{"id":596,"date":"2017-12-14T21:41:55","date_gmt":"2017-12-14T21:41:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/strong-and-weak-acids-and-bases-and-their-salts\/"},"modified":"2017-12-19T15:42:14","modified_gmt":"2017-12-19T15:42:14","slug":"strong-and-weak-acids-and-bases-and-their-salts","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-introductory-chemistry\/chapter\/strong-and-weak-acids-and-bases-and-their-salts\/","title":{"raw":"Strong and Weak Acids and Bases and Their Salts","rendered":"Strong and Weak Acids and Bases and Their Salts"},"content":{"raw":"<div id=\"ball-ch12_s04\" class=\"section\" lang=\"en\">\r\n<div id=\"ball-ch12_s04_n01\" class=\"learning_objectives editable block\">\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ol id=\"ball-ch12_s04_l01\">\r\n \t<li>Define a strong and a weak acid and base.<\/li>\r\n \t<li>Recognize an acid or a base as strong or weak.<\/li>\r\n \t<li>Determine if a salt produces an acidic or a basic solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<p id=\"ball-ch12_s04_p01\" class=\"para editable block\">Except for their names and formulas, so far we have treated all acids as equals, especially in a chemical reaction. However, acids can be very different in a very important way. Consider HCl(aq). When HCl is dissolved in H<sub class=\"subscript\">2<\/sub>O, it completely dissociates into H<sup class=\"superscript\">+<\/sup>(aq) and Cl<sup class=\"superscript\">\u2212<\/sup>(aq) ions; all the HCl molecules become ions:<\/p>\r\n<span class=\"informalequation block\">HCl\u00a0\u2192 H<sup>+<\/sup>(aq)\u00a0+\u00a0Cl<sup>\u2212<\/sup>(aq) (100%)<\/span>\r\n<p id=\"ball-ch12_s04_p02\" class=\"para editable block\">Any acid that dissociates 100% into ions is called a <span class=\"margin_term\"><a class=\"glossterm\">strong acid<\/a><\/span>. If it does not dissociate 100%, it is a <span class=\"margin_term\"><a class=\"glossterm\">weak acid<\/a><\/span>. HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> is an example of a weak acid:<\/p>\r\n<span class=\"informalequation block\">HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>\u2192 H<sup>+<\/sup>(aq)\u00a0+\u00a0C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup>(aq) (~5%)<\/span>\r\n<p id=\"ball-ch12_s04_p03\" class=\"para editable block\">Because this reaction does not go 100% to completion, it is more appropriate to write it as an equilibrium:<\/p>\r\n<span class=\"informalequation block\">HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2\u00a0<\/sub>\u21c4 H<sup>+<\/sup>(aq)\u00a0+\u00a0C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup>(aq)<\/span>\r\n<p id=\"ball-ch12_s04_p04\" class=\"para editable block\">As it turns out, there are very few strong acids, which are given in <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 \"Strong Acids and Bases\"<\/a>. If an acid is not listed here, it is a weak acid. It may be 1% ionized or 99% ionized, but it is still classified as a weak acid.<\/p>\r\n<p id=\"ball-ch12_s04_p05\" class=\"para editable block\">The issue is similar with bases: a <span class=\"margin_term\"><a class=\"glossterm\">strong base<\/a><\/span>\u00a0is a base that is 100% ionized in solution. If it is less than 100% ionized in solution, it is a <span class=\"margin_term\"><a class=\"glossterm\">weak base<\/a><\/span>. There are very few strong bases (see <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 \"Strong Acids and Bases\"<\/a>); any base not listed is a weak base. All strong bases are OH<sup class=\"superscript\">\u2013<\/sup> compounds. So a base based on some other mechanism, such as NH<sub class=\"subscript\">3<\/sub> (which does not contain OH<sup class=\"superscript\">\u2212<\/sup> ions as part of its formula), will be a weak base.<\/p>\r\n\r\n<div id=\"ball-ch12_s04_t01\" class=\"table block\">\r\n<p class=\"title\"><span class=\"title-prefix\">Table 12.2<\/span> Strong Acids and Bases<\/p>\r\n\r\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\r\n<thead>\r\n<tr>\r\n<th>Acids<\/th>\r\n<th>Bases<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>HCl<\/td>\r\n<td>LiOH<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HBr<\/td>\r\n<td>NaOH<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HI<\/td>\r\n<td>KOH<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HNO<sub class=\"subscript\">3<\/sub><\/td>\r\n<td>RbOH<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub><\/td>\r\n<td>CsOH<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HClO<sub class=\"subscript\">3<\/sub><\/td>\r\n<td>Mg(OH)<sub class=\"subscript\">2<\/sub><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HClO<sub class=\"subscript\">4<\/sub><\/td>\r\n<td>Ca(OH)<sub class=\"subscript\">2<\/sub><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>Sr(OH)<sub class=\"subscript\">2<\/sub><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>Ba(OH)<sub class=\"subscript\">2<\/sub><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 6<\/h3>\r\n<p id=\"ball-ch12_s04_p06\" class=\"para\">Identify each acid or base as strong or weak.<\/p>\r\n\r\n<ol id=\"ball-ch12_s04_l02\" class=\"orderedlist\">\r\n \t<li>HCl<\/li>\r\n \t<li>Mg(OH)<sub class=\"subscript\">2<\/sub><\/li>\r\n \t<li>C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>N<\/li>\r\n<\/ol>\r\n<p class=\"simpara\">Solution<\/p>\r\n\r\n<ol id=\"ball-ch12_s04_l03\" class=\"orderedlist\">\r\n \t<li>Because HCl is listed in <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 \"Strong Acids and Bases\"<\/a>, it is a strong acid.<\/li>\r\n \t<li>Because Mg(OH)<sub class=\"subscript\">2<\/sub> is listed in <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 \"Strong Acids and Bases\"<\/a>, it is a strong base.<\/li>\r\n \t<li>The nitrogen in C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>N would act as a proton acceptor and therefore can be considered a base, but because it does not contain an OH compound, it cannot be considered a strong base; it is a weak base.<\/li>\r\n<\/ol>\r\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\r\n<p id=\"ball-ch12_s04_p07\" class=\"para\">Identify each acid or base as strong or weak.<\/p>\r\n\r\n<ol id=\"ball-ch12_s04_l04\" class=\"orderedlist\">\r\n \t<li>RbOH<\/li>\r\n \t<li>HNO<sub class=\"subscript\">2<\/sub><\/li>\r\n<\/ol>\r\n<p class=\"simpara\"><em class=\"emphasis\">Answers<\/em><\/p>\r\n\r\n<ol id=\"ball-ch12_s04_l05\" class=\"orderedlist\">\r\n \t<li>strong base<\/li>\r\n \t<li>weak acid<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 7<\/h3>\r\n<p id=\"ball-ch12_s04_p08\" class=\"para\">Write the balanced chemical equation for the dissociation of Ca(OH)<sub class=\"subscript\">2<\/sub> and indicate whether it proceeds 100% to products or not.