{H}}_{2}\\text{S}\\right)=1.0\\times {10}^{\\text{-26}}[\/latex]<\/p>\r\nAs a saturated solution of H<sub>2<\/sub>S is 0.10 <em>M<\/em>, this later expression becomes:\r\n<p style=\"text-align: center;\">[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left[{\\text{S}}^{2-}\\right]=1.0\\times {10}^{-27}[\/latex]<\/p>\r\nFrom the equilibrium of MnS, the minimum concentration of S<sup>2\u2013<\/sup> required to cause precipitation is calculated as:\r\n<p style=\"text-align: center;\">[latex]\\text{MnS}\\left(s\\right)\\longrightarrow {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Mn<sup>2+<\/sup>][S<sup>2\u2013<\/sup>] = 4.3 [latex]\\times [\/latex] 10<sup>\u201322<\/sup><\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{S}}^{2-}\\right]=\\frac{4.3\\times {10}^{-22}}{0.125}=3.44\\times {10}^{-21}[\/latex]<\/p>\r\nThis amount of S<sup>2\u2013<\/sup> will exist in solution at a pH defined by the H<sub>2<\/sub>S equilibrium:\r\n<ul>\r\n \t<li>[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left(3.44\\times {10}^{-21}\\right)=1.0\\times {10}^{-27}[\/latex]<\/li>\r\n \t<li>[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}=2.91\\times {10}^{-7}[\/latex]<\/li>\r\n \t<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=5.39\\times {10}^{-4}M[\/latex]<\/li>\r\n \t<li>[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=3.27[\/latex]<\/li>\r\n<\/ul>\r\n18.\u00a0Three equilibria are involved:\r\n<ul>\r\n \t<li>[latex]{\\text{H}}_{2}{\\text{CO}}_{3}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{HCO}}_{3}{}^{-}{K}_{{\\text{a}}_{1}}=4.3\\times {10}^{-7}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{HCO}}_{3}{}^{-}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CO}}_{3}{}^{2-}{K}_{{\\text{a}}_{2}}=7\\times {10}^{-11}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{CdCO}}_{3}\\longrightarrow {\\text{Cd}}^{\\text{2+}}+{\\text{CO}}_{3}{}^{2-}{K}_{\\text{sp}}=2.5\\times {10}^{-14}[\/latex]<\/li>\r\n<\/ul>\r\nFirst, find the pH of the buffer from the Henderson-Hasselbach equation. Then find [H<sub>3<\/sub>O<sup>+<\/sup>]:\r\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\mathrm{log}\\frac{\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{-}\\right]}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">10.155 + log [latex]\\frac{0.115}{0.120}[\/latex] = 10.155 \u2013 0.018 = 10.137<\/p>\r\nIn this case, several more significant figures are carried than justified so that the value of the log ratio is meaningful:\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=7.3\\times {10}^{-11}[\/latex]<\/p>\r\nNow, find the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] present in the buffer solution. Next, using [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] and <em>K<\/em><sub>sp<\/sub>, calculate the concentration of Cd<sup>2+<\/sup>. This latter value represents the molar solubility. From [latex]{K}_{{a}_{1}}[\/latex] determine [H<sub>2<\/sub>CO<sub>3<\/sub>]:\r\n<p style=\"text-align: center;\">[latex]{K}_{{\\text{a}}_{1}}=4.3\\times {10}^{-7}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCO}}_{3}{}^{-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left(7.3\\times {10}^{-11}\\right)\\left(0.120\\right)}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[H<sub>2<\/sub>CO<sub>3<\/sub>] = 2.05 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><\/p>\r\nFrom [latex]{K}_{{\\text{a}}_{1}}{K}_{{\\text{a}}_{2}}[\/latex] , find the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] :\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}{K}_{{\\text{a}}_{1}}{K}_{{\\text{a}}_{2}}&amp;=&amp;\\left(4.3\\times 10 - 7\\right)\\left(7\\times 10 - 11\\right)=\\frac{\\left[{\\text{HCO}}_{3}{}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}\\times \\frac{\\left[{\\text{CO}}_{3}{}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{{\\left(7.3\\times {10}^{-11}\\right)}^{2}\\times \\left[{\\text{CO}}_{3}{}^{2-}\\right]}{2.04\\times {10}^{-5}}\\\\&amp; =&amp;3\\times {10}^{-17}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 0.115 <em>M<\/em><\/p>\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = [Cd<sup>2+<\/sup>](0.115) = 3 [latex]\\times [\/latex] 10<sup>\u201313<\/sup><\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{3\\times {10}^{-13}}{0.115}=3\\times {10}^{-12}M[\/latex]<\/p>\r\n20.\u00a0For <em>K<\/em><sub>sp<\/sub>, (CdCO<sub>3<\/sub>) = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 2.5 [latex]\\times [\/latex] 10<sup>\u201314<\/sup>, the amount of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] is governed by <em>K<\/em><sub>a<\/sub> of H<sub>2<\/sub>CO<sub>3<\/sub>, 4.3 [latex]\\times [\/latex] 10<sup>\u20137<\/sup>, and <em>K<\/em><sub>a<\/sub> of [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] , 7 [latex]\\times [\/latex] 10<sup>\u201311<\/sup>, and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] by <em>K<\/em><sub>a<\/sub> of acetic acid. First, calculate the [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] from <em>K<\/em><sub>a<\/sub> of acetic acid (HOAc):\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{a}}\\left[\\text{HOAc}\\right]}{\\left[{\\text{OAc}}^{-}\\right]}=\\frac{1.8\\times {10}^{-5}\\left(0.250\\right)}{\\left(0.375\\right)}=1.2\\times {10}^{-5}M[\/latex]<\/p>\r\nFrom this and <em>K<\/em><sub>a<\/sub> for H<sub>2<\/sub>CO<sub>3<\/sub>, calculate [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] present. As [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] is fixed by <em>K<\/em><sub>a<\/sub> of acetic acid:\r\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\left({\\text{H}}_{2}{\\text{CO}}_{3}\\right)=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCP}}_{3}{}^{-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}{}^{-}\\right]}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{2}{\\text{CO}}_{3}{}^{-}\\right]=\\frac{{K}_{\\text{a}}\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{4.3\\times {10}^{-7}\\left[0.010\\right]}{\\left[1.2\\times {10}^{-5}\\right]}=3.58\\times {10}^{-4}M[\/latex]<\/p>\r\nFrom [latex]{K}_{\\text{a}}\\left({\\text{HCO}}_{3}{}^{-}\\right)[\/latex] we obtain [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] :\r\n<p style=\"text-align: center;\">[latex]{K}_{a}=7\\times {10}^{-11}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{-}\\right]}=\\frac{1.2\\times {10}^{-17}\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[3.58\\times {10}^{-4}\\right]}\\text{ }[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]=\\frac{7\\times {10}^{-11}\\times 3.58\\times {10}^{-4}}{1.2\\times {10}^{-5}}=2.09\\times {10}^{-9}[\/latex]<\/p>\r\nFrom the solubility product:\r\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 2.5 [latex]\\times [\/latex] 10<sup>\u201314<\/sup> = Cd<sup>2+<\/sup> (2.09 [latex]\\times [\/latex] 10<sup>\u20139<\/sup>)<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{2.5\\times {10}^{-14}}{2.09\\times {10}^{-9}}=1\\times {10}^{-5}M[\/latex]<\/p>\r\n22.\u00a0[latex]{\\text{CaCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{100.09\\cancel{\\text{g}}}=3.50\\times {10}^{-3}\\text{mol}[\/latex]\r\n\r\n[latex]{\\text{SrCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{147.63\\cancel{\\text{g}}}=2.37\\times {10}^{-3}\\text{mol}[\/latex]\r\n\r\n[latex]{\\text{BaCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{197.34\\cancel{\\text{g}}}=1.77\\times {10}^{-3}\\text{mol}[\/latex]\r\n<ul>\r\n \t<li>Total: 7.64 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol<\/li>\r\n \t<li>CaCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 4.8 [latex]\\times [\/latex] 10<sup>\u20139<\/sup><\/li>\r\n \t<li>SrCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 9.42 [latex]\\times [\/latex] 10<sup>\u201310<\/sup><\/li>\r\n \t<li>BaCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 8.1 [latex]\\times [\/latex] 10<sup>\u20139<\/sup><\/li>\r\n<\/ul>\r\nSolubilities are approximately equal.\r\n\r\nWhen the three solids dissolve, 7.64 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of metal ions and 7.64 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] are initially present. The [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] reacts with CH<sub>3<\/sub>CO<sub>2<\/sub>H to form [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] :\r\n<p style=\"text-align: center;\">[latex]{\\text{ CO}}_{3}{}^{2-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)\\rightleftharpoons {\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]K=\\frac{\\left[{\\text{HCO}}_{3}{}^{-}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right]}{\\left[{\\text{CO}}_{3}{}^{2-}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}\\times \\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}{{K}_{\\text{a}}\\left(\\text{for}{\\text{HCO}}_{3}{}^{-}\\right)}=\\frac{1.8\\times {10}^{-5}}{7\\times {10}^{-11}}=2.6\\times {10}^{5}[\/latex]<\/p>\r\nThis <em>K<\/em> value is large, so virtually all [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] undergoes this reaction and approximately 7.64 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol of [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] forms. The [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] reacts with CH<sub>3<\/sub>CO<sub>2<\/sub>H:\r\n<p style=\"text-align: center;\">[latex]{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]K=\\frac{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}{{K}_{\\text{a}}\\left(\\text{for}{\\text{HCO}}_{3}{}^{-}\\right)}=\\frac{1.8\\times {10}^{-5}}{4.3\\times {10}^{-7}}=42[\/latex]<\/p>\r\nThis reaction is virtually complete as well (<em>K<\/em> is large).