{"id":848,"date":"2018-03-20T16:16:37","date_gmt":"2018-03-20T16:16:37","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/?post_type=chapter&#038;p=848"},"modified":"2018-08-08T15:36:54","modified_gmt":"2018-08-08T15:36:54","slug":"10-1-arrhenius-definition-of-acids-and-bases","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/chapter\/10-1-arrhenius-definition-of-acids-and-bases\/","title":{"raw":"10.1 Arrhenius Definition of Acids and Bases","rendered":"10.1 Arrhenius Definition of Acids and Bases"},"content":{"raw":"<div id=\"navbar-top\" class=\"navbar\">\r\n<div class=\"navbar-part left\"><\/div>\r\n<\/div>\r\n<div id=\"book-content\">\r\n<div id=\"gob-ch10_s01\" class=\"section\" xml:lang=\"en\">\r\n<h2 class=\"title editable block\"><span class=\"title-prefix\">10.1<\/span> Arrhenius Definition of Acids and Bases<\/h2>\r\n<div id=\"gob-ch10_s01_n01\" class=\"learning_objectives editable block\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3 class=\"title\">Learning Objective<\/h3>\r\n<ol id=\"gob-ch10_s01_l01\" class=\"orderedlist\">\r\n \t<li>Recognize a compound as an Arrhenius acid or an Arrhenius base.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<p id=\"gob-ch10_s01_p01\" class=\"para editable block\">One way to define a class of compounds is by describing the various characteristics its members have in common. In the case of the compounds known as acids, the common characteristics include a sour taste, the ability to change the color of the vegetable dye <em class=\"emphasis\">litmus<\/em> to red, and the ability to dissolve certain metals and simultaneously produce hydrogen gas. For the compounds called bases, the common characteristics are a slippery texture, a bitter taste, and the ability to change the color of litmus to blue. Acids and bases also react with each other to form compounds generally known as salts.<\/p>\r\n\r\n<div id=\"gob-ch10_s01_n02\" class=\"callout editable block\">\r\n<div class=\"textbox\">\r\n<h3 class=\"title\">Note<\/h3>\r\n<p id=\"gob-ch10_s01_p02\" class=\"para\">Although we include their tastes among the common characteristics of acids and bases, we never advocate tasting an unknown chemical!<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"gob-ch10_s01_p03\" class=\"para editable block\">Chemists prefer, however, to have definitions for acids and bases in chemical terms. The Swedish chemist Svante Arrhenius developed the first chemical definitions of acids and bases in the late 1800s. Arrhenius defined an <span class=\"margin_term\"><span class=\"glossterm\">acid<\/span><\/span>\u00a0as a compound that increases the concentration of hydrogen ion (H<sup class=\"superscript\">+<\/sup>) in aqueous solution. Many acids are simple compounds that release a hydrogen cation into solution when they dissolve. Similarly, Arrhenius defined a <span class=\"margin_term\"><span class=\"glossterm\">base<\/span><\/span>\u00a0as a compound that increases the concentration of hydroxide ion (OH<sup class=\"superscript\">\u2212<\/sup>) in aqueous solution. Many bases are ionic compounds that have the hydroxide ion as their anion, which is released when the base dissolves in water.<\/p>\r\n<p id=\"gob-ch10_s01_p04\" class=\"para editable block\">Many bases and their aqueous solutions are named using the normal rules of ionic compounds that were presented in <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-3\">Chapter 3 \"Ionic Bonding and Simple Ionic Compounds\"<\/a>, Section 3.4 \"Ionic Nomenclature\"; that is, they are named as hydroxide compounds. For example, the base sodium hydroxide (NaOH) is both an ionic compound and an aqueous solution. However, aqueous solutions of acids have their own naming rules. The names of <em class=\"emphasis\">binary acids<\/em> (compounds with hydrogen and one other element in their formula) are based on the root of the name of the other element preceded by the prefix <em class=\"emphasis\">hydro<\/em>- and followed by the suffix -<em class=\"emphasis\">ic acid<\/em>. Thus, an aqueous solution of HCl [designated \u201cHCl(aq)\u201d] is called hydrochloric acid, H<sub class=\"subscript\">2<\/sub>S(aq) is called hydrosulfuric acid, and so forth. Acids composed of more than two elements (typically hydrogen and oxygen and some other element) have names based on the name of the other element, followed by the suffix -<em class=\"emphasis\">ic acid<\/em> or <em class=\"emphasis\">-ous acid<\/em>, depending on the number of oxygen atoms in the acid\u2019s formula. Other prefixes, like <em class=\"emphasis\">per-<\/em> and <em class=\"emphasis\">hypo-<\/em>, also appear in the names for some acids. Unfortunately, there is no strict rule for the number of oxygen atoms that are associated with the -<em class=\"emphasis\">ic acid<\/em> suffix; the names of these acids are best memorized. <a class=\"xref\" href=\"#gob-ch10_s01_t01\">Table 10.1 \"Formulas and Names for Some Acids and Bases\"<\/a> lists some acids and bases and their names. Note that acids have hydrogen written first, as if it were the cation, while most bases have the negative hydroxide ion, if it appears in the formula, written last.<\/p>\r\n\r\n<div id=\"gob-ch10_s01_n03\" class=\"callout editable block\">\r\n<div class=\"textbox\">\r\n<h3 class=\"title\">Note<\/h3>\r\n<p id=\"gob-ch10_s01_p05\" class=\"para\">The name <em class=\"emphasis\">oxygen<\/em> comes from the Latin meaning \u201cacid producer\u201d because its discoverer, Antoine Lavoisier, thought it was the essential element in acids. Lavoisier was wrong, but it is too late to change the name now.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_t01\" class=\"table block\">\r\n<p class=\"title\"><span class=\"title-prefix\">Table 10.1<\/span> Formulas and Names for Some Acids and Bases<\/p>\r\n\r\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\r\n<thead>\r\n<tr>\r\n<th>Formula<\/th>\r\n<th>Name<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\"><strong class=\"emphasis bold\">Acids<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HCl(aq)<\/td>\r\n<td>hydrochloric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HBr(aq)<\/td>\r\n<td>hydrobromic acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HI(aq)<\/td>\r\n<td>hydriodic acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H<sub class=\"subscript\">2<\/sub>S(aq)<\/td>\r\n<td>hydrosulfuric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>(aq)<\/td>\r\n<td>acetic acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HNO<sub class=\"subscript\">3<\/sub>(aq)<\/td>\r\n<td>nitric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HNO<sub class=\"subscript\">2<\/sub>(aq)<\/td>\r\n<td>nitrous acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq)<\/td>\r\n<td>sulfuric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub>(aq)<\/td>\r\n<td>sulfurous acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HClO<sub class=\"subscript\">3<\/sub>(aq)<\/td>\r\n<td>chloric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HClO<sub class=\"subscript\">4<\/sub>(aq)<\/td>\r\n<td>perchloric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HClO<sub class=\"subscript\">2<\/sub>(aq)<\/td>\r\n<td>chlorous acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq)<\/td>\r\n<td>phosphoric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">3<\/sub>(aq)<\/td>\r\n<td>phosphorous acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\"><strong class=\"emphasis bold\">Bases<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>NaOH(aq)<\/td>\r\n<td>sodium hydroxide<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>KOH(aq)<\/td>\r\n<td>potassium hydroxide<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Mg(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/td>\r\n<td>magnesium hydroxide<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Ca(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/td>\r\n<td>calcium hydroxide<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>NH<sub class=\"subscript\">3<\/sub>(aq)<\/td>\r\n<td>ammonia<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_n04\" class=\"exercises editable block\">\r\n<h3 class=\"title\">Example 1<\/h3>\r\n<p id=\"gob-ch10_s01_p06\" class=\"para\">Name each substance.