<\/p>\r\n<p class=\"simpara\">Solution<\/p>\r\n<p id=\"ball-ch12_s04_p09\" class=\"para\">This is an ionic compound of Ca<sup class=\"superscript\">2+<\/sup> ions and OH<sup class=\"superscript\">\u2212<\/sup> ions. When an ionic compound dissolves, it separates into its constituent ions:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">Ca(OH)<sub class=\"subscript\">2<\/sub> \u2192\u00a0Ca<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p10\" class=\"para\">Because Ca(OH)<sub class=\"subscript\">2<\/sub> is listed in <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 \"Strong Acids and Bases\"<\/a>, this reaction proceeds 100% to products.<\/p>\r\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\r\n<p id=\"ball-ch12_s04_p11\" class=\"para\">Write the balanced chemical equation for the dissociation of hydrazoic acid (HN<sub class=\"subscript\">3<\/sub>) and indicate whether it proceeds 100% to products or not.<\/p>\r\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\r\n<p id=\"ball-ch12_s04_p12\" class=\"para\">The reaction is as follows:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">HN<sub class=\"subscript\">3<\/sub> \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0N<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p13\" class=\"para\">It does not proceed 100% to products because hydrazoic acid is not a strong acid.<\/p>\r\n\r\n<\/div>\r\n<p id=\"ball-ch12_s04_p14\" class=\"para editable block\">Certain salts will also affect the acidity or basicity of aqueous solutions because some of the ions will undergo hydrolysis, just like NH<sub class=\"subscript\">3<\/sub> does to make a basic solution. The general rule is that salts with ions that are part of strong acids or bases will not hydrolyze, while salts with ions that are part of weak acids or bases will hydrolyze.<\/p>\r\n<p id=\"ball-ch12_s04_p15\" class=\"para editable block\">Consider NaCl. When it dissolves in an aqueous solution, it separates into Na<sup class=\"superscript\">+<\/sup> ions and Cl<sup class=\"superscript\">\u2212<\/sup> ions:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">NaCl \u2192\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p16\" class=\"para editable block\">Will the Na<sup class=\"superscript\">+<\/sup>(aq) ion hydrolyze? If it does, it will interact with the OH<sup class=\"superscript\">\u2212<\/sup> ion to make NaOH:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0NaOH +\u00a0H<sup class=\"superscript\">+<\/sup>(aq)<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p17\" class=\"para editable block\">However, NaOH is a strong base, which means that it is 100% ionized in solution:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">NaOH \u2192\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p18\" class=\"para editable block\">The free OH<sup class=\"superscript\">\u2212<\/sup>(aq) ion reacts with the H<sup class=\"superscript\">+<\/sup>(aq) ion to remake a water molecule:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) \u2192\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p19\" class=\"para editable block\">The net result? There is no change, so there is no effect on the acidity or basicity of the solution from the Na<sup class=\"superscript\">+<\/sup>(aq) ion. What about the Cl<sup class=\"superscript\">\u2212<\/sup> ion? Will it hydrolyze? If it does, it will take an H<sup class=\"superscript\">+<\/sup> ion from a water molecule:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0HCl +\u00a0OH<sup class=\"superscript\">\u2212<\/sup><\/span><\/span>\r\n<p id=\"ball-ch12_s04_p20\" class=\"para editable block\">However, HCl is a strong acid, which means that it is 100% ionized in solution:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">HCl \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p21\" class=\"para editable block\">The free H<sup class=\"superscript\">+<\/sup>(aq) ion reacts with the OH<sup class=\"superscript\">\u2212<\/sup>(aq) ion to remake a water molecule:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) \u2192\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p22\" class=\"para editable block\">The net result? There is no change, so there is no effect on the acidity or basicity of the solution from the Cl<sup class=\"superscript\">\u2212<\/sup>(aq) ion. Because neither ion in NaCl affects the acidity or basicity of the solution, NaCl is an example of a <span class=\"margin_term\"><a class=\"glossterm\">neutral salt<\/a><\/span>.<\/p>\r\n<p id=\"ball-ch12_s04_p23\" class=\"para editable block\">Things change, however, when we consider a salt like NaC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>. We already know that the Na<sup class=\"superscript\">+<\/sup> ion won\u2019t affect the acidity of the solution. What about the acetate ion? If it hydrolyzes, it will take an H<sup class=\"superscript\">+<\/sup> from a water molecule:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p24\" class=\"para editable block\">Does this happen? Yes, it does. Why? <em class=\"emphasis\">Because HC<\/em><sub class=\"subscript\"><em class=\"emphasis\">2<\/em><\/sub><em class=\"emphasis\">H<\/em><sub class=\"subscript\"><em class=\"emphasis\">3<\/em><\/sub><em class=\"emphasis\">O<\/em><sub class=\"subscript\"><em class=\"emphasis\">2<\/em><\/sub><em class=\"emphasis\">is a weak acid.<\/em> Any chance a weak acid has to form, it will (the same with a weak base). As some C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> ions hydrolyze with H<sub class=\"subscript\">2<\/sub>O to make the molecular weak acid, OH<sup class=\"superscript\">\u2212<\/sup> ions are produced. OH<sup class=\"superscript\">\u2212<\/sup> ions make solutions basic. Thus NaC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> solutions are slightly basic, so such a salt is called a <span class=\"margin_term\"><a class=\"glossterm\">basic salt<\/a><\/span>.