{"id":245,"date":"2017-06-20T20:33:11","date_gmt":"2017-06-20T20:33:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/?post_type=chapter&#038;p=245"},"modified":"2017-06-20T21:24:08","modified_gmt":"2017-06-20T21:24:08","slug":"lab-10-introduction","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/chapter\/lab-10-introduction\/","title":{"raw":"Lab 10 Introduction","rendered":"Lab 10 Introduction"},"content":{"raw":"<b>Chesapeake Campus \u2013 Chemistry 112 Laboratory<\/b>\r\n\r\nLAB #10 \u2013The Common Ion Effect\r\n\r\n<b>Objectives<\/b>\r\n\r\nAfter completing this lab you will be able to:\r\n<ul>\r\n \t<li>Calculate the [OH<sup>-<\/sup>] using a pH probe.<\/li>\r\n \t<li>Calculate the molar solubility of Ca(OH)<sub>2 <\/sub>and Mg(OH)<sub>2<\/sub>.<\/li>\r\n \t<li>Calculate the solubility constants of Ca(OH)<sub>2 <\/sub>and Mg(OH)<sub>2<\/sub>.<\/li>\r\n<\/ul>\r\n<b>Introduction<\/b>\r\n\r\nSolubility\r\n\r\nIn CHM 111 you were introduced to the idea that some ionic compounds are soluble while others are insoluble.\u00a0 There are many salts that are classified as \u201cslightly soluble\u201d or \u201cmarginally soluble.\u201d\u00a0 Slightly soluble salts are in a constant equilibrium in solution.\u00a0 They typically have low K values indicating they are reactant favored.\u00a0 For example, lead II bromide ionizes in solution as\r\n\r\nwhere the equilibrium expression is depicted as:\r\n\r\nNotice that the equilibrium expression does not include the lead II bromide as this is a solid.\u00a0 The equilibrium constant for a slightly soluble salt is given as K<sub>sp <\/sub>or the <b>solubility constant<\/b>.\r\n\r\nThe <b>molar solubility<\/b> of a compound is the amount of that compound that can be dissolved to make a saturated solution.\u00a0 Lead II bromide has a molar solubility of 4.4 x 10<sup>-2<\/sup> .\u00a0 This means that only 0.044 mol of lead II bromide can be dissolved in a 1 L solution.\u00a0 According to the equation above, we can tell that every lead II bromide that dissolves produces 1 lead ion and 2 bromide ions.\u00a0 Therefore in solution we can determine the concentration of both ions as:\r\n\r\nThis indicates the [Pb<sup>2+<\/sup>] concentration is 4.4 x 10<sup>-2<\/sup> M and the bromide ion concentration is 8.8 x 10<sup>-2<\/sup> M.\r\n\r\nThis can be plugged into the Ksp equation to find the solubility constant:\r\n\r\nLeCh\u00e2telier\u2019s Principle\r\n\r\nLeCh\u00e2telier\u2019s Principle states that the equilibrium will shift to alleviate the stress on a system.\u00a0 Therefore, if NaBr was added to the lead II bromide solution above, it would add in Br<sup>-<\/sup>. This addition of \u201cproduct\u201d ion would cause the system to shift to the left to produce more solid lead II bromide.\u00a0 Since an ion common to the original salt was added, this is often called the <b>common ion effect<\/b>.\u00a0 The shift to the left in equilibrium is accompanied by a reduction in the amount of salt that remains in solution.\u00a0 Thus the molar solubility is decreased with the addition of a common ion.\r\n\r\nDetermining the Molar Solubility Using pH\r\n\r\nIn today\u2019s lab, you will be calculating the molar solubility of two slightly soluble hydroxide salts; Ca(OH)<sub>2<\/sub> and Mg(OH)<sub>2<\/sub>.\u00a0 Calcium hydroxide ionizes according to the following equation:\r\n\r\nEquation 1\r\n\r\nfrom this equation we can see that\r\n\r\nEquation 2\r\n\r\nThe hydroxide ion concentration will be determined by taking a pH reading for the solution.\u00a0 The pH allows the pOH to be calculated using Equation 3.\r\n\r\nEquation 3\r\n\r\nThe hydroxide ion concentration can then be calculated by solving the pOH equation for [OH<sup>=<\/sup>].\r\n\r\nEquation 4\r\n\r\nEquation 5\r\n\r\nOnce the hydroxide ion concentration is calculated, it is possible to use Equation 2 to determine the concentration of calcium in solution and the molar solubility of calcium hydroxide.\u00a0 The concentrations of calcium and hydroxide ions can be plugged into the K<sub>sp<\/sub> equation for calcium hydroxide:\r\n\r\nEquation 6\r\n\r\nEquation 6 can also be rewritten with the calcium ion concentration substituted according to Equation 2 to give Equation 7 below.\r\n\r\nEquation 7","rendered":"<p><b>Chesapeake Campus \u2013 Chemistry 112 Laboratory<\/b><\/p>\n<p>LAB #10 \u2013The Common Ion Effect<\/p>\n<p><b>Objectives<\/b><\/p>\n<p>After completing this lab you will be able to:<\/p>\n<ul>\n<li>Calculate the [OH<sup>&#8211;<\/sup>] using a pH probe.<\/li>\n<li>Calculate the molar solubility of Ca(OH)<sub>2 <\/sub>and Mg(OH)<sub>2<\/sub>.