<\/p>\r\n\r\n<ol id=\"gob-ch10_s01_l02\" class=\"orderedlist\">\r\n \t<li>HF(aq)<\/li>\r\n \t<li>Sr(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/li>\r\n<\/ol>\r\n<p class=\"simpara\">Solution<\/p>\r\n\r\n<ol id=\"gob-ch10_s01_l03\" class=\"orderedlist\">\r\n \t<li>This acid has only two elements in its formula, so its name includes the <em class=\"emphasis\">hydro<\/em>- prefix. The stem of the other element\u2019s name, fluorine, is <em class=\"emphasis\">fluor<\/em>, and we must also include the -<em class=\"emphasis\">ic acid<\/em> ending. Its name is hydrofluoric acid.<\/li>\r\n \t<li>This base is named as an ionic compound between the strontium ion and the hydroxide ion: strontium hydroxide.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_qs01\" class=\"qandaset block\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Skill-Building Exercise<\/h3>\r\n<ol id=\"gob-ch10_s01_qs01_qd01\" class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s01_qs01_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs01_p02\" class=\"para\">H<sub class=\"subscript\">2<\/sub>Se(aq)<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs01_qd01_qa02\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs01_p04\" class=\"para\">Ba(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<p id=\"gob-ch10_s01_p10\" class=\"para editable block\">Notice that one base listed in <a class=\"xref\" href=\"#gob-ch10_s01_t01\">Table 10.1 \"Formulas and Names for Some Acids and Bases\"<\/a>\u2014ammonia\u2014does not have hydroxide as part of its formula. How does this compound increase the amount of hydroxide ion in aqueous solution? Instead of dissociating into hydroxide ions, ammonia molecules react with water molecules by taking a hydrogen ion from the water molecule to produce an ammonium ion and a hydroxide ion:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">NH<sub class=\"subscript\">3<\/sub>(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113) \u2192 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) + OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"gob-ch10_s01_p11\" class=\"para editable block\">Because this reaction of ammonia with water causes an increase in the concentration of hydroxide ions in solution, ammonia satisfies the Arrhenius definition of a base. Many other nitrogen-containing compounds are bases because they too react with water to produce hydroxide ions in aqueous solution.<\/p>\r\n<p id=\"gob-ch10_s01_p12\" class=\"para editable block\">As we noted previously, acids and bases react chemically with each other to form <em class=\"emphasis\">salts<\/em>. A salt is a general chemical term for any ionic compound formed from an acid and a base. In reactions where the acid is a hydrogen ion containing compound and the base is a hydroxide ion containing compound, water is also a product. The general reaction is as follows:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">acid + base \u2192 water + salt<\/span><\/span>\r\n<p id=\"gob-ch10_s01_p13\" class=\"para editable block\">The reaction of acid and base to make water and a salt is called <span class=\"margin_term\"><span class=\"glossterm\">neutralization<\/span><\/span>. Like any chemical equation, a neutralization chemical equation must be properly balanced. For example, the neutralization reaction between sodium hydroxide and hydrochloric acid is as follows:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">NaOH(aq) + HCl(aq) \u2192 NaCl(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span>\r\n<p id=\"gob-ch10_s01_p14\" class=\"para editable block\">with coefficients all understood to be one. The neutralization reaction between sodium hydroxide and sulfuric acid is as follows:<\/p>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">2NaOH(aq) + H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) \u2192 Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) + 2H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span>\r\n<p id=\"gob-ch10_s01_p15\" class=\"para editable block\">Once a neutralization reaction is properly balanced, we can use it to perform stoichiometry calculations, such as the ones we practiced in <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-5\">Chapter 5 \"Introduction to Chemical Reactions\"<\/a> and <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-6\">Chapter 6 \"Quantities in Chemical Reactions\"<\/a>.<\/p>\r\n\r\n<div id=\"gob-ch10_s01_n06\" class=\"exercises block\">\r\n<h3 class=\"title\">Example 2<\/h3>\r\n<p id=\"gob-ch10_s01_p16\" class=\"para\">Nitric acid [HNO<sub class=\"subscript\">3<\/sub>(aq)] can be neutralized by calcium hydroxide [Ca(OH)<sub class=\"subscript\">2<\/sub>(aq)].<\/p>\r\n\r\n<ol id=\"gob-ch10_s01_l05\" class=\"orderedlist\">\r\n \t<li>Write a balanced chemical equation for the reaction between these two compounds and identify the salt it produces.<\/li>\r\n \t<li>For one reaction, 16.8 g of HNO<sub class=\"subscript\">3<\/sub> is present initially. How many grams of Ca(OH)<sub class=\"subscript\">2<\/sub> are needed to neutralize that much HNO<sub class=\"subscript\">3<\/sub>?<\/li>\r\n \t<li>In a second reaction, 805 mL of 0.672 M Ca(OH)<sub class=\"subscript\">2<\/sub> is present initially. What volume of 0.432 M HNO<sub class=\"subscript\">3<\/sub> solution is necessary to neutralize the Ca(OH)<sub class=\"subscript\">2<\/sub> solution?<\/li>\r\n<\/ol>\r\n<p class=\"simpara\">Solution<\/p>\r\n\r\n<ol id=\"gob-ch10_s01_l06\" class=\"orderedlist\">\r\n \t<li>\r\n<p class=\"para\">Because there are two OH<sup class=\"superscript\">\u2212<\/sup> ions in the formula for Ca(OH)<sub class=\"subscript\">2<\/sub>, we need two moles of HNO<sub class=\"subscript\">3<\/sub> to provide H<sup class=\"superscript\">+<\/sup> ions. The balanced chemical equation is as follows:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">Ca(OH)<sub class=\"subscript\">2<\/sub>(aq) + 2HNO<sub class=\"subscript\">3<\/sub>(aq) \u2192 Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) + 2H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span>\r\n<p id=\"gob-ch10_s01_p17\" class=\"para\">The salt formed is calcium nitrate.