<\/p>\r\n<p id=\"ball-ch12_s04_p25\" class=\"para editable block\">There are also salts whose aqueous solutions are slightly acidic. NH<sub class=\"subscript\">4<\/sub>Cl is an example. When NH<sub class=\"subscript\">4<\/sub>Cl is dissolved in H<sub class=\"subscript\">2<\/sub>O, it separates into NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ions and Cl<sup class=\"superscript\">\u2212<\/sup> ions. We have already seen that the Cl<sup class=\"superscript\">\u2212<\/sup> ion does not hydrolyze. However, the NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ion will:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0NH<sub class=\"subscript\">3<\/sub>(aq) +\u00a0H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq)<\/span><\/span>\r\n<p id=\"ball-ch12_s04_p26\" class=\"para editable block\">Recall from <a class=\"xref\" href=\"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/arrhenius-acids-and-bases\/\">Section 12.1 \"Arrhenius Acids and Bases\"<\/a> that H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> ion is the hydronium ion, the more chemically proper way to represent the H<sup class=\"superscript\">+<\/sup> ion. This is the classic acid species in solution, so a solution of NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) ions is slightly acidic. NH<sub class=\"subscript\">4<\/sub>Cl is an example of an <span class=\"margin_term\"><a class=\"glossterm\">acid salt<\/a><\/span>. The molecule NH<sub class=\"subscript\">3<\/sub> is a weak base, and it will form when it can, just like a weak acid will form when it can.<\/p>\r\n<p id=\"ball-ch12_s04_p27\" class=\"para editable block\">So there are two general rules: (1) If an ion derives from a strong acid or base, it will not affect the acidity of the solution. (2) If an ion derives from a weak acid, it will make the solution basic; if an ion derives from a weak base, it will make the solution acidic.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 8<\/h3>\r\n<p id=\"ball-ch12_s04_p28\" class=\"para\">Identify each salt as acidic, basic, or neutral.<\/p>\r\n\r\n<ol id=\"ball-ch12_s04_l06\" class=\"orderedlist\">\r\n \t<li>KCl<\/li>\r\n \t<li>KNO<sub class=\"subscript\">2<\/sub><\/li>\r\n \t<li>NH<sub class=\"subscript\">4<\/sub>Br<\/li>\r\n<\/ol>\r\n<p class=\"simpara\">Solution<\/p>\r\n\r\n<ol id=\"ball-ch12_s04_l07\" class=\"orderedlist\">\r\n \t<li>The ions from KCl derive from a strong acid (HCl) and a strong base (KOH). Therefore, neither ion will affect the acidity of the solution, so KCl is a neutral salt.<\/li>\r\n \t<li>Although the K<sup class=\"superscript\">+<\/sup> ion derives from a strong base (KOH), the NO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> ion derives from a weak acid (HNO<sub class=\"subscript\">2<\/sub>). Therefore the solution will be basic, and KNO<sub class=\"subscript\">2<\/sub> is a basic salt.<\/li>\r\n \t<li>Although the Br<sup class=\"superscript\">\u2212<\/sup> ions derive from a strong acid (HBr), the NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ion derives from a weak base (NH<sub class=\"subscript\">3<\/sub>), so the solution will be acidic, and NH<sub class=\"subscript\">4<\/sub>Br is an acidic salt.<\/li>\r\n<\/ol>\r\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\r\n<p id=\"ball-ch12_s04_p29\" class=\"para\">Identify each salt as acidic, basic, or neutral.<\/p>\r\n\r\n<ol id=\"ball-ch12_s04_l08\" class=\"orderedlist\">\r\n \t<li>(C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NH)Cl<\/li>\r\n \t<li>Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub><\/li>\r\n<\/ol>\r\n<p class=\"simpara\"><em class=\"emphasis\">Answers<\/em><\/p>\r\n\r\n<ol id=\"ball-ch12_s04_l09\" class=\"orderedlist\">\r\n \t<li>acidic<\/li>\r\n \t<li>basic<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"ball-ch12_s04_p30\" class=\"para editable block\">Some salts are composed of ions that come from both weak acids and weak bases. The overall effect on an aqueous solution depends on which ion exerts more influence on the overall acidity. We will not consider such salts here.<\/p>\r\n\r\n<div id=\"ball-ch12_s04_n05\" class=\"key_takeaways editable block\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Key Takeaways<\/h3>\r\n<ul id=\"ball-ch12_s04_l10\" class=\"itemizedlist\">\r\n \t<li>Strong acids and bases are 100% ionized in aqueous solution.<\/li>\r\n \t<li>Weak acids and bases are less than 100% ionized in aqueous solution.<\/li>\r\n \t<li>Salts of weak acids or bases can affect the acidity or basicity of their aqueous solutions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"ball-ch12_s04_qs01\" class=\"qandaset block\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n<ol id=\"ball-ch12_s04_qs01_qd01\" class=\"qandadiv\">\r\n \t<li id=\"ball-ch12_s04_qs01_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p01\" class=\"para\">Differentiate between a strong acid and a weak acid.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch12_s04_qs01_qd01_qa02\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p03\" class=\"para\">Differentiate between a strong base and a weak base.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"ball-ch12_s04_qs01_qd01_qa03\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p05\" class=\"para\">Identify each as a strong acid or a weak acid. Assume aqueous solutions.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\na) \u00a0HF\r\n\r\nb) \u00a0HCl\r\n\r\nc) \u00a0HC<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p06\" class=\"para\">4. \u00a0Identify each as a strong base or a weak base. Assume aqueous solutions.<\/p>\r\na) \u00a0NaOH\r\n\r\nb) \u00a0Al(OH)<sub class=\"subscript\">3<\/sub>\r\n\r\nc) \u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>NH<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p07\" class=\"para\">5. \u00a0Write a chemical equation for the ionization of each acid and indicate whether it proceeds 100% to products or not.<\/p>\r\na) \u00a0HNO<sub class=\"subscript\">3<\/sub>\r\n\r\nb) \u00a0HNO<sub class=\"subscript\">2<\/sub>\r\n\r\nc) \u00a0HI<sub class=\"subscript\">3<\/sub>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p08\" class=\"para\">6. \u00a0Write a chemical equation for the ionization of each base and indicate whether it proceeds 100% to products or not.<\/p>\r\na) \u00a0NH<sub class=\"subscript\">3<\/sub>\r\n\r\nb) \u00a0(CH<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub>N\r\n\r\nc) \u00a0Mg(OH)<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p09\" class=\"para\">7. \u00a0Write the balanced chemical equation for the reaction of each acid and base pair.<\/p>\r\na) \u00a0HCl +\u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>N\r\n\r\nb) \u00a0H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a0NH<sub class=\"subscript\">3<\/sub>\r\n\r\nc) \u00a0HNO<sub class=\"subscript\">2<\/sub> +\u00a0C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">9<\/sub>N\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p10\" class=\"para\">8. \u00a0Write the balanced chemical equation for the reaction of each acid and base pair.