\r\n\r\nFor each mol of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] produced, 2 mol of CH<sub>3<\/sub>CO<sub>2<\/sub>H is required for conversion to H<sub>2<\/sub>CO<sub>3<\/sub>. Thus 15.3 [latex]\\times [\/latex] 10<sup>\u20133<\/sup> mol CH<sub>3<\/sub>CO<sub>2<\/sub>H is needed:\r\n<p style=\"text-align: center;\">[latex]\\text{Volume}=\\frac{15.3\\times {10}^{-3}\\text{mol}}{1.50\\text{mol}{\\text{L}}^{-1}}=0.0102\\text{L}\\left(10.2\\text{mL}\\right)[\/latex]<\/p>\r\n24.\u00a0There are two equilibria involved: [latex]\\begin{array}{l}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right){K}_{\\text{sp}}=1.5\\times {10}^{-11}\\\\ {\\text{CN}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HCN}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{1.0\\times {10}^{-14}}{4\\times {10}^{-10}}=2.5\\times {10}^{-5}\\right)\\end{array}[\/latex] The Mg(NO<sub>3<\/sub>)<sub>2<\/sub> dissolves, and [Mg<sup>2+<\/sup>] = 0.010 <em>M<\/em>.\r\n\r\n[Mg<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = <em>K<\/em><sub>sp<\/sub> = 1.5 [latex]\\times [\/latex] 10<sup>\u201311<\/sup>\r\n\r\n(0.010)[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 1.5 [latex]\\times [\/latex] 10<sup>\u201311<\/sup>\r\n\r\n[OH<sup>\u2013<\/sup>] = 3.9 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em>\r\n\r\nWe need to add enough CN<sup>\u2013<\/sup> to make [OH<sup>\u2013<\/sup>] = 3.9 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em>. Both OH<sup>\u2013<\/sup> and HCN come from CN<sup>\u2013<\/sup>, so [OH<sup>\u2013<\/sup>] = [HCN]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{\\left[\\text{HCN}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{CN}}^{-}\\right]}={K}_{\\text{b}}=2\\times {10}^{-5}\\\\ \\frac{{\\left(3.9\\times {10}^{-5}\\right)}^{2}}{\\left[{\\text{CN}}^{-}\\right]}=2.5\\times {10}^{-5}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[CN<sup>\u2013<\/sup>] = 6.1 [latex]\\times [\/latex] 10<sup>\u20135<\/sup><em>M<\/em><\/p>\r\nmol NaCN = mol HCN + mol CN<sup>\u2013<\/sup> = 3.9 [latex]\\times [\/latex] 10<sup>\u20135<\/sup> + 6.1 [latex]\\times [\/latex] 10<sup>\u20135<\/sup> = 1.0 [latex]\\times [\/latex] 10<sup>\u20134<\/sup>\r\n\r\n[latex]\\text{mass}\\left(\\text{NaCN}\\right)=1.0\\times {10}^{-4}\\text{mol}\\times \\frac{49.007\\text{g}}{1\\text{mol}}=5\\times {10}^{-3}\\text{g}[\/latex]\r\n\r\n26. The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = (1.21 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)(2 [latex]\\times [\/latex] 1.21 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 7.09 [latex]\\times [\/latex] 10<sup>\u20139<\/sup><\/li>\r\n \t<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = [<em>x<\/em>][0.100 + 2<em>x<\/em>]<sup>2<\/sup> = 7.09 [latex]\\times [\/latex] 10<sup>\u20139\r\n<\/sup>Assume that 2<em>x<\/em> is small when compared with 0.100 <em>M<\/em>.\r\n0.100<em>x<\/em> = 7.09 [latex]\\times [\/latex] 10<sup>\u20139\r\n<\/sup><em>x<\/em> = [MgF<sub>2<\/sub>] = 7.09 [latex]\\times [\/latex] 10<sup>\u20137<\/sup><em>M\r\n<\/em>The value 7.09 [latex]\\times [\/latex] 10<sup>\u20137<\/sup><em>M<\/em> is quite small when compared with 0.100 <em>M<\/em>, so the assumption is valid.<\/li>\r\n \t<li>Determine the concentration of Mg<sup>2+<\/sup> and F<sup>\u2013<\/sup> that will be present in the final volume. Compare the value of the ion product [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> with <em>K<\/em><sub>sp<\/sub>. If this value is larger than <em>K<\/em><sub>sp<\/sub>, precipitation will occur.\r\n0.1000 L [latex]\\times [\/latex] 3.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> = 0.3000 L [latex]\\times [\/latex] <em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2\r\n<\/sub><em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> = 1.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M\r\n<\/em>0.2000 L [latex]\\times [\/latex] 2.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M<\/em> NaF = 0.3000 L [latex]\\times [\/latex] <em>M<\/em> NaF\r\n<em>M<\/em> NaF = 1.33 [latex]\\times [\/latex] 10<sup>\u20133<\/sup><em>M\r\n<\/em>ion product = (1.00 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)(1.33 [latex]\\times [\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 1.77 [latex]\\times [\/latex] 10<sup>\u20139\r\n<\/sup>This value is smaller than <em>K<\/em><sub>sp<\/sub>, so no precipitation will occur.<\/li>\r\n \t<li>MgF<sub>2<\/sub> is less soluble at 27 \u00b0C than at 18 \u00b0C. Because added heat acts like an added reagent, when it appears on the product side, the Le Ch\u00e2telier\u2019s principle states that the equilibrium will shift to the reactants\u2019 side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.<\/li>\r\n<\/ol>\r\n28.\u00a0Effect on amount of solid Mg(OH)<sub>2<\/sub>, [Mg<sup>2+<\/sup>], [OH<sup>\u2013<\/sup>]:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>increase, increase, decrease;<\/li>\r\n \t<li>increase, decrease, increase;<\/li>\r\n \t<li>decrease, increase, decrease;<\/li>\r\n \t<li>no effect predicted;<\/li>\r\n \t<li>increase, no effect, no effect<\/li>\r\n<\/ol>\r\n30.\u00a0[H<sub>2<\/sub>O] &gt; [Ca<sup>2+<\/sup>] &gt; [latex]\\left[{\\text{PO}}_{4}{}^{3-}\\right][\/latex] &gt; [latex]{\\text{HPO}}_{4}{}^{2-}[\/latex] = [OH<sup>\u2013<\/sup>] &gt; [latex]\\left[{\\text{HPO}}_{4}{}^{-}\\right][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<b>coupled equilibrium: <\/b>system characterized by more than one state of balance between a slightly soluble ionic solid and an aqueous solution of ions working simultaneously","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe examples of systems involving two (or more) coupled chemical equilibria<\/li>\n<li>Calculate reactant and product concentrations for coupled equilibrium systems<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idm703568\">As discussed in preceding chapters on equilibrium,\u00a0<em>coupled equilibria<\/em>\u00a0involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.<\/p>\n<p id=\"fs-idp11174336\">An environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean\u2019s coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates).<\/p>\n<p>\u00a0Carbon dioxide in the air dissolves in sea water, forming carbonic acid (H<sub>2<\/sub>CO<sub>3<\/sub>). The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions [latex]\\left({\\text{HCO}}_{3}{}^{-}\\right)[\/latex], which can further ionize into more hydrogen ions and carbonate ions [latex]\\left({\\text{CO}}_{3}{}^{2-}\\right):[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}{\\text{CO}}_{2}\\left(g\\right)&\\rightleftharpoons&{\\text{CO}}_{2}\\left(aq\\right)\\\\{\\text{CO}}_{2}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}&\\rightleftharpoons&{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)\\\\{\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)&\\rightleftharpoons&{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)\\\\{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)&\\rightleftharpoons&{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{CO}}_{3}{}^{2-}\\left(aq\\right)\\end{array}[\/latex]<\/p>\n<p>The excess H<sup>+<\/sup> ions make seawater more acidic. Increased ocean acidification can then have negative impacts on reef-building coral, as they cannot absorb the calcium carbonate they need to grow and maintain their skeletons (Figure\u00a01). This in turn disrupts the local biosystem that depends upon the health of the reefs for its survival. If enough local reefs are similarly affected, the disruptions to sea life can be felt globally. The world\u2019s oceans are presently in the midst of a period of intense acidification, believed to have begun in the mid-nineteenth century, and which is now accelerating at a rate faster than any change to oceanic pH in the last 20 million years.<\/p>\n<div id=\"attachment_5431\" style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5431\" class=\"size-full wp-image-5431\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042406\/CNX_Chem_15_03_CoralReef.jpg\" alt=\"This figure contains two photographs of coral reefs. In a, a colorful reef that includes hues of purple and pink corals is shown in blue green water with fish swimming in the background. In b, grey-green mossy looking coral is shown in a blue aquatic environment. This photo does not have the colorful appearance or fish that were shown in figure a.