<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"para\">This calculation is much like the calculations we did in <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-6\">Chapter 6 \"Quantities in Chemical Reactions\"<\/a>. First we convert the mass of HNO<sub class=\"subscript\">3<\/sub> to moles using its molar mass of 1.01 + 14.00 + 3(16.00) = 63.01 g\/mol; then we use the balanced chemical equation to determine the related number of moles of Ca(OH)<sub class=\"subscript\">2<\/sub> needed to neutralize it; and then we convert that number of moles of Ca(OH)<sub class=\"subscript\">2<\/sub> to the mass of Ca(OH)<sub class=\"subscript\">2<\/sub> using its molar mass of 40.08 + 2(1.01) + 2(16.00) = 74.10 g\/mol.<\/p>\r\n<span class=\"informalequation\"> [latex]16.8\\,g\\,\\rlap{\\text{---------}}HNO_3\\,\u00d7\\,\\frac{1\\,mol\\,\\rlap{\\text{---------}}HNO_3}{63.01\\,g\\,\\rlap{\\text{---------}}HNO_3}\\,\u00d7\\,\\frac{1\\,mol\\,\\rlap{\\text{------------}}Ca(OH)_2}{2\\,mol\\,\\rlap{\\text{---------}}HNO_3}\\,\u00d7\\,\\frac{74.10\\,g\\,\\rlap{\\text{------------}}Ca(OH)_2}{1\\,mol\\,\\rlap{\\text{------------}}Ca(OH)_2}\\,=\\,9.88\\,g\\,Ca(OH)_2\\,needed[\/latex] <\/span><\/li>\r\n \t<li>\r\n<p class=\"para\">Having concentration information allows us to employ the skills we developed in <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-9\">Chapter 9 \"Solutions\"<\/a>. First, we use the concentration and volume data to determine the number of moles of Ca(OH)<sub class=\"subscript\">2<\/sub> present. Recognizing that 805 mL = 0.805 L,<\/p>\r\n<span class=\"informalequation\"> [latex]0.672\\,M\\,Ca(OH)_2\\,=\\,\\frac{mol\\,Ca(OH)_2}{0.805\\,L\\,soln}[\/latex] <\/span>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">(0.672 M CaOH)<sub class=\"subscript\">2<\/sub> \u00d7 (0.805 L soln) = mol Ca(OH)<sub class=\"subscript\">2<\/sub> = 0.541 mol Ca(OH)<sub class=\"subscript\">2<\/sub><\/span><\/span>\r\n<p id=\"gob-ch10_s01_p18\" class=\"para\">We combine this information with the proper ratio from the balanced chemical equation to determine the number of moles of HNO<sub class=\"subscript\">3<\/sub> needed:<\/p>\r\n<span class=\"informalequation\"> [latex]0.541\\,\\rlap{\\text{----------------}}mol\\,Ca(OH)_2\\,\u00d7\\,\\frac{2\\,mol\\,HNO_3}{1\\,\\rlap{\\text{-----------------}}mol\\,Ca(OH)_2}\\,=\\,1.08\\,mol\\,HNO_3[\/latex]<\/span>\r\n<p id=\"gob-ch10_s01_p19\" class=\"para\">Now, using the definition of molarity one more time, we determine the volume of acid solution needed:<\/p>\r\n<span class=\"informalequation\"> [latex]0.432\\,M\\,HNO_3\\,=\\,\\frac{1.08\\,mol\\,HNO_3}{volume\\,of\\,HNO_3}[\/latex]\u00a0<\/span>\r\n<span class=\"informalequation\"> [latex]volume\\,of\\,HNO_3\\,=\\,\\frac{1.08\\,mol\\,HNO_3}{0.432\\,M\\,HNO_3}\\,=\\,2.50\\,L\\,=\\,2.50\\,\u00d7\\,10^3\\,mL\\,HNO_3[\/latex] <\/span><\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_qs02\" class=\"qandaset block\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Skill-Building Exercise<\/h3>\r\n<ol id=\"gob-ch10_s01_qs02_qd01\" class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s01_qs02_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs02_p02\" class=\"para\">Write a balanced chemical equation for the reaction between these two compounds and identify the salt it produces.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs02_qd01_qa02\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs02_p04\" class=\"para\">For one reaction, 37.5 g of HCN is present initially. How many grams of KOH are needed to neutralize that much HCN?<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs02_qd01_qa03\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs02_p06\" class=\"para\">In a second reaction, 43.0 mL of 0.0663 M KOH is present initially. What volume of 0.107 M HCN solution is necessary to neutralize the KOH solution?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_n08\" class=\"callout editable block\">\r\n<div class=\"textbox\">\r\n<div id=\"gob-ch10_s01_n08\" class=\"callout editable block\">\r\n<h3 class=\"title\">Note<\/h3>\r\n<p id=\"gob-ch10_s01_p24\" class=\"para\">Hydrocyanic acid (HCN) is one exception to the acid-naming rules that specify using the prefix <em class=\"emphasis\">hydro-<\/em> for binary acids (acids composed of hydrogen and only one other element).<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<div id=\"gob-ch10_s01_n08\" class=\"callout editable block\">\r\n<h3 class=\"title\">Concept Review Exercises<\/h3>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_qs03\" class=\"qandaset block\">\r\n<ol id=\"gob-ch10_s01_qs03_qd01\" class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s01_qs03_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs03_p01\" class=\"para\">Give the Arrhenius definitions of an acid and a base.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs03_qd01_qa02\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs03_p03\" class=\"para\">What is neutralization?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_qs03_ans\" class=\"qandaset block\">\r\n<div class=\"textbox exercises\">\r\n<h3 class=\"title\">Answers<\/h3>\r\n<ol class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s01_qs03_qd01_qa01_ans\" class=\"qandaentry\">\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s01_qs03_p02_ans\" class=\"para\">Arrhenius acid: a compound that increases the concentration of hydrogen ion (H<sup class=\"superscript\">+<\/sup>) in aqueous solution; Arrhenius base: a compound that increases the concentration of hydroxide ion (OH<sup class=\"superscript\">\u2212<\/sup>) in aqueous solution.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs03_qd01_qa02_ans\" class=\"qandaentry\">\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s01_qs03_p04_ans\" class=\"para\">the reaction of an acid and a base<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_n10\" class=\"key_takeaways editable block\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3 class=\"title\">Key Takeaway<\/h3>\r\n<ul id=\"gob-ch10_s01_l09\" class=\"itemizedlist\">\r\n \t<li>An Arrhenius acid increases the H<sup class=\"superscript\">+<\/sup> ion concentration in water, while an Arrhenius base increases the OH<sup class=\"superscript\">\u2212<\/sup> ion concentration in water.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_qs04\" class=\"qandaset block\">\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol id=\"gob-ch10_s01_qs04_qd01\" class=\"qandadiv\">\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa01\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p01\" class=\"para\">Give two examples of Arrhenius acids.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa02\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p03\" class=\"para\">Give two examples of Arrhenius bases.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa03\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p05\" class=\"para\">List the general properties of acids.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa04\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p07\" class=\"para\">List the general properties of bases.