<\/li>\n<li>Calculate the solubility constants of Ca(OH)<sub>2 <\/sub>and Mg(OH)<sub>2<\/sub>.<\/li>\n<\/ul>\n<p><b>Introduction<\/b><\/p>\n<p>Solubility<\/p>\n<p>In CHM 111 you were introduced to the idea that some ionic compounds are soluble while others are insoluble.\u00a0 There are many salts that are classified as \u201cslightly soluble\u201d or \u201cmarginally soluble.\u201d\u00a0 Slightly soluble salts are in a constant equilibrium in solution.\u00a0 They typically have low K values indicating they are reactant favored.\u00a0 For example, lead II bromide ionizes in solution as<\/p>\n<p>where the equilibrium expression is depicted as:<\/p>\n<p>Notice that the equilibrium expression does not include the lead II bromide as this is a solid.\u00a0 The equilibrium constant for a slightly soluble salt is given as K<sub>sp <\/sub>or the <b>solubility constant<\/b>.<\/p>\n<p>The <b>molar solubility<\/b> of a compound is the amount of that compound that can be dissolved to make a saturated solution.\u00a0 Lead II bromide has a molar solubility of 4.4 x 10<sup>-2<\/sup> .\u00a0 This means that only 0.044 mol of lead II bromide can be dissolved in a 1 L solution.\u00a0 According to the equation above, we can tell that every lead II bromide that dissolves produces 1 lead ion and 2 bromide ions.\u00a0 Therefore in solution we can determine the concentration of both ions as:<\/p>\n<p>This indicates the [Pb<sup>2+<\/sup>] concentration is 4.4 x 10<sup>-2<\/sup> M and the bromide ion concentration is 8.8 x 10<sup>-2<\/sup> M.<\/p>\n<p>This can be plugged into the Ksp equation to find the solubility constant:<\/p>\n<p>LeCh\u00e2telier\u2019s Principle<\/p>\n<p>LeCh\u00e2telier\u2019s Principle states that the equilibrium will shift to alleviate the stress on a system.\u00a0 Therefore, if NaBr was added to the lead II bromide solution above, it would add in Br<sup>&#8211;<\/sup>. This addition of \u201cproduct\u201d ion would cause the system to shift to the left to produce more solid lead II bromide.\u00a0 Since an ion common to the original salt was added, this is often called the <b>common ion effect<\/b>.\u00a0 The shift to the left in equilibrium is accompanied by a reduction in the amount of salt that remains in solution.\u00a0 Thus the molar solubility is decreased with the addition of a common ion.<\/p>\n<p>Determining the Molar Solubility Using pH<\/p>\n<p>In today\u2019s lab, you will be calculating the molar solubility of two slightly soluble hydroxide salts; Ca(OH)<sub>2<\/sub> and Mg(OH)<sub>2<\/sub>.\u00a0 Calcium hydroxide ionizes according to the following equation:<\/p>\n<p>Equation 1<\/p>\n<p>from this equation we can see that<\/p>\n<p>Equation 2<\/p>\n<p>The hydroxide ion concentration will be determined by taking a pH reading for the solution.\u00a0 The pH allows the pOH to be calculated using Equation 3.<\/p>\n<p>Equation 3<\/p>\n<p>The hydroxide ion concentration can then be calculated by solving the pOH equation for [OH<sup>=<\/sup>].<\/p>\n<p>Equation 4<\/p>\n<p>Equation 5<\/p>\n<p>Once the hydroxide ion concentration is calculated, it is possible to use Equation 2 to determine the concentration of calcium in solution and the molar solubility of calcium hydroxide.\u00a0 The concentrations of calcium and hydroxide ions can be plugged into the K<sub>sp<\/sub> equation for calcium hydroxide:<\/p>\n<p>Equation 6<\/p>\n<p>Equation 6 can also be rewritten with the calcium ion concentration substituted according to Equation 2 to give Equation 7 below.<\/p>\n<p>Equation 7<\/p>\n","protected":false},"author":23588,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-245","chapter","type-chapter","status-publish","hentry"],"part":28,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/pressbooks\/v2\/chapters\/245","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/wp\/v2\/users\/23588"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/pressbooks\/v2\/chapters\/245\/revisions"}],"predecessor-version":[{"id":246,"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/pressbooks\/v2\/chapters\/245\/revisions\/246"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/pressbooks\/v2\/parts\/28"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/pressbooks\/v2\/chapters\/245\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/wp\/v2\/media?parent=245"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/pressbooks\/v2\/chapter-type?post=245"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/wp\/v2\/contributor?post=245"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistry2labs\/wp-json\/wp\/v2\/license?post=245"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}