<\/p>\r\na) \u00a0H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub> +\u00a0Mg(OH)<sub class=\"subscript\">2<\/sub>\r\n\r\nb) \u00a0HC<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a0(CH<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub>N\r\n\r\nc) \u00a0HBr +\u00a0Fe(OH)<sub class=\"subscript\">3<\/sub>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p11\" class=\"para\">9. \u00a0Identify each salt as neutral, acidic, or basic.<\/p>\r\na) \u00a0NaBr\r\n\r\nb) \u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>\r\n\r\nc) \u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p12\" class=\"para\">10. \u00a0Identify each salt as neutral, acidic, or basic.<\/p>\r\na) \u00a0NH<sub class=\"subscript\">4<\/sub>I\r\n\r\nb) \u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">3<\/sub>Cl\r\n\r\nc) \u00a0KI\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p13\" class=\"para\">11. \u00a0Identify each salt as neutral, acidic, or basic.<\/p>\r\na) \u00a0NaNO<sub class=\"subscript\">2<\/sub>\r\n\r\nb) \u00a0NaNO<sub class=\"subscript\">3<\/sub>\r\n\r\nc) \u00a0NH<sub class=\"subscript\">4<\/sub>NO<sub class=\"subscript\">3<\/sub>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p14\" class=\"para\">12. \u00a0Identify each salt as neutral, acidic, or basic.<\/p>\r\na) \u00a0KC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>\r\n\r\nb) \u00a0KHSO<sub class=\"subscript\">4<\/sub>\r\n\r\nc) \u00a0KClO<sub class=\"subscript\">3<\/sub>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p15\" class=\"para\">13. \u00a0Write the hydrolysis reaction that occurs, if any, when each salt dissolves in water.<\/p>\r\na) \u00a0K<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub>\r\n\r\nb) \u00a0KI\r\n\r\nc) \u00a0NH<sub class=\"subscript\">4<\/sub>ClO<sub class=\"subscript\">3<\/sub>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p16\" class=\"para\">14. \u00a0Write the hydrolysis reaction that occurs, if any, when each salt dissolves in water.<\/p>\r\na) \u00a0NaNO<sub class=\"subscript\">3<\/sub>\r\n\r\nb) \u00a0CaC<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>\r\n\r\nc) \u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NHCl\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p17\" class=\"para\">15. \u00a0When NH<sub class=\"subscript\">4<\/sub>NO<sub class=\"subscript\">2<\/sub> dissolves in H<sub class=\"subscript\">2<\/sub>O, both ions hydrolyze. Write chemical equations for both reactions. Can you tell if the solution will be acidic or basic overall?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p19\" class=\"para\">16. \u00a0When pyridinium acetate (C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NHC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>) dissolves in H<sub class=\"subscript\">2<\/sub>O, both ions hydrolyze. Write chemical equations for both reactions. Can you tell if the solution will be acidic or basic overall?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p21\" class=\"para\">17. \u00a0A lab technician mixes a solution of 0.015 M Mg(OH)<sub class=\"subscript\">2<\/sub>. Is the resulting OH<sup class=\"superscript\">\u2212<\/sup> concentration greater than, equal to, or less than 0.015 M? Explain your answer.<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch12_s04_qs01_p23\" class=\"para\">18. \u00a0A lab technician mixes a solution of 0.55 M HNO<sub class=\"subscript\">3<\/sub>. Is the resulting H<sup class=\"superscript\">+<\/sup> concentration greater than, equal to, or less than 0.55 M? Explain your answer.<\/p>\r\n\r\n<\/div>\r\n<b>Answers<\/b>\r\n\r\n<strong>1.<\/strong>\r\n\r\nA strong acid is 100% ionized in aqueous solution, whereas a weak acid is not 100% ionized.\r\n\r\n<strong>3.<\/strong>\r\n\r\na) \u00a0weak acid\r\n\r\nb) \u00a0strong acid\r\n\r\nc) \u00a0weak acid\r\n\r\n<strong>5.<\/strong>\r\n\r\na) \u00a0HNO<sub class=\"subscript\">3<\/sub>(aq) \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq); proceeds 100%\r\n\r\nb) \u00a0HNO<sub class=\"subscript\">2<\/sub>(aq) \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0NO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq); does not proceed 100%\r\n\r\nc) \u00a0HI<sub class=\"subscript\">3<\/sub>(aq) \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0I<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq); does not proceed 100%\r\n\r\n<strong>9.<\/strong>\r\n\r\na) \u00a0HCl +\u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>N \u2192\u00a0Cl<sup class=\"superscript\">\u2212<\/sup> +\u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sup class=\"superscript\">+<\/sup>\r\n\r\nb) \u00a0H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a02NH<sub class=\"subscript\">3<\/sub> \u2192\u00a0C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup> +\u00a02NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>\r\n\r\nc) \u00a0HNO<sub class=\"subscript\">2<\/sub> +\u00a0C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">9<\/sub>N \u2192\u00a0NO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> +\u00a0C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">9<\/sub>NH<sup class=\"superscript\">+<\/sup><strong>11.<\/strong>\r\n\r\na) \u00a0neutral\r\n\r\nb) \u00a0acidic\r\n\r\nc) \u00a0acidic\r\n\r\n<strong>13.<\/strong>\r\n\r\na) \u00a0basic\r\n\r\nb) \u00a0neutral\r\n\r\nc) \u00a0acidic\r\n\r\n<strong>15.<\/strong>\r\n\r\na) \u00a0SO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">2\u2212<\/sup> +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0HSO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup> +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>\r\n\r\nb) \u00a0no reaction\r\n\r\nc) \u00a0NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0NH<sub class=\"subscript\">3<\/sub> +\u00a0H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup><strong>17.<\/strong>\r\n\r\nNH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0NH<sub class=\"subscript\">3<\/sub> +\u00a0H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>; NO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0HNO<sub class=\"subscript\">2<\/sub> +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>; it is not possible to determine whether the solution will be acidic or basic.\r\n\r\n<strong>19.