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa05\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p09\" class=\"para\">Name each compound.<\/p>\r\n\r\n<ol id=\"gob-ch10_s01_qs04_l01\" class=\"orderedlist\">\r\n \t<li>HBr(aq)<\/li>\r\n \t<li>Ca(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/li>\r\n \t<li>HNO<sub class=\"subscript\">3<\/sub>(aq)<\/li>\r\n \t<li>Fe(OH)<sub class=\"subscript\">3<\/sub>(aq)<\/li>\r\n<\/ol>\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa06\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p10\" class=\"para\">Name each compound.<\/p>\r\n\r\n<ol id=\"gob-ch10_s01_qs04_l03\" class=\"orderedlist\">\r\n \t<li>HI(aq)<\/li>\r\n \t<li>Cu(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/li>\r\n \t<li>H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq)<\/li>\r\n \t<li>CsOH(aq)<\/li>\r\n<\/ol>\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa07\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p11\" class=\"para\">Propose a name for water (H<sub class=\"subscript\">2<\/sub>O) using the rules for naming acids.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa08\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p13\" class=\"para\">Propose a name for hydrogen peroxide (H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub>) using the rules for naming acids.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa09\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p15\" class=\"para\">Write a balanced chemical equation for the neutralization of Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) with HNO<sub class=\"subscript\">3<\/sub>(aq).<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa10\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p17\" class=\"para\">Write a balanced chemical equation for the neutralization of H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) with Cr(OH)<sub class=\"subscript\">3<\/sub>(aq).<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa11\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p19\" class=\"para\">How many moles of sodium hydroxide (NaOH) are needed to neutralize 0.844 mol of acetic acid (HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>)? (Hint: begin by writing a balanced chemical equation for the process.)<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa12\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p21\" class=\"para\">How many moles of perchloric acid (HClO<sub class=\"subscript\">4<\/sub>) are needed to neutralize 0.052 mol of calcium hydroxide [Ca(OH)<sub class=\"subscript\">2<\/sub>]? (Hint: begin by writing a balanced chemical equation for the process.)<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa13\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p23\" class=\"para\">Hydrazoic acid (HN<sub class=\"subscript\">3<\/sub>) can be neutralized by a base.<\/p>\r\n\r\n<ol id=\"gob-ch10_s01_qs04_l05\" class=\"orderedlist\">\r\n \t<li>Write the balanced chemical equation for the reaction between hydrazoic acid and calcium hydroxide.<\/li>\r\n \t<li>How many milliliters of 0.0245 M Ca(OH)<sub class=\"subscript\">2<\/sub> are needed to neutralize 0.564 g of HN<sub class=\"subscript\">3<\/sub>?<\/li>\r\n<\/ol>\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa14\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p24\" class=\"para\">Citric acid (H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub>) has three hydrogen atoms that can form hydrogen ions in solution.<\/p>\r\n\r\n<ol id=\"gob-ch10_s01_qs04_l07\" class=\"orderedlist\">\r\n \t<li>Write the balanced chemical equation for the reaction between citric acid and sodium hydroxide.<\/li>\r\n \t<li>If an orange contains 0.0675 g of H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub>, how many milliliters of 0.00332 M NaOH solution are needed to neutralize the acid?<\/li>\r\n<\/ol>\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa15\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p25\" class=\"para\">Magnesium hydroxide [Mg(OH)<sub class=\"subscript\">2<\/sub>] is an ingredient in some antacids. How many grams of Mg(OH)<sub class=\"subscript\">2<\/sub> are needed to neutralize the acid in 158 mL of 0.106 M HCl(aq)? It might help to write the balanced chemical equation first.<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li id=\"gob-ch10_s01_qs04_qd01_qa16\" class=\"qandaentry\">\r\n<div class=\"question\">\r\n<p id=\"gob-ch10_s01_qs04_p27\" class=\"para\">Aluminum hydroxide [Al(OH)<sub class=\"subscript\">3<\/sub>] is an ingredient in some antacids. How many grams of Al(OH)<sub class=\"subscript\">3<\/sub> are needed to neutralize the acid in 96.5 mL of 0.556 M H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq)? It might help to write the balanced chemical equation first.<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"gob-ch10_s01_qs04_ans\" class=\"qandaset block\">\r\n<div class=\"textbox exercises\">\r\n<div id=\"book-content\">\r\n<div id=\"gob-ch10_s01\" class=\"section\" xml:lang=\"en\">\r\n<div id=\"gob-ch10_s01_qs04_ans\" class=\"qandaset block\">\r\n<h3 class=\"title\">Answers<\/h3>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s01_qs04_p02_ans\" class=\"para\">1. HCl and HNO<sub class=\"subscript\">3<\/sub> (answers will vary)<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s01_qs04_p06_ans\" class=\"para\">3. sour taste, react with metals, react with bases, and turn litmus red<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n\r\n5. a. hydrobromic acid\r\n\r\nb. calcium hydroxide\r\n\r\nc. nitric acid\r\n\r\nd. iron(III) hydroxide\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s01_qs04_p12_ans\" class=\"para\">7. perhaps hydroxic acid<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s01_qs04_p16_ans\" class=\"para\">9. 2HNO<sub class=\"subscript\">3<\/sub>(aq) + Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) \u2192 Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) + 2H<sub class=\"subscript\">2<\/sub>O<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s01_qs04_p20_ans\" class=\"para\">11. 0.844 mol<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n\r\n13. a. 2HN<sub class=\"subscript\">3<\/sub>(aq) + Ca(OH)<sub class=\"subscript\">2<\/sub> \u2192 Ca(N<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> + 2H<sub class=\"subscript\">2<\/sub>O\r\n\r\nb. 268 mL\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<div class=\"answer\">\r\n<p id=\"gob-ch10_s01_qs04_p26_ans\" class=\"para\">15. 0.488 g<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"navbar-bottom\" class=\"navbar\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div id=\"navbar-top\" class=\"navbar\">\n<div class=\"navbar-part left\"><\/div>\n<\/div>\n<div id=\"book-content\">\n<div id=\"gob-ch10_s01\" class=\"section\" xml:lang=\"en\">\n<h2 class=\"title editable block\"><span class=\"title-prefix\">10.1<\/span> Arrhenius Definition of Acids and Bases<\/h2>\n<div id=\"gob-ch10_s01_n01\" class=\"learning_objectives editable block\">\n<div class=\"textbox learning-objectives\">\n<h3 class=\"title\">Learning Objective<\/h3>\n<ol id=\"gob-ch10_s01_l01\" class=\"orderedlist\">\n<li>Recognize a compound as an Arrhenius acid or an Arrhenius base.