\" width=\"975\" height=\"386\" \/><\/p>\n<p id=\"caption-attachment-5431\" class=\"wp-caption-text\">Figure\u00a01. Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonite skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by \u201cprilfish\u201d\/Flickr)<\/p>\n<\/div>\n<div class=\"textbox\">\n<p>Learn more about ocean acidification and how it affects other marine creatures.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=5559071&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=kxPwbhFeZSw&amp;video_target=tpm-plugin-0cebssbe-kxPwbhFeZSw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/OceanAcidification_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Ocean Acidification&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<div class=\"textbox\">This site has detailed information about <a href=\"http:\/\/www.teachoceanscience.net\/teaching_resources\/education_modules\/coral_reefs_and_climate_change\/how_does_climate_change_affect_coral_reefs\/\" target=\"_blank\" rel=\"noopener\">how ocean acidification specifically affects coral reefs<\/a>.<\/div>\n<p>The dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (Figure 2), a sparingly soluble ionic compound whose dissolution equilibrium is<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\left(s\\right)\\longrightarrow 5{\\text{Ca}}^{\\text{2+}}\\left(aq\\right)+3{\\text{PO}}_{4}^{3-}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)[\/latex]<\/p>\n<div id=\"attachment_5434\" style=\"width: 660px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5434\" class=\"wp-image-5434 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042527\/CNX_Chem_15_03_Apatite.jpg\" alt=\"This figure includes an image of two large light blue apatite crystals in a mineral conglomerate that includes white, grey, and tan crystals. The blue apatite crystals have a dull, dusty, or powdered appearance.\" width=\"650\" height=\"383\" \/><\/p>\n<p id=\"caption-attachment-5434\" class=\"wp-caption-text\">Figure\u00a02. Crystal of the mineral hydroxylapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>OH, is shown here. Pure apatite is white, but like many other minerals, this sample is colored because of the presence of impurities.<\/p>\n<\/div>\n<p>This compound dissolved to yield two different basic ions: triprotic phosphate ions<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{PO}}_{4}{}^{3-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{2-}+{\\text{H}}_{2}\\text{O}[\/latex]<br \/>\n[latex]{\\text{PO}}_{4}{}^{2-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}+{\\text{H}}_{2}\\text{O}[\/latex]<br \/>\n[latex]{\\text{H}}_{2}{\\text{PO}}_{4}{}^{-}+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow {\\text{H}}_{3}{\\text{PO}}_{4}+{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\n<p>and monoprotic hydroxide ions:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{OH}}^{-}\\left(aq\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\longrightarrow 2{\\text{H}}_{2}\\text{O}[\/latex]<\/p>\n<p>Of the two basic productions, the hydroxide is, of course, by far the stronger base (it\u2019s the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added NaF or SnF<sub>2<\/sub>\u00a0that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{NaF}+{\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{OH}\\rightleftharpoons {\\text{Ca}}_{5}{\\left({\\text{PO}}_{4}\\right)}_{3}\\text{F}+{\\text{Na}}^{+}+{\\text{OH}}^{-}[\/latex]<\/p>\n<p>The weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.<\/p>\n<div class=\"textbox shaded\">\n<h3>Role of Fluoride in Preventing Tooth Decay<\/h3>\n<p>As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca<sub>5<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure\u00a03).<\/p>\n<div id=\"attachment_5435\" style=\"width: 660px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5435\" class=\"wp-image-5435 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042617\/CNX_Chem_15_03_Toothpaste.jpg\" alt=\"A tube of toothpaste\" width=\"650\" height=\"269\" \/><\/p>\n<p id=\"caption-attachment-5435\" class=\"wp-caption-text\">Figure 3. Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).<\/p>\n<\/div>\n<p>Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm\u00a0(4 mg\/L)\u00a0of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.<\/p>\n<\/div>\n<p id=\"fs-idp19645632\">The solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion\u00a0<span class=\"os-math-in-para\"><span id=\"MathJax-Element-554-Frame\" class=\"MathJax\" style=\"overflow: initial; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; letter-spacing: normal; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px;\" role=\"presentation\"><span id=\"MathJax-Span-18138\" class=\"math\"><span id=\"MathJax-Span-18139\" class=\"mrow\"><span id=\"MathJax-Span-18140\" class=\"semantics\"><span id=\"MathJax-Span-18141\" class=\"mrow\"><span id=\"MathJax-Span-18142\" class=\"mrow\"><span id=\"MathJax-Span-18143\" class=\"mtext\">Al<\/span><span id=\"MathJax-Span-18144\" class=\"msub\"><span id=\"MathJax-Span-18145\" class=\"mrow\"><span id=\"MathJax-Span-18146\" class=\"mo\">(<\/span><span id=\"MathJax-Span-18147\" class=\"mtext\">OH<\/span><span id=\"MathJax-Span-18148\" class=\"mo\">)<\/span><\/span><span id=\"MathJax-Span-18149\" class=\"mn\">4<\/span><\/span><span id=\"MathJax-Span-18150\" class=\"msup\"><span id=\"MathJax-Span-18151\" class=\"mrow\"><\/span><span id=\"MathJax-Span-18152\" class=\"mtext\">\u2212<\/span><\/span><span id=\"MathJax-Span-18153\" class=\"mo\">.<\/span><\/span><\/span><\/span><\/span><\/span><span class=\"MJX_Assistive_MathML\" role=\"presentation\">Al(OH)4\u2212.<\/span><\/span><\/span><\/p>\n<p><span id=\"fs-idp137004736\" class=\"scaled-down\"><img decoding=\"async\" id=\"7\" src=\"https:\/\/openstax.org\/resources\/ec0583eb0fb6238971af3b9efbce4e52b7ee02c8\" alt=\"An H atom is bonded to an O atom. The O atom has 2 dots above it and 2 dots below it. The O atom is bonded to an A l atom, which has three additional O atoms bonded to it as well. Each of these additional O atoms has 4 dots arranged around it, and is bonded to an H atom. This entire molecule is contained in brackets, to the right of which is a superscripted negative sign.\" \/><\/span><\/p>\n<p id=\"fs-idm55619648\">The equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of Al(OH)<sub>3<\/sub>.<\/p>\n<div style=\"text-align: center;\">[latex]\\text{Al(OH)}_3 (s) \\rightleftharpoons \\text{Al}^{3+} (aq) + 3\\text{OH}^- (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_{sp} = 2 \\times 10^{-32}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{Al}^{3+} (aq) + 4\\text{OH}^- (aq) \\rightleftharpoons \\text{Al(OH)}_4^{-} (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_{f} = 1.1 \\times 10^{33}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{Net: Al(OH)}_3 (s) + \\text{OH}^- (aq) \\rightleftharpoons \\text{Al(OH)}_4^{-} (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K = K_{sp} K_f = 22[\/latex]<\/div>\n<div class=\"textbox examples\">\n<h3>Example 1: Increased solubility in acidic solutions<\/h3>\n<p>Compute and compare the molar solublities for aluminum hydroxide, Al(OH)<sub>3<\/sub>, dissolved in (a) pure water and (b) a buffer containing 0.100\u00a0<em>M<\/em>\u00a0acetic acid and 0.100\u00a0<em>M<\/em>\u00a0sodium acetate.<\/p>\n<h4>Solution<\/h4>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q576838\">Show Solution<\/span><\/p>\n<div id=\"q576838\" class=\"hidden-answer\" style=\"display: none\">\n(a) The molar solubility of aluminum hydroxide in water is computed considering the dissolution equilibrium only as demonstrated in several previous examples:<\/p>\n<div style=\"text-align: center;\">[latex]\\text{Al(OH)}_3 (s) \\rightleftharpoons \\text{Al}^{3+} (aq) + 3\\text{OH}^- (aq)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]K_{sp} = 2 \\times 10^{-32}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{molar solubility in water} = [\\text{Al}^{3+}] = (2\\times 10^{-32} \/ 27)^{1\/4} = 5\\times 10^{-9} M[\/latex]<\/div>\n<p id=\"fs-idm324374096\">(b) The concentration of hydroxide ion of the buffered solution is conveniently calculated by the Henderson-Hasselbalch equation:<\/p>\n<div style=\"text-align: center;\">[latex]\\text{pH} = \\text{pKa} +\\text{log}[\\text{CH}_3\\text{COO}^-] \/ [\\text{CH}_3\\text{COOH}][\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{pH} = 4.74 +\\text{log} (0.100\/0.100) = 4.74[\/latex]<\/div>\n<p id=\"fs-idm378269776\">At this pH, the concentration of hydroxide ion is<\/p>\n<div style=\"text-align: center;\">[latex]\\text{pOH} = 14.00-4.74=9.26[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex][\\text{OH}^-] = 10^{-9.26} = 5.5\\times 10^{-10}[\/latex]<\/div>\n<p id=\"fs-idm373349040\">The solubility of Al(OH)<sub>3<\/sub>\u00a0in this buffer is then calculated from its solubility product expressions:<\/p>\n<div style=\"text-align: center;\">[latex]K_{sp} = [\\text{Al}^{3+}][\\text{OH}^{-}]^3[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\text{molar solubility in buffer} = [\\text{Al}^{3+}] = K_{sp}\/[\\text{OH}^{-}]^3 = (2\\times 10^{-32}) \/ (5\\times 10^{-10})^3 = 1.2\\times 10^{-4} M[\/latex]<\/div>\n<p id=\"fs-idm329760656\">Compared to pure water, the solubility of aluminum hydroxide in this mildly acidic buffer is approximately ten million times greater (though still relatively low).