<\/strong>\r\n\r\ngreater than 0.015 M because there are two OH<sup class=\"superscript\">\u2212<\/sup> ions per formula unit of Mg(OH)<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div id=\"ball-ch12_s04\" class=\"section\" lang=\"en\">\n<div id=\"ball-ch12_s04_n01\" class=\"learning_objectives editable block\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ol id=\"ball-ch12_s04_l01\">\n<li>Define a strong and a weak acid and base.<\/li>\n<li>Recognize an acid or a base as strong or weak.<\/li>\n<li>Determine if a salt produces an acidic or a basic solution.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p id=\"ball-ch12_s04_p01\" class=\"para editable block\">Except for their names and formulas, so far we have treated all acids as equals, especially in a chemical reaction. However, acids can be very different in a very important way. Consider HCl(aq). When HCl is dissolved in H<sub class=\"subscript\">2<\/sub>O, it completely dissociates into H<sup class=\"superscript\">+<\/sup>(aq) and Cl<sup class=\"superscript\">\u2212<\/sup>(aq) ions; all the HCl molecules become ions:<\/p>\n<p><span class=\"informalequation block\">HCl\u00a0\u2192 H<sup>+<\/sup>(aq)\u00a0+\u00a0Cl<sup>\u2212<\/sup>(aq) (100%)<\/span><\/p>\n<p id=\"ball-ch12_s04_p02\" class=\"para editable block\">Any acid that dissociates 100% into ions is called a <span class=\"margin_term\"><a class=\"glossterm\">strong acid<\/a><\/span>. If it does not dissociate 100%, it is a <span class=\"margin_term\"><a class=\"glossterm\">weak acid<\/a><\/span>. HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> is an example of a weak acid:<\/p>\n<p><span class=\"informalequation block\">HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub>\u2192 H<sup>+<\/sup>(aq)\u00a0+\u00a0C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup>(aq) (~5%)<\/span><\/p>\n<p id=\"ball-ch12_s04_p03\" class=\"para editable block\">Because this reaction does not go 100% to completion, it is more appropriate to write it as an equilibrium:<\/p>\n<p><span class=\"informalequation block\">HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2\u00a0<\/sub>\u21c4 H<sup>+<\/sup>(aq)\u00a0+\u00a0C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup>(aq)<\/span><\/p>\n<p id=\"ball-ch12_s04_p04\" class=\"para editable block\">As it turns out, there are very few strong acids, which are given in <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 &#8220;Strong Acids and Bases&#8221;<\/a>. If an acid is not listed here, it is a weak acid. It may be 1% ionized or 99% ionized, but it is still classified as a weak acid.<\/p>\n<p id=\"ball-ch12_s04_p05\" class=\"para editable block\">The issue is similar with bases: a <span class=\"margin_term\"><a class=\"glossterm\">strong base<\/a><\/span>\u00a0is a base that is 100% ionized in solution. If it is less than 100% ionized in solution, it is a <span class=\"margin_term\"><a class=\"glossterm\">weak base<\/a><\/span>. There are very few strong bases (see <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 &#8220;Strong Acids and Bases&#8221;<\/a>); any base not listed is a weak base. All strong bases are OH<sup class=\"superscript\">\u2013<\/sup> compounds. So a base based on some other mechanism, such as NH<sub class=\"subscript\">3<\/sub> (which does not contain OH<sup class=\"superscript\">\u2212<\/sup> ions as part of its formula), will be a weak base.<\/p>\n<div id=\"ball-ch12_s04_t01\" class=\"table block\">\n<p class=\"title\"><span class=\"title-prefix\">Table 12.2<\/span> Strong Acids and Bases<\/p>\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\n<thead>\n<tr>\n<th>Acids<\/th>\n<th>Bases<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>HCl<\/td>\n<td>LiOH<\/td>\n<\/tr>\n<tr>\n<td>HBr<\/td>\n<td>NaOH<\/td>\n<\/tr>\n<tr>\n<td>HI<\/td>\n<td>KOH<\/td>\n<\/tr>\n<tr>\n<td>HNO<sub class=\"subscript\">3<\/sub><\/td>\n<td>RbOH<\/td>\n<\/tr>\n<tr>\n<td>H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub><\/td>\n<td>CsOH<\/td>\n<\/tr>\n<tr>\n<td>HClO<sub class=\"subscript\">3<\/sub><\/td>\n<td>Mg(OH)<sub class=\"subscript\">2<\/sub><\/td>\n<\/tr>\n<tr>\n<td>HClO<sub class=\"subscript\">4<\/sub><\/td>\n<td>Ca(OH)<sub class=\"subscript\">2<\/sub><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>Sr(OH)<sub class=\"subscript\">2<\/sub><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>Ba(OH)<sub class=\"subscript\">2<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 6<\/h3>\n<p id=\"ball-ch12_s04_p06\" class=\"para\">Identify each acid or base as strong or weak.<\/p>\n<ol id=\"ball-ch12_s04_l02\" class=\"orderedlist\">\n<li>HCl<\/li>\n<li>Mg(OH)<sub class=\"subscript\">2<\/sub><\/li>\n<li>C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>N<\/li>\n<\/ol>\n<p class=\"simpara\">Solution<\/p>\n<ol id=\"ball-ch12_s04_l03\" class=\"orderedlist\">\n<li>Because HCl is listed in <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 &#8220;Strong Acids and Bases&#8221;<\/a>, it is a strong acid.<\/li>\n<li>Because Mg(OH)<sub class=\"subscript\">2<\/sub> is listed in <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 &#8220;Strong Acids and Bases&#8221;<\/a>, it is a strong base.<\/li>\n<li>The nitrogen in C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>N would act as a proton acceptor and therefore can be considered a base, but because it does not contain an OH compound, it cannot be considered a strong base; it is a weak base.<\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s04_p07\" class=\"para\">Identify each acid or base as strong or weak.<\/p>\n<ol id=\"ball-ch12_s04_l04\" class=\"orderedlist\">\n<li>RbOH<\/li>\n<li>HNO<sub class=\"subscript\">2<\/sub><\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis\">Answers<\/em><\/p>\n<ol id=\"ball-ch12_s04_l05\" class=\"orderedlist\">\n<li>strong base<\/li>\n<li>weak acid<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 7<\/h3>\n<p id=\"ball-ch12_s04_p08\" class=\"para\">Write the balanced chemical equation for the dissociation of Ca(OH)<sub class=\"subscript\">2<\/sub> and indicate whether it proceeds 100% to products or not.<\/p>\n<p class=\"simpara\">Solution<\/p>\n<p id=\"ball-ch12_s04_p09\" class=\"para\">This is an ionic compound of Ca<sup class=\"superscript\">2+<\/sup> ions and OH<sup class=\"superscript\">\u2212<\/sup> ions. When an ionic compound dissolves, it separates into its constituent ions:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">Ca(OH)<sub class=\"subscript\">2<\/sub> \u2192\u00a0Ca<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p10\" class=\"para\">Because Ca(OH)<sub class=\"subscript\">2<\/sub> is listed in <a class=\"xref\" href=\"#ball-ch12_s04_t01\">Table 12.2 &#8220;Strong Acids and Bases&#8221;<\/a>, this reaction proceeds 100% to products.