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p id=\"gob-ch10_s01_p01\" class=\"para editable block\">One way to define a class of compounds is by describing the various characteristics its members have in common. In the case of the compounds known as acids, the common characteristics include a sour taste, the ability to change the color of the vegetable dye <em class=\"emphasis\">litmus<\/em> to red, and the ability to dissolve certain metals and simultaneously produce hydrogen gas. For the compounds called bases, the common characteristics are a slippery texture, a bitter taste, and the ability to change the color of litmus to blue. Acids and bases also react with each other to form compounds generally known as salts.<\/p>\n<div id=\"gob-ch10_s01_n02\" class=\"callout editable block\">\n<div class=\"textbox\">\n<h3 class=\"title\">Note<\/h3>\n<p id=\"gob-ch10_s01_p02\" class=\"para\">Although we include their tastes among the common characteristics of acids and bases, we never advocate tasting an unknown chemical!<\/p>\n<\/div>\n<\/div>\n<p id=\"gob-ch10_s01_p03\" class=\"para editable block\">Chemists prefer, however, to have definitions for acids and bases in chemical terms. The Swedish chemist Svante Arrhenius developed the first chemical definitions of acids and bases in the late 1800s. Arrhenius defined an <span class=\"margin_term\"><span class=\"glossterm\">acid<\/span><\/span>\u00a0as a compound that increases the concentration of hydrogen ion (H<sup class=\"superscript\">+<\/sup>) in aqueous solution. Many acids are simple compounds that release a hydrogen cation into solution when they dissolve. Similarly, Arrhenius defined a <span class=\"margin_term\"><span class=\"glossterm\">base<\/span><\/span>\u00a0as a compound that increases the concentration of hydroxide ion (OH<sup class=\"superscript\">\u2212<\/sup>) in aqueous solution. Many bases are ionic compounds that have the hydroxide ion as their anion, which is released when the base dissolves in water.<\/p>\n<p id=\"gob-ch10_s01_p04\" class=\"para editable block\">Many bases and their aqueous solutions are named using the normal rules of ionic compounds that were presented in <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-3\">Chapter 3 &#8220;Ionic Bonding and Simple Ionic Compounds&#8221;<\/a>, Section 3.4 &#8220;Ionic Nomenclature&#8221;; that is, they are named as hydroxide compounds. For example, the base sodium hydroxide (NaOH) is both an ionic compound and an aqueous solution. However, aqueous solutions of acids have their own naming rules. The names of <em class=\"emphasis\">binary acids<\/em> (compounds with hydrogen and one other element in their formula) are based on the root of the name of the other element preceded by the prefix <em class=\"emphasis\">hydro<\/em>&#8211; and followed by the suffix &#8211;<em class=\"emphasis\">ic acid<\/em>. Thus, an aqueous solution of HCl [designated \u201cHCl(aq)\u201d] is called hydrochloric acid, H<sub class=\"subscript\">2<\/sub>S(aq) is called hydrosulfuric acid, and so forth. Acids composed of more than two elements (typically hydrogen and oxygen and some other element) have names based on the name of the other element, followed by the suffix &#8211;<em class=\"emphasis\">ic acid<\/em> or <em class=\"emphasis\">-ous acid<\/em>, depending on the number of oxygen atoms in the acid\u2019s formula. Other prefixes, like <em class=\"emphasis\">per-<\/em> and <em class=\"emphasis\">hypo-<\/em>, also appear in the names for some acids. Unfortunately, there is no strict rule for the number of oxygen atoms that are associated with the &#8211;<em class=\"emphasis\">ic acid<\/em> suffix; the names of these acids are best memorized. <a class=\"xref\" href=\"#gob-ch10_s01_t01\">Table 10.1 &#8220;Formulas and Names for Some Acids and Bases&#8221;<\/a> lists some acids and bases and their names. Note that acids have hydrogen written first, as if it were the cation, while most bases have the negative hydroxide ion, if it appears in the formula, written last.<\/p>\n<div id=\"gob-ch10_s01_n03\" class=\"callout editable block\">\n<div class=\"textbox\">\n<h3 class=\"title\">Note<\/h3>\n<p id=\"gob-ch10_s01_p05\" class=\"para\">The name <em class=\"emphasis\">oxygen<\/em> comes from the Latin meaning \u201cacid producer\u201d because its discoverer, Antoine Lavoisier, thought it was the essential element in acids. Lavoisier was wrong, but it is too late to change the name now.<\/p>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s01_t01\" class=\"table block\">\n<p class=\"title\"><span class=\"title-prefix\">Table 10.1<\/span> Formulas and Names for Some Acids and Bases<\/p>\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\n<thead>\n<tr>\n<th>Formula<\/th>\n<th>Name<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td colspan=\"2\"><strong class=\"emphasis bold\">Acids<\/strong><\/td>\n<\/tr>\n<tr>\n<td>HCl(aq)<\/td>\n<td>hydrochloric acid<\/td>\n<\/tr>\n<tr>\n<td>HBr(aq)<\/td>\n<td>hydrobromic acid<\/td>\n<\/tr>\n<tr>\n<td>HI(aq)<\/td>\n<td>hydriodic acid<\/td>\n<\/tr>\n<tr>\n<td>H<sub class=\"subscript\">2<\/sub>S(aq)<\/td>\n<td>hydrosulfuric acid<\/td>\n<\/tr>\n<tr>\n<td>HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>(aq)<\/td>\n<td>acetic acid<\/td>\n<\/tr>\n<tr>\n<td>HNO<sub class=\"subscript\">3<\/sub>(aq)<\/td>\n<td>nitric acid<\/td>\n<\/tr>\n<tr>\n<td>HNO<sub class=\"subscript\">2<\/sub>(aq)<\/td>\n<td>nitrous acid<\/td>\n<\/tr>\n<tr>\n<td>H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq)<\/td>\n<td>sulfuric acid<\/td>\n<\/tr>\n<tr>\n<td>H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub>(aq)<\/td>\n<td>sulfurous acid<\/td>\n<\/tr>\n<tr>\n<td>HClO<sub class=\"subscript\">3<\/sub>(aq)<\/td>\n<td>chloric acid<\/td>\n<\/tr>\n<tr>\n<td>HClO<sub class=\"subscript\">4<\/sub>(aq)<\/td>\n<td>perchloric acid<\/td>\n<\/tr>\n<tr>\n<td>HClO<sub class=\"subscript\">2<\/sub>(aq)<\/td>\n<td>chlorous acid<\/td>\n<\/tr>\n<tr>\n<td>H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq)<\/td>\n<td>phosphoric acid<\/td>\n<\/tr>\n<tr>\n<td>H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">3<\/sub>(aq)<\/td>\n<td>phosphorous acid<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\"><strong class=\"emphasis bold\">Bases<\/strong><\/td>\n<\/tr>\n<tr>\n<td>NaOH(aq)<\/td>\n<td>sodium hydroxide<\/td>\n<\/tr>\n<tr>\n<td>KOH(aq)<\/td>\n<td>potassium hydroxide<\/td>\n<\/tr>\n<tr>\n<td>Mg(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/td>\n<td>magnesium hydroxide<\/td>\n<\/tr>\n<tr>\n<td>Ca(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/td>\n<td>calcium hydroxide<\/td>\n<\/tr>\n<tr>\n<td>NH<sub class=\"subscript\">3<\/sub>(aq)<\/td>\n<td>ammonia<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"gob-ch10_s01_n04\" class=\"exercises editable block\">\n<h3 class=\"title\">Example 1<\/h3>\n<p id=\"gob-ch10_s01_p06\" class=\"para\">Name each substance.<\/p>\n<ol id=\"gob-ch10_s01_l02\" class=\"orderedlist\">\n<li>HF(aq)<\/li>\n<li>Sr(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/li>\n<\/ol>\n<p class=\"simpara\">Solution<\/p>\n<ol id=\"gob-ch10_s01_l03\" class=\"orderedlist\">\n<li>This acid has only two elements in its formula, so its name includes the <em class=\"emphasis\">hydro<\/em>&#8211; prefix. The stem of the other element\u2019s name, fluorine, is <em class=\"emphasis\">fluor<\/em>, and we must also include the &#8211;<em class=\"emphasis\">ic acid<\/em> ending. Its name is hydrofluoric acid.