<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>What is the solubility of aluminum hydroxide in a buffer comprised of 0.100\u00a0<em>M<\/em>\u00a0formic acid and 0.100\u00a0<em>M<\/em>\u00a0sodium formate?<\/p>\n<div id=\"fs-idp14780320\" class=\"ui-has-child-title\">\n<header>\n<h4 class=\"os-title\"><span id=\"8\" class=\"os-title-label\">ANSWER:<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q576839\">Show Solution<\/span><\/p>\n<div id=\"q576839\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"os-note-body\">\n<p id=\"fs-idp15140704\">0.1\u00a0<em>M<\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0Multiple Equilibria<\/h3>\n<p>Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub>, called hypo) to form the complex ion [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}[\/latex] (<em>K<\/em><sub>f<\/sub> = 4.7 [latex]\\times[\/latex] 10<sup>13<\/sup>).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-5437\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/219\/2016\/08\/09042721\/CNX_Chem_15_03_AgBr_img.jpg\" alt=\"A chemical reaction is shown using structural formulas. On the left, A g superscript plus is followed by a plus sign, the number 2, and a structure in brackets. The structure is composed of a central S atom which has O atoms single bonded above, right, and below. A second S atom is single bonded to the left. Each of these bonded atoms has 6 dots around it. Outside the brackets is a superscript 2 negative. Following a bidirectional arrow is a structure in brackets with a central A g atom. To the left and right, S atoms are single bonded to the A g atom. Each of these S atoms has four dots around it, and an S atom connected with a single bond moving out from the central A g atom, forming the ends of the structure. Each of these atoms has three O atoms attached with single bonds above, below, and at the end of the structure. Each O atom has six dots around it. Outside the brackets is a superscript 3 negative.\" width=\"1000\" height=\"176\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>What mass of Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of [latex]\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q693371\">Show Solution<\/span><\/p>\n<div id=\"q693371\" class=\"hidden-answer\" style=\"display: none\">\n<p>Two equilibria are involved when AgBr dissolves in a solution containing the [latex]{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}[\/latex] ion:<\/p>\n<ul>\n<li>Reaction 1 (dissolution): [latex]\\text{AgBr}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Br}}^{-}\\left(aq\\right)\\qquad{K}_{\\text{sp}}=5.0\\times {10}^{\\text{-13}}[\/latex]<\/li>\n<li>Reaction 2 (complexation): [latex]{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\left(aq\\right)\\rightleftharpoons \\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{3-}\\left(aq\\right)\\qquad{K}_{\\text{f}}=4.7\\times {10}^{13}[\/latex]<\/li>\n<\/ul>\n<p>First, calculate the concentration of bromide that will result when the 1.00 g of AgBr is completely dissolved via the cited complexation reaction:<\/p>\n<p style=\"text-align: center;\">[latex]1.00\\text{ g AgBr }\/(187.77\\text{ g\/mol})(1\\text{ mol Br}^{-}\/1\\text{ mol AgBr})=0.00532\\text{ mol Br}^{-}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]0.00532\\text{ mol Br}^{-}\/1.00\\text{ L}=0.00532 M\\text{ Br}^{-}[\/latex]<\/p>\n<p style=\"text-align: left;\">Next, use this bromide molarity and the solubility product for silver bromide to calculate the silver ion molarity in the solution:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Ag}}^{+}\\right]=K_{\\text{sp}}\/ \\left[\\text{Br}^{-}\\right]=5.0\\times{10}^{-13}\/0.00532=9.4\\times {10}^{-11}M[\/latex]<\/p>\n<p>Based on the stoichiometry of the complex ion formation, the concentration of complex ion produced is<\/p>\n<p style=\"text-align: center;\">[latex]0.00532-9.4\\times{10}^{-11}=0.00521M[\/latex]<\/p>\n<p>Use the silver ion and complex ion concentrations and the formation constant for the complex ion to compute the concentration of thiosulfate ion.<\/p>\n<p style=\"padding-left: 60px;\">[latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]^{2}=\\left[\\text{Ag}{\\left({\\text{S}}_{2}{\\text{O}}_{3}\\right)}_{2}{}^{\\text{3-}}\\right]\/\\left[\\text{Ag}^{+}\\right]K_{\\text{f}}=0.00521\/(9.6\\times{10}^{-11})(4.7\\times{10}^{13})=1.15\\times{10}^{-6}[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">[latex]\\left[{\\text{S}}_{2}{\\text{O}}_{3}{}^{2-}\\right]=1.1\\times{10}^{-3}M[\/latex]<\/p>\n<p>Finally, use this molar concentration to derive the required mass of sodium thiosulfate:<\/p>\n<p style=\"text-align: center;\">[latex](1.1\\times {10}^{\\text{-3}}\\text{ mol}{\\text{ S}}_{2}{\\text{O}}_{3}{}^{2-}\/\\text{L})\\times(1\\text{ mol}{\\text{ Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}\/1\\text{ mol}{\\text{ S}}_{2}{\\text{O}}_{3}{}^{2-})\\times(158.1\\text{ g}{\\text{ Na}}_{2}{\\text{S}}_{2}{\\text{O}}_{3}\/\\text{mol})=1.7\\text{ g}[\/latex]<\/p>\n<p>Thus, 1.00 L of a solution prepared from 1.7 g Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>3<\/sub> dissolves 1.0 g of AgBr.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>AgCl(s), silver chloride, is well known to have a very low solubility:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{AgCl}(\\text{s})\\rightleftharpoons\\text{Ag}^{+}(\\text{aq}) + \\text{Cl}^{-}(\\text{aq})\\qquad{K}_{\\text{sp}}=1.77\\times{10}^{-10}[\/latex]<\/p>\n<p>Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Ag}^{+}(\\text{aq})+2\\text{NH}_{3}(\\text{aq})\\rightleftharpoons\\text{Ag(NH}_{3})_{2}{}^{+}(\\text{aq})\\qquad{K}_{\\text{f}}=1.7\\times{10}^{7}[\/latex]<\/p>\n<p>What mass of NH<sub>3<\/sub> is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of Ag(NH<sub>3<\/sub>)<sub>2<\/sub><sup>+<\/sup>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q88870\">Show Solution<\/span><\/p>\n<div id=\"q88870\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.00 L of a solution prepared with 4.81 g NH<sub>3<\/sub> dissolves 2.0 g of AgCl.<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<div class=\"PageContent-ny9bj0-0 iapMdy\">\n<div id=\"main-content\" class=\"MainContent__HideOutline-sc-6yy1if-0 bdVAq\">\n<div id=\"composite-page-59\" class=\"os-eoc os-summary-container\">\n<section id=\"fs-idm305648\" class=\"summary\">\n<p id=\"fs-idm304720\">Systems involving two or more chemical equilibria that share one or more reactant or product are called coupled equilibria. Common examples of coupled equilibria include the increased solubility of some compounds in acidic solutions (coupled dissolution and neutralization equilibria) and in solutions containing ligands (coupled dissolution and complex formation). The equilibrium tools from other chapters may be applied to describe and perform calculations on these systems.<\/p>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"PrevNextBar__BarWrapper-sc-13m2i12-3 fEZPiF\"><\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li id=\"fs-idm301296\">A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?<\/li>\n<li>Calculate the equilibrium concentration of Ni<sup>2+<\/sup> in a 1.0-<em>M<\/em> solution [Ni(NH<sub>3<\/sub>)<sub>6<\/sub>](NO<sub>3<\/sub>)<sub>2<\/sub>.<\/li>\n<li>Calculate the equilibrium concentration of Zn<sup>2+<\/sup> in a 0.30-<em>M<\/em> solution of [latex]\\text{Zn}{\\left(\\text{CN}\\right)}_{4}{}^{2-}[\/latex].<\/li>\n<li>Calculate the equilibrium concentration of Cu<sup>2+<\/sup> in a solution initially with 0.050 <em>M<\/em> Cu<sup>2+<\/sup> and 1.00 <em>M<\/em> NH<sub>3<\/sub>.<\/li>\n<li>Calculate the equilibrium concentration of Zn<sup>2+<\/sup> in a solution initially with 0.150 <em>M<\/em> Zn<sup>2+<\/sup> and 2.50 <em>M<\/em> CN<sup>\u2013<\/sup>.<\/li>\n<li>Calculate the Fe<sup>3+<\/sup> equilibrium concentration when 0.0888 mole of K<sub>3<\/sub>[Fe(CN)<sub>6<\/sub>] is added to a solution with 0.0.00010 <em>M<\/em> CN<sup>\u2013<\/sup>.<\/li>\n<li>Calculate the Co<sup>2+<\/sup> equilibrium concentration when 0.100 mole of [Co(NH<sub>3<\/sub>)<sub>6<\/sub>](NO<sub>3<\/sub>)<sub>2<\/sub> is added to a solution with 0.025 <em>M<\/em> NH<sub>3<\/sub>. Assume the volume is 1.00 L.