<\/p>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s04_p11\" class=\"para\">Write the balanced chemical equation for the dissociation of hydrazoic acid (HN<sub class=\"subscript\">3<\/sub>) and indicate whether it proceeds 100% to products or not.<\/p>\n<p class=\"simpara\"><em class=\"emphasis\">Answer<\/em><\/p>\n<p id=\"ball-ch12_s04_p12\" class=\"para\">The reaction is as follows:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">HN<sub class=\"subscript\">3<\/sub> \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0N<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p13\" class=\"para\">It does not proceed 100% to products because hydrazoic acid is not a strong acid.<\/p>\n<\/div>\n<p id=\"ball-ch12_s04_p14\" class=\"para editable block\">Certain salts will also affect the acidity or basicity of aqueous solutions because some of the ions will undergo hydrolysis, just like NH<sub class=\"subscript\">3<\/sub> does to make a basic solution. The general rule is that salts with ions that are part of strong acids or bases will not hydrolyze, while salts with ions that are part of weak acids or bases will hydrolyze.<\/p>\n<p id=\"ball-ch12_s04_p15\" class=\"para editable block\">Consider NaCl. When it dissolves in an aqueous solution, it separates into Na<sup class=\"superscript\">+<\/sup> ions and Cl<sup class=\"superscript\">\u2212<\/sup> ions:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">NaCl \u2192\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p16\" class=\"para editable block\">Will the Na<sup class=\"superscript\">+<\/sup>(aq) ion hydrolyze? If it does, it will interact with the OH<sup class=\"superscript\">\u2212<\/sup> ion to make NaOH:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0NaOH +\u00a0H<sup class=\"superscript\">+<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p17\" class=\"para editable block\">However, NaOH is a strong base, which means that it is 100% ionized in solution:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">NaOH \u2192\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p18\" class=\"para editable block\">The free OH<sup class=\"superscript\">\u2212<\/sup>(aq) ion reacts with the H<sup class=\"superscript\">+<\/sup>(aq) ion to remake a water molecule:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) \u2192\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p19\" class=\"para editable block\">The net result? There is no change, so there is no effect on the acidity or basicity of the solution from the Na<sup class=\"superscript\">+<\/sup>(aq) ion. What about the Cl<sup class=\"superscript\">\u2212<\/sup> ion? Will it hydrolyze? If it does, it will take an H<sup class=\"superscript\">+<\/sup> ion from a water molecule:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0HCl +\u00a0OH<sup class=\"superscript\">\u2212<\/sup><\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p20\" class=\"para editable block\">However, HCl is a strong acid, which means that it is 100% ionized in solution:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">HCl \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p21\" class=\"para editable block\">The free H<sup class=\"superscript\">+<\/sup>(aq) ion reacts with the OH<sup class=\"superscript\">\u2212<\/sup>(aq) ion to remake a water molecule:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) \u2192\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p22\" class=\"para editable block\">The net result? There is no change, so there is no effect on the acidity or basicity of the solution from the Cl<sup class=\"superscript\">\u2212<\/sup>(aq) ion. Because neither ion in NaCl affects the acidity or basicity of the solution, NaCl is an example of a <span class=\"margin_term\"><a class=\"glossterm\">neutral salt<\/a><\/span>.<\/p>\n<p id=\"ball-ch12_s04_p23\" class=\"para editable block\">Things change, however, when we consider a salt like NaC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>. We already know that the Na<sup class=\"superscript\">+<\/sup> ion won\u2019t affect the acidity of the solution. What about the acetate ion? If it hydrolyzes, it will take an H<sup class=\"superscript\">+<\/sup> from a water molecule:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p24\" class=\"para editable block\">Does this happen? Yes, it does. Why? <em class=\"emphasis\">Because HC<\/em><sub class=\"subscript\"><em class=\"emphasis\">2<\/em><\/sub><em class=\"emphasis\">H<\/em><sub class=\"subscript\"><em class=\"emphasis\">3<\/em><\/sub><em class=\"emphasis\">O<\/em><sub class=\"subscript\"><em class=\"emphasis\">2<\/em><\/sub><em class=\"emphasis\">is a weak acid.<\/em> Any chance a weak acid has to form, it will (the same with a weak base). As some C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> ions hydrolyze with H<sub class=\"subscript\">2<\/sub>O to make the molecular weak acid, OH<sup class=\"superscript\">\u2212<\/sup> ions are produced. OH<sup class=\"superscript\">\u2212<\/sup> ions make solutions basic. Thus NaC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub> solutions are slightly basic, so such a salt is called a <span class=\"margin_term\"><a class=\"glossterm\">basic salt<\/a><\/span>.<\/p>\n<p id=\"ball-ch12_s04_p25\" class=\"para editable block\">There are also salts whose aqueous solutions are slightly acidic. NH<sub class=\"subscript\">4<\/sub>Cl is an example. When NH<sub class=\"subscript\">4<\/sub>Cl is dissolved in H<sub class=\"subscript\">2<\/sub>O, it separates into NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ions and Cl<sup class=\"superscript\">\u2212<\/sup> ions. We have already seen that the Cl<sup class=\"superscript\">\u2212<\/sup> ion does not hydrolyze. However, the NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ion will:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0NH<sub class=\"subscript\">3<\/sub>(aq) +\u00a0H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch12_s04_p26\" class=\"para editable block\">Recall from <a class=\"xref\" href=\"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/arrhenius-acids-and-bases\/\">Section 12.1 &#8220;Arrhenius Acids and Bases&#8221;<\/a> that H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> ion is the hydronium ion, the more chemically proper way to represent the H<sup class=\"superscript\">+<\/sup> ion. This is the classic acid species in solution, so a solution of NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) ions is slightly acidic. NH<sub class=\"subscript\">4<\/sub>Cl is an example of an <span class=\"margin_term\"><a class=\"glossterm\">acid salt<\/a><\/span>. The molecule NH<sub class=\"subscript\">3<\/sub> is a weak base, and it will form when it can, just like a weak acid will form when it can.