<\/li>\n<li>This base is named as an ionic compound between the strontium ion and the hydroxide ion: strontium hydroxide.<\/li>\n<\/ol>\n<\/div>\n<div id=\"gob-ch10_s01_qs01\" class=\"qandaset block\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Skill-Building Exercise<\/h3>\n<ol id=\"gob-ch10_s01_qs01_qd01\" class=\"qandadiv\">\n<li id=\"gob-ch10_s01_qs01_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs01_p02\" class=\"para\">H<sub class=\"subscript\">2<\/sub>Se(aq)<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs01_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs01_p04\" class=\"para\">Ba(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p id=\"gob-ch10_s01_p10\" class=\"para editable block\">Notice that one base listed in <a class=\"xref\" href=\"#gob-ch10_s01_t01\">Table 10.1 &#8220;Formulas and Names for Some Acids and Bases&#8221;<\/a>\u2014ammonia\u2014does not have hydroxide as part of its formula. How does this compound increase the amount of hydroxide ion in aqueous solution? Instead of dissociating into hydroxide ions, ammonia molecules react with water molecules by taking a hydrogen ion from the water molecule to produce an ammonium ion and a hydroxide ion:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">NH<sub class=\"subscript\">3<\/sub>(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113) \u2192 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) + OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"gob-ch10_s01_p11\" class=\"para editable block\">Because this reaction of ammonia with water causes an increase in the concentration of hydroxide ions in solution, ammonia satisfies the Arrhenius definition of a base. Many other nitrogen-containing compounds are bases because they too react with water to produce hydroxide ions in aqueous solution.<\/p>\n<p id=\"gob-ch10_s01_p12\" class=\"para editable block\">As we noted previously, acids and bases react chemically with each other to form <em class=\"emphasis\">salts<\/em>. A salt is a general chemical term for any ionic compound formed from an acid and a base. In reactions where the acid is a hydrogen ion containing compound and the base is a hydroxide ion containing compound, water is also a product. The general reaction is as follows:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">acid + base \u2192 water + salt<\/span><\/span><\/p>\n<p id=\"gob-ch10_s01_p13\" class=\"para editable block\">The reaction of acid and base to make water and a salt is called <span class=\"margin_term\"><span class=\"glossterm\">neutralization<\/span><\/span>. Like any chemical equation, a neutralization chemical equation must be properly balanced. For example, the neutralization reaction between sodium hydroxide and hydrochloric acid is as follows:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">NaOH(aq) + HCl(aq) \u2192 NaCl(aq) + H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"gob-ch10_s01_p14\" class=\"para editable block\">with coefficients all understood to be one. The neutralization reaction between sodium hydroxide and sulfuric acid is as follows:<\/p>\n<p><span class=\"informalequation block\"><span class=\"mathphrase\">2NaOH(aq) + H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) \u2192 Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) + 2H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"gob-ch10_s01_p15\" class=\"para editable block\">Once a neutralization reaction is properly balanced, we can use it to perform stoichiometry calculations, such as the ones we practiced in <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-5\">Chapter 5 &#8220;Introduction to Chemical Reactions&#8221;<\/a> and <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-6\">Chapter 6 &#8220;Quantities in Chemical Reactions&#8221;<\/a>.<\/p>\n<div id=\"gob-ch10_s01_n06\" class=\"exercises block\">\n<h3 class=\"title\">Example 2<\/h3>\n<p id=\"gob-ch10_s01_p16\" class=\"para\">Nitric acid [HNO<sub class=\"subscript\">3<\/sub>(aq)] can be neutralized by calcium hydroxide [Ca(OH)<sub class=\"subscript\">2<\/sub>(aq)].<\/p>\n<ol id=\"gob-ch10_s01_l05\" class=\"orderedlist\">\n<li>Write a balanced chemical equation for the reaction between these two compounds and identify the salt it produces.<\/li>\n<li>For one reaction, 16.8 g of HNO<sub class=\"subscript\">3<\/sub> is present initially. How many grams of Ca(OH)<sub class=\"subscript\">2<\/sub> are needed to neutralize that much HNO<sub class=\"subscript\">3<\/sub>?<\/li>\n<li>In a second reaction, 805 mL of 0.672 M Ca(OH)<sub class=\"subscript\">2<\/sub> is present initially. What volume of 0.432 M HNO<sub class=\"subscript\">3<\/sub> solution is necessary to neutralize the Ca(OH)<sub class=\"subscript\">2<\/sub> solution?<\/li>\n<\/ol>\n<p class=\"simpara\">Solution<\/p>\n<ol id=\"gob-ch10_s01_l06\" class=\"orderedlist\">\n<li>\n<p class=\"para\">Because there are two OH<sup class=\"superscript\">\u2212<\/sup> ions in the formula for Ca(OH)<sub class=\"subscript\">2<\/sub>, we need two moles of HNO<sub class=\"subscript\">3<\/sub> to provide H<sup class=\"superscript\">+<\/sup> ions. The balanced chemical equation is as follows:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">Ca(OH)<sub class=\"subscript\">2<\/sub>(aq) + 2HNO<sub class=\"subscript\">3<\/sub>(aq) \u2192 Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) + 2H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"gob-ch10_s01_p17\" class=\"para\">The salt formed is calcium nitrate.<\/p>\n<\/li>\n<li>\n<p class=\"para\">This calculation is much like the calculations we did in <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-6\">Chapter 6 &#8220;Quantities in Chemical Reactions&#8221;<\/a>. First we convert the mass of HNO<sub class=\"subscript\">3<\/sub> to moles using its molar mass of 1.01 + 14.00 + 3(16.00) = 63.01 g\/mol; then we use the balanced chemical equation to determine the related number of moles of Ca(OH)<sub class=\"subscript\">2<\/sub> needed to neutralize it; and then we convert that number of moles of Ca(OH)<sub class=\"subscript\">2<\/sub> to the mass of Ca(OH)<sub class=\"subscript\">2<\/sub> using its molar mass of 40.08 + 2(1.01) + 2(16.00) = 74.10 g\/mol.<\/p>\n<p><span class=\"informalequation\"> [latex]16.8\\,g\\,\\rlap{\\text{---------}}HNO_3\\,\u00d7\\,\\frac{1\\,mol\\,\\rlap{\\text{---------}}HNO_3}{63.01\\,g\\,\\rlap{\\text{---------}}HNO_3}\\,\u00d7\\,\\frac{1\\,mol\\,\\rlap{\\text{------------}}Ca(OH)_2}{2\\,mol\\,\\rlap{\\text{---------}}HNO_3}\\,\u00d7\\,\\frac{74.10\\,g\\,\\rlap{\\text{------------}}Ca(OH)_2}{1\\,mol\\,\\rlap{\\text{------------}}Ca(OH)_2}\\,=\\,9.88\\,g\\,Ca(OH)_2\\,needed[\/latex] <\/span><\/li>\n<li>\n<p class=\"para\">Having concentration information allows us to employ the skills we developed in <a class=\"xref\" href=\"..