<\/li>\n<li>The equilibrium constant for the reaction [latex]{\\text{Hg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{Cl}}^{-}\\left(aq\\right)\\rightleftharpoons {\\text{HgCl}}_{2}\\left(aq\\right)[\/latex] is 1.6 [latex]\\times[\/latex] 10<sup>13<\/sup>. Is HgCl<sub>2<\/sub> a strong electrolyte or a weak electrolyte? What are the concentrations of Hg<sup>2+<\/sup> and Cl<sup>\u2013<\/sup> in a 0.015-<em>M<\/em> solution of HgCl<sub>2<\/sub>?<\/li>\n<li>Calculate the molar solubility of Sn(OH)<sub>2<\/sub> in a buffer solution containing equal concentrations of NH<sub>3<\/sub> and [latex]{\\text{NH}}_{4}{}^{+}[\/latex].<\/li>\n<li>Calculate the molar solubility of Al(OH)<sub>3<\/sub> in a buffer solution with 0.100 <em>M<\/em> NH<sub>3<\/sub> and 0.400 <em>M<\/em> [latex]{\\text{NH}}_{4}{}^{+}[\/latex].<\/li>\n<li>What is the molar solubility of CaF<sub>2<\/sub> in a 0.100-<em>M<\/em> solution of HF? <em>K<\/em><sub>a<\/sub> for HF = 7.2 [latex]\\times[\/latex] 10<sup>\u20134<\/sup>.<\/li>\n<li>What is the molar solubility of BaSO<sub>4<\/sub> in a 0.250-<em>M<\/em> solution of NaHSO<sub>4<\/sub>? <em>K<\/em><sub>a<\/sub> for [latex]{\\text{HSO}}_{4}{}^{-}[\/latex] = 1.2 [latex]\\times[\/latex] 10<sup>\u20132<\/sup>.<\/li>\n<li>What is the molar solubility of Tl(OH)<sub>3<\/sub> in a 0.10-<em>M<\/em> solution of NH<sub>3<\/sub>?<\/li>\n<li>What is the molar solubility of Pb(OH)<sub>2<\/sub> in a 0.138-<em>M<\/em> solution of CH<sub>3<\/sub>NH<sub>2<\/sub>?<\/li>\n<li>A solution of 0.075 <em>M<\/em> CoBr<sub>2<\/sub> is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). What is the minimum pH at which CoS begins to precipitate?<br \/>\n[latex]\\text{CoS}\\left(s\\right)\\rightleftharpoons {\\text{Co}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{sp}}=4.5\\times {10}^{-27}[\/latex][latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times {10}^{\\text{-26}}[\/latex]<\/li>\n<li>A 0.125-<em>M<\/em> solution of Mn(NO<sub>3<\/sub>)<sub>2<\/sub> is saturated with H<sub>2<\/sub>S ([H<sub>2<\/sub>S] = 0.10 <em>M<\/em>). At what pH does MnS begin to precipitate?[latex]\\text{MnS}\\left(s\\right)\\rightleftharpoons {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right){K}_{\\text{sp}}=4.3\\times {10}^{-22}[\/latex][latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)K=1.0\\times {10}^{-26}[\/latex]<\/li>\n<li>Calculate the molar solubility of BaF<sub>2<\/sub> in a buffer solution containing 0.20 <em>M<\/em> HF and 0.20 <em>M<\/em> NaF.<\/li>\n<li>Calculate the molar solubility of CdCO<sub>3<\/sub> in a buffer solution containing 0.115 <em>M<\/em> Na<sub>2<\/sub>CO<sub>3<\/sub> and 0.120 <em>M<\/em> NaHCO<sub>3<\/sub><\/li>\n<li>To a 0.10-<em>M<\/em> solution of Pb(NO<sub>3<\/sub>)<sub>2<\/sub> is added enough HF(<em>g<\/em>) to make [HF] = 0.10 <em>M<\/em>.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Does PbF<sub>2<\/sub> precipitate from this solution? Show the calculations that support your conclusion.<\/li>\n<li>What is the minimum pH at which PbF<sub>2<\/sub> precipitates?<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the concentration of Cd<sup>2+<\/sup> resulting from the dissolution of CdCO<sub>3<\/sub> in a solution that is 0.250 <em>M<\/em> in CH<sub>3<\/sub>CO<sub>2<\/sub>H, 0.375 <em>M<\/em> in NaCH<sub>3<\/sub>CO<sub>2<\/sub>, and 0.010 <em>M<\/em> in H<sub>2<\/sub>CO<sub>3<\/sub>.<\/li>\n<li>Both AgCl and AgI dissolve in NH<sub>3<\/sub>.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>What mass of AgI dissolves in 1.0 L of 1.0 <em>M<\/em> NH<sub>3<\/sub>?<\/li>\n<li>What mass of AgCl dissolves in 1.0 L of 1.0 <em>M<\/em> NH<sub>3<\/sub>?<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the volume of 1.50 <em>M<\/em> CH<sub>3<\/sub>CO<sub>2<\/sub>H required to dissolve a precipitate composed of 350 mg each of CaCO<sub>3<\/sub>, SrCO<sub>3<\/sub>, and BaCO<sub>3<\/sub>.<\/li>\n<li>Even though Ca(OH)<sub>2<\/sub> is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)<sub>2<\/sub>?<\/li>\n<li>What mass of NaCN must be added to 1 L of 0.010 <em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> in order to produce the first trace of Mg(OH)<sub>2<\/sub>?<\/li>\n<li>Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)<\/li>\n<li>The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Solve the following problem: [latex]{\\text{MgF}}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{F}}^{-}\\left(aq\\right)[\/latex]In a saturated solution of MgF<sub>2<\/sub> at 18 \u00b0C, the concentration of Mg<sup>2+<\/sup> is 1.21 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em>. The equilibrium is represented by the equation above.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Write the expression for the solubility-product constant, <em>K<\/em><sub>sp<\/sub>, and calculate its value at 18 \u00b0C.<\/li>\n<li>Calculate the equilibrium concentration of Mg<sup>2+<\/sup> in 1.000 L of saturated MgF<sub>2<\/sub> solution at 18 \u00b0C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.<\/li>\n<li>Predict whether a precipitate of MgF<sub>2<\/sub> will form when 100.0 mL of a 3.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>&#8211;<em>M<\/em> solution of Mg(NO<sub>3<\/sub>)<sub>2<\/sub> is mixed with 200.0 mL of a 2.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>&#8211;<em>M<\/em> solution of NaF at 18 \u00b0C. Show the calculations to support your prediction.<\/li>\n<li>At 27 \u00b0C the concentration of Mg<sup>2+<\/sup> in a saturated solution of MgF<sub>2<\/sub> is 1.17 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em>. Is the dissolving of MgF<sub>2<\/sub> in water an endothermic or an exothermic process? Give an explanation to support your conclusion.<\/li>\n<\/ol>\n<\/li>\n<li>Which of the following compounds, when dissolved in a 0.01-<em>M<\/em> solution of HClO<sub>4<\/sub>, has a solubility greater than in pure water: AgBr, BaF<sub>2<\/sub>, Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>3<\/sub>, ZnS, PbI<sub>2<\/sub>? Explain your answer.<\/li>\n<li>What is the effect on the amount of solid Mg(OH)<sub>2<\/sub> that dissolves and the concentrations of Mg<sup>2+<\/sup> and OH<sup>\u2013<\/sup> when each of the following are added to a mixture of solid Mg(OH)<sub>2<\/sub> and water at equilibrium?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>MgCl<sub>2<\/sub><\/li>\n<li>KOH<\/li>\n<li>HClO<sub>4<\/sub><\/li>\n<li>NaNO<sub>3<\/sub><\/li>\n<li>Mg(OH)<sub>2<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>What is the effect on the amount of CaHPO<sub>4<\/sub> that dissolves and the concentrations of Ca<sup>2+<\/sup> and [latex]{\\text{HPO}}_{4}{}^{-}[\/latex] when each of the following are added to a mixture of solid CaHPO<sub>4<\/sub> and water at equilibrium?\n<ol style=\"list-style-type: lower-alpha;\">\n<li>CaCl<sub>2<\/sub><\/li>\n<li>HCl<\/li>\n<li>KClO<sub>4<\/sub><\/li>\n<li>NaOH<\/li>\n<li>CaHPO<sub>4<\/sub><\/li>\n<\/ol>\n<\/li>\n<li>Identify all chemical species present in an aqueous solution of Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> and list these species in decreasing order of their concentrations. (Hint: Remember that the [latex]{\\text{PO}}_{4}{}^{\\text{3-}}[\/latex] ion is a weak base.)<\/li>\n<li>A volume of 50 mL of 1.8 <em>M<\/em> NH<sub>3<\/sub> is mixed with an equal volume of a solution containing 0.95 g of MgCl<sub>2<\/sub>. What mass of NH<sub>4<\/sub>Cl must be added to the resulting solution to prevent the precipitation of Mg(OH)<sub>2<\/sub>?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348957\">Show Selected Solutions<\/span><\/p>\n<div id=\"q348957\" class=\"hidden-answer\" style=\"display: none\">\n<p>2. [latex]{\\text{Ni}}^{\\text{2+}}\\left(aq\\right)+6{\\text{NH}}_{3}\\left(aq\\right)\\rightleftharpoons {\\left[\\text{Ni}{\\left({\\text{NH}}_{3}\\right)}_{6}\\right]}^{\\text{2+}}\\left(aq\\right)\\,\\,\\,\\,\\,\\,\\,{;}\\,\\,\\,\\,\\,\\,\\,{K}_{\\text{f}}=1.8\\times {10}^{8}[\/latex]<\/p>\n<p>Let <em>x<\/em> be the change in concentration as Ni<sup>2+<\/sup> dissociates. Because the initial Ni<sup>2+<\/sup> concentration is 0, the concentration at any times is <em>x<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]1.8\\times {10}^{8}=\\frac{{\\left[\\text{Ni}{\\left({\\text{NH}}_{3}\\right)}_{6}\\right]}^{\\text{2+}}}{\\left[{\\text{Ni}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{6}}=\\frac{\\left(1.0-x\\right)}{x{\\left(6x\\right)}^{6}}[\/latex]<\/p>\n<p style=\"text-align: center;\">1.8 [latex]\\times[\/latex] 10<sup>8<\/sup>(46656<em>x<\/em><sup>7<\/sup>) = 1.0 \u2013 <em>x<\/em><\/p>\n<p style=\"text-align: center;\">8.40 [latex]\\times[\/latex] 10<sup>12<\/sup>(<em>x<\/em><sup>2<\/sup>) = 1.0 \u2013 <em>x<\/em><\/p>\n<p>Since <em>x<\/em> is small in comparison with 1.0, drop <em>x<\/em>:<\/p>\n<p style=\"text-align: center;\">8.40 [latex]\\times[\/latex] 10<sup>12<\/sup>(<em>x<\/em><sup>7<\/sup>) = 1.0<\/p>\n<p style=\"text-align: center;\"><em>x<\/em><sup>7<\/sup> = 1.19 [latex]\\times[\/latex] 10<sup>\u201313<\/sup><\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 0.014 <em>M<\/em><\/p>\n<p>4.