<\/p>\n<p id=\"ball-ch12_s04_p27\" class=\"para editable block\">So there are two general rules: (1) If an ion derives from a strong acid or base, it will not affect the acidity of the solution. (2) If an ion derives from a weak acid, it will make the solution basic; if an ion derives from a weak base, it will make the solution acidic.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 8<\/h3>\n<p id=\"ball-ch12_s04_p28\" class=\"para\">Identify each salt as acidic, basic, or neutral.<\/p>\n<ol id=\"ball-ch12_s04_l06\" class=\"orderedlist\">\n<li>KCl<\/li>\n<li>KNO<sub class=\"subscript\">2<\/sub><\/li>\n<li>NH<sub class=\"subscript\">4<\/sub>Br<\/li>\n<\/ol>\n<p class=\"simpara\">Solution<\/p>\n<ol id=\"ball-ch12_s04_l07\" class=\"orderedlist\">\n<li>The ions from KCl derive from a strong acid (HCl) and a strong base (KOH). Therefore, neither ion will affect the acidity of the solution, so KCl is a neutral salt.<\/li>\n<li>Although the K<sup class=\"superscript\">+<\/sup> ion derives from a strong base (KOH), the NO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> ion derives from a weak acid (HNO<sub class=\"subscript\">2<\/sub>). Therefore the solution will be basic, and KNO<sub class=\"subscript\">2<\/sub> is a basic salt.<\/li>\n<li>Although the Br<sup class=\"superscript\">\u2212<\/sup> ions derive from a strong acid (HBr), the NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> ion derives from a weak base (NH<sub class=\"subscript\">3<\/sub>), so the solution will be acidic, and NH<sub class=\"subscript\">4<\/sub>Br is an acidic salt.<\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/p>\n<p id=\"ball-ch12_s04_p29\" class=\"para\">Identify each salt as acidic, basic, or neutral.<\/p>\n<ol id=\"ball-ch12_s04_l08\" class=\"orderedlist\">\n<li>(C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NH)Cl<\/li>\n<li>Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub><\/li>\n<\/ol>\n<p class=\"simpara\"><em class=\"emphasis\">Answers<\/em><\/p>\n<ol id=\"ball-ch12_s04_l09\" class=\"orderedlist\">\n<li>acidic<\/li>\n<li>basic<\/li>\n<\/ol>\n<\/div>\n<p id=\"ball-ch12_s04_p30\" class=\"para editable block\">Some salts are composed of ions that come from both weak acids and weak bases. The overall effect on an aqueous solution depends on which ion exerts more influence on the overall acidity. We will not consider such salts here.<\/p>\n<div id=\"ball-ch12_s04_n05\" class=\"key_takeaways editable block\">\n<div class=\"bcc-box bcc-success\">\n<h3>Key Takeaways<\/h3>\n<ul id=\"ball-ch12_s04_l10\" class=\"itemizedlist\">\n<li>Strong acids and bases are 100% ionized in aqueous solution.<\/li>\n<li>Weak acids and bases are less than 100% ionized in aqueous solution.<\/li>\n<li>Salts of weak acids or bases can affect the acidity or basicity of their aqueous solutions.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"ball-ch12_s04_qs01\" class=\"qandaset block\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<ol id=\"ball-ch12_s04_qs01_qd01\" class=\"qandadiv\">\n<li id=\"ball-ch12_s04_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p01\" class=\"para\">Differentiate between a strong acid and a weak acid.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s04_qs01_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p03\" class=\"para\">Differentiate between a strong base and a weak base.<\/p>\n<\/div>\n<\/li>\n<li id=\"ball-ch12_s04_qs01_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p05\" class=\"para\">Identify each as a strong acid or a weak acid. Assume aqueous solutions.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<p>a) \u00a0HF<\/p>\n<p>b) \u00a0HCl<\/p>\n<p>c) \u00a0HC<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub><\/p>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p06\" class=\"para\">4. \u00a0Identify each as a strong base or a weak base. Assume aqueous solutions.<\/p>\n<p>a) \u00a0NaOH<\/p>\n<p>b) \u00a0Al(OH)<sub class=\"subscript\">3<\/sub><\/p>\n<p>c) \u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>NH<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p07\" class=\"para\">5. \u00a0Write a chemical equation for the ionization of each acid and indicate whether it proceeds 100% to products or not.<\/p>\n<p>a) \u00a0HNO<sub class=\"subscript\">3<\/sub><\/p>\n<p>b) \u00a0HNO<sub class=\"subscript\">2<\/sub><\/p>\n<p>c) \u00a0HI<sub class=\"subscript\">3<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p08\" class=\"para\">6. \u00a0Write a chemical equation for the ionization of each base and indicate whether it proceeds 100% to products or not.<\/p>\n<p>a) \u00a0NH<sub class=\"subscript\">3<\/sub><\/p>\n<p>b) \u00a0(CH<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub>N<\/p>\n<p>c) \u00a0Mg(OH)<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p09\" class=\"para\">7. \u00a0Write the balanced chemical equation for the reaction of each acid and base pair.<\/p>\n<p>a) \u00a0HCl +\u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>N<\/p>\n<p>b) \u00a0H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a0NH<sub class=\"subscript\">3<\/sub><\/p>\n<p>c) \u00a0HNO<sub class=\"subscript\">2<\/sub> +\u00a0C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">9<\/sub>N<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p10\" class=\"para\">8. \u00a0Write the balanced chemical equation for the reaction of each acid and base pair.<\/p>\n<p>a) \u00a0H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub> +\u00a0Mg(OH)<sub class=\"subscript\">2<\/sub><\/p>\n<p>b) \u00a0HC<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a0(CH<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub>N<\/p>\n<p>c) \u00a0HBr +\u00a0Fe(OH)<sub class=\"subscript\">3<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p11\" class=\"para\">9. \u00a0Identify each salt as neutral, acidic, or basic.<\/p>\n<p>a) \u00a0NaBr<\/p>\n<p>b) \u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub><\/p>\n<p>c) \u00a0Fe(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">3<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p12\" class=\"para\">10. \u00a0Identify each salt as neutral, acidic, or basic.<\/p>\n<p>a) \u00a0NH<sub class=\"subscript\">4<\/sub>I<\/p>\n<p>b) \u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sub class=\"subscript\">3<\/sub>Cl<\/p>\n<p>c) \u00a0KI<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p13\" class=\"para\">11. \u00a0Identify each salt as neutral, acidic, or basic.