\/suny-orgbiochemistry\/chapter\/introduction-9\">Chapter 9 &#8220;Solutions&#8221;<\/a>. First, we use the concentration and volume data to determine the number of moles of Ca(OH)<sub class=\"subscript\">2<\/sub> present. Recognizing that 805 mL = 0.805 L,<\/p>\n<p><span class=\"informalequation\"> [latex]0.672\\,M\\,Ca(OH)_2\\,=\\,\\frac{mol\\,Ca(OH)_2}{0.805\\,L\\,soln}[\/latex] <\/span><br \/>\n<span class=\"informalequation\"><span class=\"mathphrase\">(0.672 M CaOH)<sub class=\"subscript\">2<\/sub> \u00d7 (0.805 L soln) = mol Ca(OH)<sub class=\"subscript\">2<\/sub> = 0.541 mol Ca(OH)<sub class=\"subscript\">2<\/sub><\/span><\/span><\/p>\n<p id=\"gob-ch10_s01_p18\" class=\"para\">We combine this information with the proper ratio from the balanced chemical equation to determine the number of moles of HNO<sub class=\"subscript\">3<\/sub> needed:<\/p>\n<p><span class=\"informalequation\"> [latex]0.541\\,\\rlap{\\text{----------------}}mol\\,Ca(OH)_2\\,\u00d7\\,\\frac{2\\,mol\\,HNO_3}{1\\,\\rlap{\\text{-----------------}}mol\\,Ca(OH)_2}\\,=\\,1.08\\,mol\\,HNO_3[\/latex]<\/span><\/p>\n<p id=\"gob-ch10_s01_p19\" class=\"para\">Now, using the definition of molarity one more time, we determine the volume of acid solution needed:<\/p>\n<p><span class=\"informalequation\"> [latex]0.432\\,M\\,HNO_3\\,=\\,\\frac{1.08\\,mol\\,HNO_3}{volume\\,of\\,HNO_3}[\/latex]\u00a0<\/span><br \/>\n<span class=\"informalequation\"> [latex]volume\\,of\\,HNO_3\\,=\\,\\frac{1.08\\,mol\\,HNO_3}{0.432\\,M\\,HNO_3}\\,=\\,2.50\\,L\\,=\\,2.50\\,\u00d7\\,10^3\\,mL\\,HNO_3[\/latex] <\/span><\/li>\n<\/ol>\n<\/div>\n<div id=\"gob-ch10_s01_qs02\" class=\"qandaset block\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Skill-Building Exercise<\/h3>\n<ol id=\"gob-ch10_s01_qs02_qd01\" class=\"qandadiv\">\n<li id=\"gob-ch10_s01_qs02_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs02_p02\" class=\"para\">Write a balanced chemical equation for the reaction between these two compounds and identify the salt it produces.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs02_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs02_p04\" class=\"para\">For one reaction, 37.5 g of HCN is present initially. How many grams of KOH are needed to neutralize that much HCN?<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs02_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs02_p06\" class=\"para\">In a second reaction, 43.0 mL of 0.0663 M KOH is present initially. What volume of 0.107 M HCN solution is necessary to neutralize the KOH solution?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s01_n08\" class=\"callout editable block\">\n<div class=\"textbox\">\n<div id=\"gob-ch10_s01_n08\" class=\"callout editable block\">\n<h3 class=\"title\">Note<\/h3>\n<p id=\"gob-ch10_s01_p24\" class=\"para\">Hydrocyanic acid (HCN) is one exception to the acid-naming rules that specify using the prefix <em class=\"emphasis\">hydro-<\/em> for binary acids (acids composed of hydrogen and only one other element).<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<div id=\"gob-ch10_s01_n08\" class=\"callout editable block\">\n<h3 class=\"title\">Concept Review Exercises<\/h3>\n<\/div>\n<div id=\"gob-ch10_s01_qs03\" class=\"qandaset block\">\n<ol id=\"gob-ch10_s01_qs03_qd01\" class=\"qandadiv\">\n<li id=\"gob-ch10_s01_qs03_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs03_p01\" class=\"para\">Give the Arrhenius definitions of an acid and a base.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs03_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs03_p03\" class=\"para\">What is neutralization?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s01_qs03_ans\" class=\"qandaset block\">\n<div class=\"textbox exercises\">\n<h3 class=\"title\">Answers<\/h3>\n<ol class=\"qandadiv\">\n<li id=\"gob-ch10_s01_qs03_qd01_qa01_ans\" class=\"qandaentry\">\n<div class=\"answer\">\n<p id=\"gob-ch10_s01_qs03_p02_ans\" class=\"para\">Arrhenius acid: a compound that increases the concentration of hydrogen ion (H<sup class=\"superscript\">+<\/sup>) in aqueous solution; Arrhenius base: a compound that increases the concentration of hydroxide ion (OH<sup class=\"superscript\">\u2212<\/sup>) in aqueous solution.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs03_qd01_qa02_ans\" class=\"qandaentry\">\n<div class=\"answer\">\n<p id=\"gob-ch10_s01_qs03_p04_ans\" class=\"para\">the reaction of an acid and a base<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s01_n10\" class=\"key_takeaways editable block\">\n<div class=\"textbox key-takeaways\">\n<h3 class=\"title\">Key Takeaway<\/h3>\n<ul id=\"gob-ch10_s01_l09\" class=\"itemizedlist\">\n<li>An Arrhenius acid increases the H<sup class=\"superscript\">+<\/sup> ion concentration in water, while an Arrhenius base increases the OH<sup class=\"superscript\">\u2212<\/sup> ion concentration in water.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s01_qs04\" class=\"qandaset block\">\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol id=\"gob-ch10_s01_qs04_qd01\" class=\"qandadiv\">\n<li id=\"gob-ch10_s01_qs04_qd01_qa01\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p01\" class=\"para\">Give two examples of Arrhenius acids.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa02\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p03\" class=\"para\">Give two examples of Arrhenius bases.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa03\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p05\" class=\"para\">List the general properties of acids.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa04\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p07\" class=\"para\">List the general properties of bases.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa05\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p09\" class=\"para\">Name each compound.<\/p>\n<ol id=\"gob-ch10_s01_qs04_l01\" class=\"orderedlist\">\n<li>HBr(aq)<\/li>\n<li>Ca(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/li>\n<li>HNO<sub class=\"subscript\">3<\/sub>(aq)<\/li>\n<li>Fe(OH)<sub class=\"subscript\">3<\/sub>(aq)<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa06\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p10\" class=\"para\">Name each compound.<\/p>\n<ol id=\"gob-ch10_s01_qs04_l03\" class=\"orderedlist\">\n<li>HI(aq)<\/li>\n<li>Cu(OH)<sub class=\"subscript\">2<\/sub>(aq)<\/li>\n<li>H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq)<\/li>\n<li>CsOH(aq)<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa07\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p11\" class=\"para\">Propose a name for water (H<sub class=\"subscript\">2<\/sub>O) using the rules for naming acids.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa08\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p13\" class=\"para\">Propose a name for hydrogen peroxide (H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub>) using the rules for naming acids.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa09\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p15\" class=\"para\">Write a balanced chemical equation for the neutralization of Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) with HNO<sub class=\"subscript\">3<\/sub>(aq).