\u00a0Assume that all Cu<sup>2+<\/sup> forms the complex whose concentration is 0.050 <em>M<\/em> and the remaining NH<sub>3<\/sub> has a concentration of 1.00 <em>M<\/em> \u2013 4(0.050 <em>M<\/em>) = 0.80 <em>M<\/em>. The complex dissociates:<\/p>\n<p style=\"text-align: center;\">[latex]{\\left[\\text{Cu}{\\left({\\text{NH}}_{3}\\right)}_{4}\\right]}^{\\text{2+}}\\rightleftharpoons \\left[{\\text{Cu}}^{\\text{2+}}\\right]+4\\left[{\\text{NH}}_{3}\\right][\/latex]<\/p>\n<p>Let <em>x<\/em> be the change in concentration of Cu<sup>2+<\/sup> that dissociates:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[Cu(NH<sub>3<\/sub>)<sub>4<\/sub><sup>2+<\/sup>]<\/th>\n<th>[Cu<sup>2+<\/sup>]<\/th>\n<th>[NH<sub>3<\/sub>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.050<\/td>\n<td>0<\/td>\n<td>0.80<\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.050\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>4<em>x<\/em> + 0.80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Cu}{\\left({\\text{NH}}_{3}\\right)}_{4}{}^{\\text{2+}}\\right]}{\\left[{\\text{Cu}}^{\\text{2+}}\\right]{\\left[{\\text{NH}}_{3}\\right]}^{4}}=1.2\\times {10}^{12}=\\frac{0.050-x}{x{\\left(4x+0.80\\right)}^{4}}[\/latex]<\/p>\n<p>Assume that 4<em>x<\/em> is small when compared with 0.80 and that <em>x<\/em> is small when compared with 0.050:<\/p>\n<p style=\"text-align: center;\">(0.80)<sup>4<\/sup> [latex]\\times[\/latex] 1.2 [latex]\\times[\/latex] 10<sup>12<\/sup><em>x<\/em> = 0.050<\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 1.0 [latex]\\times[\/latex] 10<sup>\u201313<\/sup><em>M<\/em><\/p>\n<p>6.\u00a0Set up a table listing initial and equilibrium concentrations for the reaction:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{Fe}}^{\\text{3+}}+6{\\text{CN}}^{-}\\rightleftharpoons {\\left(\\text{Fe}{\\left(\\text{CN}\\right)}_{6}\\right]}^{3-}{K}_{\\text{f}}=1\\times {10}^{44}[\/latex]<\/p>\n<p>Let <em>x<\/em> be the concentration of Fe<sup>3+<\/sup> that dissociates when 0.0888 mol dissolves in 1.00 L of 0.00010 <em>M<\/em> CN<sup>\u2013<\/sup>. Assume no volume change upon dissolution:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[Fe(CN)<sub>6<\/sub><sup>3\u2212<\/sup>]<\/th>\n<th>[Fe<sup>3+<\/sup>]<\/th>\n<th>[CN<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.0888<\/td>\n<td>0<\/td>\n<td>0.00010<\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.0888\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>0.00010\u00a0\u2212 6<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">[latex]\\frac{\\left[\\text{Fe}{\\left(\\text{CN}\\right)}_{6}{}^{3-}\\right]}{\\left[{\\text{Fe}}^{\\text{3+}}\\right]{\\left[{\\text{CN}}^{-}\\right]}^{6}}=\\frac{0.0888-x}{x{\\left(0.000010 - 6x\\right)}^{6}}=1\\times {10}^{44}[\/latex]<\/p>\n<p>Assume that <em>x<\/em> is small when compared with the terms from which it is subtracted:<\/p>\n<p style=\"text-align: center;\">0.0888 = (0.00010)<sup>6<\/sup>(<em>x<\/em>)(1 [latex]\\times[\/latex] 10<sup>44<\/sup>)<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{0.0888}{1\\times {10}^{26}}=9\\times {10}^{-22}M[\/latex]<\/p>\n<p>8.\u00a0Let <em>x<\/em> be the change in the number of moles of Hg<sup>2+<\/sup> that form per liter:<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>[HgCl<sub>2<\/sub>]<\/th>\n<th>[Hg<sup>2+<\/sup>]<\/th>\n<th>[Cl<sup>\u2212<\/sup>]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Initial concentration (<em>M<\/em>)<\/th>\n<td>0.015<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<th>Equilibrium (<em>M<\/em>)<\/th>\n<td>0.015\u00a0\u2212\u00a0<em>x<\/em><\/td>\n<td><em>x<\/em><\/td>\n<td>2<em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{\\left[{\\text{HgCl}}_{2}\\right]}{\\left[{\\text{Hg}}^{\\text{2+}}\\right]{\\left[{\\text{Cl}}^{-}\\right]}^{2}}=K=1.6\\times {10}^{13}\\\\ \\frac{0.015-x}{\\left(x\\right){\\left(2x\\right)}^{2}}\\approx \\frac{0.015}{4{x}^{3}}=1.6\\times {10}^{13}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>x<\/em><sup>3<\/sup> = 2.3 [latex]\\times[\/latex] 10<sup>\u201316<\/sup><\/p>\n<p style=\"text-align: center;\"><em>x<\/em> = 6.2 [latex]\\times[\/latex] 10<sup>\u20136<\/sup><em>M<\/em> = [Hg<sup>2+<\/sup>]<\/p>\n<p style=\"text-align: center;\">2x = 1.2 [latex]\\times[\/latex] 10<sup>\u20135<\/sup>] <em>M<\/em> = [Cl<sup>\u2013<\/sup>]<\/p>\n<p>The substance is a weak electrolyte because very little of the initial 0.015 <em>M<\/em> HgCl<sub>2<\/sub> dissolved.<\/p>\n<p>10.\u00a0[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{NH}}_{4}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{NH}}_{3}\\right]}=\\frac{\\left(0.400\\right)\\left[{\\text{OH}}^{-}\\right]}{\\left(0.100\\right)}=1.8\\times {10}^{-5}[\/latex]<\/p>\n<p>[latex]\\left[{\\text{OH}}^{-}\\right]=\\frac{\\left(0.100\\right)\\left(1.8\\times {10}^{-5}\\right)}{0.0400}\\text{=}4.5\\times {10}^{-5}[\/latex]<\/p>\n<p><em>K<\/em><sub>sp<\/sub> = [Al<sup>3+<\/sup>][OH<sup>\u2013<\/sup>]<sup>3<\/sup> = [Al<sup>3+<\/sup>](4.5 [latex]\\times[\/latex] 10<sup>\u20135<\/sup>)<sup>3<\/sup> = 1.9 [latex]\\times[\/latex] 10<sup>\u201333<\/sup><\/p>\n<p>[Al<sup>3+<\/sup>] = 2.1 [latex]\\times[\/latex] 10<sup>\u201320<\/sup> (molar solubility)<\/p>\n<p>12.\u00a0Find the amount of [latex]{\\text{SO}}_{4}{}^{2-}[\/latex] present from <em>K<\/em><sub>a<\/sub> for the equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{HSO}}_{4}{}^{-}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{SO}}_{4}{}^{2-}[\/latex]<\/p>\n<p>Let <em>x<\/em> be the change in [latex]\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex] :<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{SO}}_{4}{}^{2-}\\right]}{\\left[{\\text{HSO}}_{4}{}^{-}\\right]}=\\frac{{x}^{2}}{0.250-x}=1.2\\times {10}^{-2}[\/latex]<\/p>\n<p>Because <em>K<\/em><sub>a<\/sub> is too large to disregard <em>x<\/em> in the expression 0.250 \u2013 <em>x<\/em>, we must solve the quadratic equation:<\/p>\n<p style=\"text-align: center;\"><em>x<\/em><sup>2<\/sup> + 1.2 [latex]\\times[\/latex] 10<sup>\u20132<\/sup><em>x<\/em> \u2013 0.250(1.2 [latex]\\times[\/latex] 10<sup>\u20132<\/sup>) = 0<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-1.2\\times {10}^{-2}\\pm \\sqrt{{\\left(1.2\\times {10}^{-2}\\right)}^{2}+4\\left(3.0\\times {10}^{\\text{-3}}\\right)}}{2}=\\frac{-1.2\\times {10}^{-2}\\pm 0.11}{2}=0.049M[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Ba<sup>2+<\/sup>] [latex]\\left[{\\text{SO}}_{4}{}^{2-}\\right][\/latex] = [Ba<sup>2+<\/sup>](0.049) = 1.08 [latex]\\times[\/latex] 10<sup>\u201310<\/sup><\/p>\n<p style=\"text-align: center;\">[Ba<sup>2+<\/sup>] = 2.2 [latex]\\times[\/latex] 10<sup>\u20139<\/sup> (molar solubility)<\/p>\n<p>14.\u00a0[latex]{\\text{CH}}_{3}{\\text{NH}}_{2}+{\\text{H}}_{2}\\text{O}\\rightleftharpoons {\\text{CH}}_{3}{\\text{NH}}_{3}{}^{+}+{\\text{OH}}^{-}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{b}}=\\frac{\\left[{\\text{CH}}_{3}{\\text{NH}}_{3}{}^{+}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{CH}}_{3}{\\text{NH}}_{2}\\right]}=\\frac{\\left(x\\right)\\left(x\\right)}{0.138-x}=4.4\\times {10}^{-4}[\/latex]<\/p>\n<p>Solve the quadratic equation using the quadratic formula:<\/p>\n<p style=\"text-align: center;\"><em>x<\/em><sup>2<\/sup> + 4.4 [latex]\\times[\/latex] 10<sup>\u20134<\/sup><em>x<\/em> \u2013 0.138(4.4 [latex]\\times[\/latex] 10<sup>\u20134<\/sup>) = 0<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4.4\\times {10}^{-4}\\pm \\sqrt{{\\left(4.4\\times {10}^{-4}\\right)}^{2}+4\\left(6.07\\times {10}^{\\text{-5}}\\right)}}{2}=\\frac{-4.4\\times {10}^{\\text{-4}}\\pm \\sqrt{2.43\\times {10}^{-4}}}{2}=\\frac{0.0152}{2}=7.6\\times {10}^{-3}M[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Pb<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = [Pb<sup>2+<\/sup>](7.6 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 2.8 [latex]\\times[\/latex] 10<sup>\u201316<\/sup><\/p>\n<p style=\"text-align: center;\">[Pb<sup>2+<\/sup>] = 4.8 [latex]\\times[\/latex] 10<sup>\u201312<\/sup> (molar solubility)<\/p>\n<p>16.