<\/p>\n<p>a) \u00a0NaNO<sub class=\"subscript\">2<\/sub><\/p>\n<p>b) \u00a0NaNO<sub class=\"subscript\">3<\/sub><\/p>\n<p>c) \u00a0NH<sub class=\"subscript\">4<\/sub>NO<sub class=\"subscript\">3<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p14\" class=\"para\">12. \u00a0Identify each salt as neutral, acidic, or basic.<\/p>\n<p>a) \u00a0KC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><\/p>\n<p>b) \u00a0KHSO<sub class=\"subscript\">4<\/sub><\/p>\n<p>c) \u00a0KClO<sub class=\"subscript\">3<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p15\" class=\"para\">13. \u00a0Write the hydrolysis reaction that occurs, if any, when each salt dissolves in water.<\/p>\n<p>a) \u00a0K<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub><\/p>\n<p>b) \u00a0KI<\/p>\n<p>c) \u00a0NH<sub class=\"subscript\">4<\/sub>ClO<sub class=\"subscript\">3<\/sub><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p16\" class=\"para\">14. \u00a0Write the hydrolysis reaction that occurs, if any, when each salt dissolves in water.<\/p>\n<p>a) \u00a0NaNO<sub class=\"subscript\">3<\/sub><\/p>\n<p>b) \u00a0CaC<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub><\/p>\n<p>c) \u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NHCl<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p17\" class=\"para\">15. \u00a0When NH<sub class=\"subscript\">4<\/sub>NO<sub class=\"subscript\">2<\/sub> dissolves in H<sub class=\"subscript\">2<\/sub>O, both ions hydrolyze. Write chemical equations for both reactions. Can you tell if the solution will be acidic or basic overall?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p19\" class=\"para\">16. \u00a0When pyridinium acetate (C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NHC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>) dissolves in H<sub class=\"subscript\">2<\/sub>O, both ions hydrolyze. Write chemical equations for both reactions. Can you tell if the solution will be acidic or basic overall?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p21\" class=\"para\">17. \u00a0A lab technician mixes a solution of 0.015 M Mg(OH)<sub class=\"subscript\">2<\/sub>. Is the resulting OH<sup class=\"superscript\">\u2212<\/sup> concentration greater than, equal to, or less than 0.015 M? Explain your answer.<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch12_s04_qs01_p23\" class=\"para\">18. \u00a0A lab technician mixes a solution of 0.55 M HNO<sub class=\"subscript\">3<\/sub>. Is the resulting H<sup class=\"superscript\">+<\/sup> concentration greater than, equal to, or less than 0.55 M? Explain your answer.<\/p>\n<\/div>\n<p><b>Answers<\/b><\/p>\n<p><strong>1.<\/strong><\/p>\n<p>A strong acid is 100% ionized in aqueous solution, whereas a weak acid is not 100% ionized.<\/p>\n<p><strong>3.<\/strong><\/p>\n<p>a) \u00a0weak acid<\/p>\n<p>b) \u00a0strong acid<\/p>\n<p>c) \u00a0weak acid<\/p>\n<p><strong>5.<\/strong><\/p>\n<p>a) \u00a0HNO<sub class=\"subscript\">3<\/sub>(aq) \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq); proceeds 100%<\/p>\n<p>b) \u00a0HNO<sub class=\"subscript\">2<\/sub>(aq) \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0NO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq); does not proceed 100%<\/p>\n<p>c) \u00a0HI<sub class=\"subscript\">3<\/sub>(aq) \u2192\u00a0H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0I<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq); does not proceed 100%<\/p>\n<p><strong>9.<\/strong><\/p>\n<p>a) \u00a0HCl +\u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>N \u2192\u00a0Cl<sup class=\"superscript\">\u2212<\/sup> +\u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">5<\/sub>NH<sup class=\"superscript\">+<\/sup><\/p>\n<p>b) \u00a0H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a02NH<sub class=\"subscript\">3<\/sub> \u2192\u00a0C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup> +\u00a02NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup><\/p>\n<p>c) \u00a0HNO<sub class=\"subscript\">2<\/sub> +\u00a0C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">9<\/sub>N \u2192\u00a0NO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> +\u00a0C<sub class=\"subscript\">7<\/sub>H<sub class=\"subscript\">9<\/sub>NH<sup class=\"superscript\">+<\/sup><strong>11.<\/strong><\/p>\n<p>a) \u00a0neutral<\/p>\n<p>b) \u00a0acidic<\/p>\n<p>c) \u00a0acidic<\/p>\n<p><strong>13.<\/strong><\/p>\n<p>a) \u00a0basic<\/p>\n<p>b) \u00a0neutral<\/p>\n<p>c) \u00a0acidic<\/p>\n<p><strong>15.<\/strong><\/p>\n<p>a) \u00a0SO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">2\u2212<\/sup> +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0HSO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup> +\u00a0OH<sup class=\"superscript\">\u2212<\/sup><\/p>\n<p>b) \u00a0no reaction<\/p>\n<p>c) \u00a0NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0NH<sub class=\"subscript\">3<\/sub> +\u00a0H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup><strong>17.<\/strong><\/p>\n<p>NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup> +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0NH<sub class=\"subscript\">3<\/sub> +\u00a0H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>; NO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> +\u00a0H<sub class=\"subscript\">2<\/sub>O \u2192\u00a0HNO<sub class=\"subscript\">2<\/sub> +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>; it is not possible to determine whether the solution will be acidic or basic.<\/p>\n<p><strong>19.<\/strong><\/p>\n<p>greater than 0.015 M because there are two OH<sup class=\"superscript\">\u2212<\/sup> ions per formula unit of Mg(OH)<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<\/div>\n<\/div>\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-596\">\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>Introductory Chemistry- 1st Canadian Edition . <strong>Authored by<\/strong>: Jessie A. Key and David W. Ball. <strong>Provided by<\/strong>: BCCampus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/opentextbc.ca\/introductorychemistry\/\">https:\/\/opentextbc.ca\/introductorychemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download this book for free at http:\/\/open.bccampus.ca<\/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":23485,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Chemistry- 1st Canadian Edition \",\"author\":\"Jessie A. Key and David W. 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