<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa10\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p17\" class=\"para\">Write a balanced chemical equation for the neutralization of H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) with Cr(OH)<sub class=\"subscript\">3<\/sub>(aq).<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa11\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p19\" class=\"para\">How many moles of sodium hydroxide (NaOH) are needed to neutralize 0.844 mol of acetic acid (HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>)? (Hint: begin by writing a balanced chemical equation for the process.)<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa12\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p21\" class=\"para\">How many moles of perchloric acid (HClO<sub class=\"subscript\">4<\/sub>) are needed to neutralize 0.052 mol of calcium hydroxide [Ca(OH)<sub class=\"subscript\">2<\/sub>]? (Hint: begin by writing a balanced chemical equation for the process.)<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa13\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p23\" class=\"para\">Hydrazoic acid (HN<sub class=\"subscript\">3<\/sub>) can be neutralized by a base.<\/p>\n<ol id=\"gob-ch10_s01_qs04_l05\" class=\"orderedlist\">\n<li>Write the balanced chemical equation for the reaction between hydrazoic acid and calcium hydroxide.<\/li>\n<li>How many milliliters of 0.0245 M Ca(OH)<sub class=\"subscript\">2<\/sub> are needed to neutralize 0.564 g of HN<sub class=\"subscript\">3<\/sub>?<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa14\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p24\" class=\"para\">Citric acid (H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub>) has three hydrogen atoms that can form hydrogen ions in solution.<\/p>\n<ol id=\"gob-ch10_s01_qs04_l07\" class=\"orderedlist\">\n<li>Write the balanced chemical equation for the reaction between citric acid and sodium hydroxide.<\/li>\n<li>If an orange contains 0.0675 g of H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub>, how many milliliters of 0.00332 M NaOH solution are needed to neutralize the acid?<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa15\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p25\" class=\"para\">Magnesium hydroxide [Mg(OH)<sub class=\"subscript\">2<\/sub>] is an ingredient in some antacids. How many grams of Mg(OH)<sub class=\"subscript\">2<\/sub> are needed to neutralize the acid in 158 mL of 0.106 M HCl(aq)? It might help to write the balanced chemical equation first.<\/p>\n<\/div>\n<\/li>\n<li id=\"gob-ch10_s01_qs04_qd01_qa16\" class=\"qandaentry\">\n<div class=\"question\">\n<p id=\"gob-ch10_s01_qs04_p27\" class=\"para\">Aluminum hydroxide [Al(OH)<sub class=\"subscript\">3<\/sub>] is an ingredient in some antacids. How many grams of Al(OH)<sub class=\"subscript\">3<\/sub> are needed to neutralize the acid in 96.5 mL of 0.556 M H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq)? It might help to write the balanced chemical equation first.<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"gob-ch10_s01_qs04_ans\" class=\"qandaset block\">\n<div class=\"textbox exercises\">\n<div id=\"book-content\">\n<div id=\"gob-ch10_s01\" class=\"section\" xml:lang=\"en\">\n<div id=\"gob-ch10_s01_qs04_ans\" class=\"qandaset block\">\n<h3 class=\"title\">Answers<\/h3>\n<div class=\"answer\">\n<p id=\"gob-ch10_s01_qs04_p02_ans\" class=\"para\">1. HCl and HNO<sub class=\"subscript\">3<\/sub> (answers will vary)<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p id=\"gob-ch10_s01_qs04_p06_ans\" class=\"para\">3. sour taste, react with metals, react with bases, and turn litmus red<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p>5. a. hydrobromic acid<\/p>\n<p>b. calcium hydroxide<\/p>\n<p>c. nitric acid<\/p>\n<p>d. iron(III) hydroxide<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p id=\"gob-ch10_s01_qs04_p12_ans\" class=\"para\">7. perhaps hydroxic acid<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p id=\"gob-ch10_s01_qs04_p16_ans\" class=\"para\">9. 2HNO<sub class=\"subscript\">3<\/sub>(aq) + Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) \u2192 Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) + 2H<sub class=\"subscript\">2<\/sub>O<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p id=\"gob-ch10_s01_qs04_p20_ans\" class=\"para\">11. 0.844 mol<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p>13. a. 2HN<sub class=\"subscript\">3<\/sub>(aq) + Ca(OH)<sub class=\"subscript\">2<\/sub> \u2192 Ca(N<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub> + 2H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p>b. 268 mL<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<div class=\"answer\">\n<p id=\"gob-ch10_s01_qs04_p26_ans\" class=\"para\">15. 0.488 g<\/p>\n<\/div>\n<div class=\"answer\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"navbar-bottom\" class=\"navbar\"><\/div>\n<\/div>\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-848\">\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>The Basics of General, Organic, and Biological Chemistry v. 1.0. <strong>Provided by<\/strong>: Saylor Academy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/saylordotorg.github.io\/text_the-basics-of-general-organic-and-biological-chemistry\/\">https:\/\/saylordotorg.github.io\/text_the-basics-of-general-organic-and-biological-chemistry\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC: Attribution-NonCommercial<\/a><\/em>. <strong>License Terms<\/strong>: This text was adapted by Saylor Academy under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License without attribution as requested by the work&#039;s original creator or licensor.<\/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":44985,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"The Basics of General, Organic, and Biological Chemistry v. 1.0\",\"author\":\"\",\"organization\":\"Saylor Academy\",\"url\":\"https:\/\/saylordotorg.github.io\/text_the-basics-of-general-organic-and-biological-chemistry\/\",\"project\":\"\",\"license\":\"cc-by-nc\",\"license_terms\":\"This text was adapted by Saylor Academy under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License without attribution as requested by the work\\'s original creator or licensor.\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-848","chapter","type-chapter","status-publish","hentry"],"part":838,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters\/848","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/users\/44985"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters\/848\/revisions"}],"predecessor-version":[{"id":3123,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters\/848\/revisions\/3123"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/parts\/838"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapters\/848\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/media?parent=848"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=848"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/contributor?post=848"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/wp-json\/wp\/v2\/license?post=848"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}