\u00a0Two equilibria are in competition for the ions and must be considered simultaneously. Precipitation of MnS will occur when the concentration of S<sup>2\u2013<\/sup> in conjunction with 0.125 <em>M<\/em> Mn<sup>2+<\/sup> exceeds the <em>K<\/em><sub>sp<\/sub> of MnS. The [S<sup>2\u2013<\/sup>] must come from the ionization of H<sub>2<\/sub>S as defined by the equilibrium:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{H}}_{2}\\text{S}\\left(aq\\right)+2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\longrightarrow 2{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left[{\\text{S}}^{2-}\\right]}{\\left[{\\text{H}}_{2}\\text{S}\\right]}={K}_{1}{K}_{2}\\left({\\text{H}}_{2}\\text{S}\\right)=1.0\\times {10}^{\\text{-26}}[\/latex]<\/p>\n<p>As a saturated solution of H<sub>2<\/sub>S is 0.10 <em>M<\/em>, this later expression becomes:<\/p>\n<p style=\"text-align: center;\">[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left[{\\text{S}}^{2-}\\right]=1.0\\times {10}^{-27}[\/latex]<\/p>\n<p>From the equilibrium of MnS, the minimum concentration of S<sup>2\u2013<\/sup> required to cause precipitation is calculated as:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{MnS}\\left(s\\right)\\longrightarrow {\\text{Mn}}^{\\text{2+}}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Mn<sup>2+<\/sup>][S<sup>2\u2013<\/sup>] = 4.3 [latex]\\times[\/latex] 10<sup>\u201322<\/sup><\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{S}}^{2-}\\right]=\\frac{4.3\\times {10}^{-22}}{0.125}=3.44\\times {10}^{-21}[\/latex]<\/p>\n<p>This amount of S<sup>2\u2013<\/sup> will exist in solution at a pH defined by the H<sub>2<\/sub>S equilibrium:<\/p>\n<ul>\n<li>[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}\\left(3.44\\times {10}^{-21}\\right)=1.0\\times {10}^{-27}[\/latex]<\/li>\n<li>[latex]{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}^{2}=2.91\\times {10}^{-7}[\/latex]<\/li>\n<li>[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=5.39\\times {10}^{-4}M[\/latex]<\/li>\n<li>[latex]\\text{pH}=-\\text{log}\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=3.27[\/latex]<\/li>\n<\/ul>\n<p>18.\u00a0Three equilibria are involved:<\/p>\n<ul>\n<li>[latex]{\\text{H}}_{2}{\\text{CO}}_{3}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{HCO}}_{3}{}^{-}{K}_{{\\text{a}}_{1}}=4.3\\times {10}^{-7}[\/latex]<\/li>\n<li>[latex]{\\text{HCO}}_{3}{}^{-}+{\\text{H}}_{2}\\text{O}\\longrightarrow {\\text{H}}_{3}{\\text{O}}^{+}+{\\text{CO}}_{3}{}^{2-}{K}_{{\\text{a}}_{2}}=7\\times {10}^{-11}[\/latex]<\/li>\n<li>[latex]{\\text{CdCO}}_{3}\\longrightarrow {\\text{Cd}}^{\\text{2+}}+{\\text{CO}}_{3}{}^{2-}{K}_{\\text{sp}}=2.5\\times {10}^{-14}[\/latex]<\/li>\n<\/ul>\n<p>First, find the pH of the buffer from the Henderson-Hasselbach equation. Then find [H<sub>3<\/sub>O<sup>+<\/sup>]:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{pH}=\\text{p}{K}_{\\text{a}}+\\mathrm{log}\\frac{\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{-}\\right]}[\/latex]<\/p>\n<p style=\"text-align: center;\">10.155 + log [latex]\\frac{0.115}{0.120}[\/latex] = 10.155 \u2013 0.018 = 10.137<\/p>\n<p>In this case, several more significant figures are carried than justified so that the value of the log ratio is meaningful:<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=7.3\\times {10}^{-11}[\/latex]<\/p>\n<p>Now, find the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] present in the buffer solution. Next, using [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] and <em>K<\/em><sub>sp<\/sub>, calculate the concentration of Cd<sup>2+<\/sup>. This latter value represents the molar solubility. From [latex]{K}_{{a}_{1}}[\/latex] determine [H<sub>2<\/sub>CO<sub>3<\/sub>]:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{{\\text{a}}_{1}}=4.3\\times {10}^{-7}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCO}}_{3}{}^{-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left(7.3\\times {10}^{-11}\\right)\\left(0.120\\right)}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}[\/latex]<\/p>\n<p style=\"text-align: center;\">[H<sub>2<\/sub>CO<sub>3<\/sub>] = 2.05 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><\/p>\n<p>From [latex]{K}_{{\\text{a}}_{1}}{K}_{{\\text{a}}_{2}}[\/latex] , find the concentration of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] :<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{}{K}_{{\\text{a}}_{1}}{K}_{{\\text{a}}_{2}}&=&\\left(4.3\\times 10 - 7\\right)\\left(7\\times 10 - 11\\right)=\\frac{\\left[{\\text{HCO}}_{3}{}^{-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}\\times \\frac{\\left[{\\text{CO}}_{3}{}^{2-}\\right]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}=\\frac{{\\left(7.3\\times {10}^{-11}\\right)}^{2}\\times \\left[{\\text{CO}}_{3}{}^{2-}\\right]}{2.04\\times {10}^{-5}}\\\\& =&3\\times {10}^{-17}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 0.115 <em>M<\/em><\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = [Cd<sup>2+<\/sup>](0.115) = 3 [latex]\\times[\/latex] 10<sup>\u201313<\/sup><\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{3\\times {10}^{-13}}{0.115}=3\\times {10}^{-12}M[\/latex]<\/p>\n<p>20.\u00a0For <em>K<\/em><sub>sp<\/sub>, (CdCO<sub>3<\/sub>) = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 2.5 [latex]\\times[\/latex] 10<sup>\u201314<\/sup>, the amount of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] is governed by <em>K<\/em><sub>a<\/sub> of H<sub>2<\/sub>CO<sub>3<\/sub>, 4.3 [latex]\\times[\/latex] 10<sup>\u20137<\/sup>, and <em>K<\/em><sub>a<\/sub> of [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] , 7 [latex]\\times[\/latex] 10<sup>\u201311<\/sup>, and [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] by <em>K<\/em><sub>a<\/sub> of acetic acid. First, calculate the [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] from <em>K<\/em><sub>a<\/sub> of acetic acid (HOAc):<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]=\\frac{{K}_{\\text{a}}\\left[\\text{HOAc}\\right]}{\\left[{\\text{OAc}}^{-}\\right]}=\\frac{1.8\\times {10}^{-5}\\left(0.250\\right)}{\\left(0.375\\right)}=1.2\\times {10}^{-5}M[\/latex]<\/p>\n<p>From this and <em>K<\/em><sub>a<\/sub> for H<sub>2<\/sub>CO<sub>3<\/sub>, calculate [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] present. As [latex]\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right][\/latex] is fixed by <em>K<\/em><sub>a<\/sub> of acetic acid:<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{\\text{a}}\\left({\\text{H}}_{2}{\\text{CO}}_{3}\\right)=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{HCP}}_{3}{}^{-}\\right]}{\\left[{\\text{H}}_{2}{\\text{CO}}_{3}{}^{-}\\right]}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{H}}_{2}{\\text{CO}}_{3}{}^{-}\\right]=\\frac{{K}_{\\text{a}}\\left[{\\text{H}}_{2}{\\text{CO}}_{3}\\right]}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{4.3\\times {10}^{-7}\\left[0.010\\right]}{\\left[1.2\\times {10}^{-5}\\right]}=3.58\\times {10}^{-4}M[\/latex]<\/p>\n<p>From [latex]{K}_{\\text{a}}\\left({\\text{HCO}}_{3}{}^{-}\\right)[\/latex] we obtain [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] :<\/p>\n<p style=\"text-align: center;\">[latex]{K}_{a}=7\\times {10}^{-11}=\\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[{\\text{HCO}}_{3}{}^{-}\\right]}=\\frac{1.2\\times {10}^{-17}\\left[{\\text{CO}}_{3}{}^{2-}\\right]}{\\left[3.58\\times {10}^{-4}\\right]}\\text{ }[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right]=\\frac{7\\times {10}^{-11}\\times 3.58\\times {10}^{-4}}{1.2\\times {10}^{-5}}=2.09\\times {10}^{-9}[\/latex]<\/p>\n<p>From the solubility product:<\/p>\n<p style=\"text-align: center;\"><em>K<\/em><sub>sp<\/sub> = [Cd<sup>2+<\/sup>] [latex]\\left[{\\text{CO}}_{3}{}^{2-}\\right][\/latex] = 2.5 [latex]\\times[\/latex] 10<sup>\u201314<\/sup> = Cd<sup>2+<\/sup> (2.09 [latex]\\times[\/latex] 10<sup>\u20139<\/sup>)<\/p>\n<p style=\"text-align: center;\">[latex]\\left[{\\text{Cd}}^{\\text{2+}}\\right]=\\frac{2.5\\times {10}^{-14}}{2.09\\times {10}^{-9}}=1\\times {10}^{-5}M[\/latex]<\/p>\n<p>22.\u00a0[latex]{\\text{CaCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{100.09\\cancel{\\text{g}}}=3.50\\times {10}^{-3}\\text{mol}[\/latex]<\/p>\n<p>[latex]{\\text{SrCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{147.63\\cancel{\\text{g}}}=2.37\\times {10}^{-3}\\text{mol}[\/latex]<\/p>\n<p>[latex]{\\text{BaCO}}_{3}:0.350\\cancel{\\text{g}}\\times \\frac{1\\text{mol}}{197.34\\cancel{\\text{g}}}=1.77\\times {10}^{-3}\\text{mol}[\/latex]<\/p>\n<ul>\n<li>Total: 7.64 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol<\/li>\n<li>CaCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 4.8 [latex]\\times[\/latex] 10<sup>\u20139<\/sup><\/li>\n<li>SrCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 9.42 [latex]\\times[\/latex] 10<sup>\u201310<\/sup><\/li>\n<li>BaCO<sub>3<\/sub>: <em>K<\/em><sub>sp<\/sub> = 8.1 [latex]\\times[\/latex] 10<sup>\u20139<\/sup><\/li>\n<\/ul>\n<p>Solubilities are approximately equal.<\/p>\n<p>When the three solids dissolve, 7.64 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of metal ions and 7.64 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] are initially present. The [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] reacts with CH<sub>3<\/sub>CO<sub>2<\/sub>H to form [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] :<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{ CO}}_{3}{}^{2-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)\\rightleftharpoons {\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]K=\\frac{\\left[{\\text{HCO}}_{3}{}^{-}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\right]}{\\left[{\\text{CO}}_{3}{}^{2-}\\right]\\left[{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right]}\\times \\frac{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}{\\left[{\\text{H}}_{3}{\\text{O}}^{+}\\right]}=\\frac{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}{{K}_{\\text{a}}\\left(\\text{for}{\\text{HCO}}_{3}{}^{-}\\right)}=\\frac{1.8\\times {10}^{-5}}{7\\times {10}^{-11}}=2.6\\times {10}^{5}[\/latex]<\/p>\n<p>This <em>K<\/em> value is large, so virtually all [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] undergoes this reaction and approximately 7.64 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol of [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] forms. The [latex]{\\text{HCO}}_{3}{}^{-}[\/latex] reacts with CH<sub>3<\/sub>CO<sub>2<\/sub>H:<\/p>\n<p style=\"text-align: center;\">[latex]{\\text{HCO}}_{3}{}^{-}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)\\longrightarrow {\\text{H}}_{2}{\\text{CO}}_{3}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{}^{-}\\left(aq\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]K=\\frac{{K}_{\\text{a}}\\left(\\text{for}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\right)}{{K}_{\\text{a}}\\left(\\text{for}{\\text{HCO}}_{3}{}^{-}\\right)}=\\frac{1.8\\times {10}^{-5}}{4.3\\times {10}^{-7}}=42[\/latex]<\/p>\n<p>This reaction is virtually complete as well (<em>K<\/em> is large).<\/p>\n<p>For each mol of [latex]{\\text{CO}}_{3}{}^{2-}[\/latex] produced, 2 mol of CH<sub>3<\/sub>CO<sub>2<\/sub>H is required for conversion to H<sub>2<\/sub>CO<sub>3<\/sub>. Thus 15.3 [latex]\\times[\/latex] 10<sup>\u20133<\/sup> mol CH<sub>3<\/sub>CO<sub>2<\/sub>H is needed:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Volume}=\\frac{15.3\\times {10}^{-3}\\text{mol}}{1.50\\text{mol}{\\text{L}}^{-1}}=0.0102\\text{L}\\left(10.2\\text{mL}\\right)[\/latex]<\/p>\n<p>24.\u00a0There are two equilibria involved: [latex]\\begin{array}{l}\\text{Mg}{\\left(\\text{OH}\\right)}_{2}\\left(s\\right)\\rightleftharpoons {\\text{Mg}}^{\\text{2+}}\\left(aq\\right)+2{\\text{OH}}^{-}\\left(aq\\right){K}_{\\text{sp}}=1.5\\times {10}^{-11}\\\\ {\\text{CN}}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons \\text{HCN}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\left({K}_{\\text{b}}=\\frac{{K}_{\\text{w}}}{{K}_{\\text{a}}}=\\frac{1.0\\times {10}^{-14}}{4\\times {10}^{-10}}=2.5\\times {10}^{-5}\\right)\\end{array}[\/latex] The Mg(NO<sub>3<\/sub>)<sub>2<\/sub> dissolves, and [Mg<sup>2+<\/sup>] = 0.010 <em>M<\/em>.<\/p>\n<p>[Mg<sup>2+<\/sup>][OH<sup>\u2013<\/sup>]<sup>2<\/sup> = <em>K<\/em><sub>sp<\/sub> = 1.5 [latex]\\times[\/latex] 10<sup>\u201311<\/sup><\/p>\n<p>(0.010)[OH<sup>\u2013<\/sup>]<sup>2<\/sup> = 1.5 [latex]\\times[\/latex] 10<sup>\u201311<\/sup><\/p>\n<p>[OH<sup>\u2013<\/sup>] = 3.9 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em><\/p>\n<p>We need to add enough CN<sup>\u2013<\/sup> to make [OH<sup>\u2013<\/sup>] = 3.9 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em>. Both OH<sup>\u2013<\/sup> and HCN come from CN<sup>\u2013<\/sup>, so [OH<sup>\u2013<\/sup>] = [HCN]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{\\left[\\text{HCN}\\right]\\left[{\\text{OH}}^{-}\\right]}{\\left[{\\text{CN}}^{-}\\right]}={K}_{\\text{b}}=2\\times {10}^{-5}\\\\ \\frac{{\\left(3.9\\times {10}^{-5}\\right)}^{2}}{\\left[{\\text{CN}}^{-}\\right]}=2.5\\times {10}^{-5}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[CN<sup>\u2013<\/sup>] = 6.1 [latex]\\times[\/latex] 10<sup>\u20135<\/sup><em>M<\/em><\/p>\n<p>mol NaCN = mol HCN + mol CN<sup>\u2013<\/sup> = 3.9 [latex]\\times[\/latex] 10<sup>\u20135<\/sup> + 6.1 [latex]\\times[\/latex] 10<sup>\u20135<\/sup> = 1.0 [latex]\\times[\/latex] 10<sup>\u20134<\/sup><\/p>\n<p>[latex]\\text{mass}\\left(\\text{NaCN}\\right)=1.0\\times {10}^{-4}\\text{mol}\\times \\frac{49.007\\text{g}}{1\\text{mol}}=5\\times {10}^{-3}\\text{g}[\/latex]<\/p>\n<p>26. The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = (1.21 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)(2 [latex]\\times[\/latex] 1.21 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 7.09 [latex]\\times[\/latex] 10<sup>\u20139<\/sup><\/li>\n<li><em>K<\/em><sub>sp<\/sub> = [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> = [<em>x<\/em>][0.100 + 2<em>x<\/em>]<sup>2<\/sup> = 7.09 [latex]\\times[\/latex] 10<sup>\u20139<br \/>\n<\/sup>Assume that 2<em>x<\/em> is small when compared with 0.100 <em>M<\/em>.<br \/>\n0.100<em>x<\/em> = 7.09 [latex]\\times[\/latex] 10<sup>\u20139<br \/>\n<\/sup><em>x<\/em> = [MgF<sub>2<\/sub>] = 7.09 [latex]\\times[\/latex] 10<sup>\u20137<\/sup><em>M<br \/>\n<\/em>The value 7.09 [latex]\\times[\/latex] 10<sup>\u20137<\/sup><em>M<\/em> is quite small when compared with 0.100 <em>M<\/em>, so the assumption is valid.<\/li>\n<li>Determine the concentration of Mg<sup>2+<\/sup> and F<sup>\u2013<\/sup> that will be present in the final volume. Compare the value of the ion product [Mg<sup>2+<\/sup>][F<sup>\u2013<\/sup>]<sup>2<\/sup> with <em>K<\/em><sub>sp<\/sub>. If this value is larger than <em>K<\/em><sub>sp<\/sub>, precipitation will occur.<br \/>\n0.1000 L [latex]\\times[\/latex] 3.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> = 0.3000 L [latex]\\times[\/latex] <em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<br \/>\n<\/sub><em>M<\/em> Mg(NO<sub>3<\/sub>)<sub>2<\/sub> = 1.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<br \/>\n<\/em>0.2000 L [latex]\\times[\/latex] 2.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<\/em> NaF = 0.3000 L [latex]\\times[\/latex] <em>M<\/em> NaF<br \/>\n<em>M<\/em> NaF = 1.33 [latex]\\times[\/latex] 10<sup>\u20133<\/sup><em>M<br \/>\n<\/em>ion product = (1.00 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)(1.33 [latex]\\times[\/latex] 10<sup>\u20133<\/sup>)<sup>2<\/sup> = 1.77 [latex]\\times[\/latex] 10<sup>\u20139<br \/>\n<\/sup>This value is smaller than <em>K<\/em><sub>sp<\/sub>, so no precipitation will occur.<\/li>\n<li>MgF<sub>2<\/sub> is less soluble at 27 \u00b0C than at 18 \u00b0C. Because added heat acts like an added reagent, when it appears on the product side, the Le Ch\u00e2telier\u2019s principle states that the equilibrium will shift to the reactants\u2019 side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.<\/li>\n<\/ol>\n<p>28.\u00a0Effect on amount of solid Mg(OH)<sub>2<\/sub>, [Mg<sup>2+<\/sup>], [OH<sup>\u2013<\/sup>]:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>increase, increase, decrease;<\/li>\n<li>increase, decrease, increase;<\/li>\n<li>decrease, increase, decrease;<\/li>\n<li>no effect predicted;<\/li>\n<li>increase, no effect, no effect<\/li>\n<\/ol>\n<p>30.\u00a0[H<sub>2<\/sub>O] &gt; [Ca<sup>2+<\/sup>] &gt; [latex]\\left[{\\text{PO}}_{4}{}^{3-}\\right][\/latex] &gt; [latex]{\\text{HPO}}_{4}{}^{2-}[\/latex] = [OH<sup>\u2013<\/sup>] &gt; [latex]\\left[{\\text{HPO}}_{4}{}^{-}\\right][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><b>coupled equilibrium: <\/b>system characterized by more than one state of balance between a slightly soluble ionic solid and an aqueous solution of ions working simultaneously<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3605\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry 2e. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/\">https:\/\/openstax.org\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Ocean Acidification. <strong>Authored by<\/strong>: NCAquariumFortFisher. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/kxPwbhFeZSw\">https:\/\/youtu.be\/kxPwbhFeZSw<\/a>. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry 2e\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction\"},{\"type\":\"copyrighted_video\",\"description\":\"Ocean Acidification\",\"author\":\"NCAquariumFortFisher\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/kxPwbhFeZSw\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"Standard YouTube License\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3605","chapter","type-chapter","status-publish","hentry"],"part":2983,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3605","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":23,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3605\/revisions"}],"predecessor-version":[{"id":7878,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3605\/revisions\/7878"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/parts\/2983"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3605\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/media?parent=3605"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapter-type?post=3605"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/contributor?post=3605"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/license?post=3605"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}