{"id":523,"date":"2017-08-21T10:01:54","date_gmt":"2017-08-21T10:01:54","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=523"},"modified":"2017-09-20T21:21:25","modified_gmt":"2017-09-20T21:21:25","slug":"relationships-between-acids-bases-and-salts","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/relationships-between-acids-bases-and-salts\/","title":{"rendered":"Relationships Between Acids, Bases and Salts"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/the-development-and-theories-of-acids-and-bases\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/gas-laws-and-solutions\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/review-5\">Next Lesson \u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Relationships Between Acids, Bases, and Salts<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will examine the relationships between acids, bases, and salts and equilibrium in both experimental<br \/>\nsituations and common calculations.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Comparing and contrasting the Arrhenius, Br\u00f8nsted-Lowry, and Lewis definitions of acids and bases<\/li>\n<li>Identifying the conjugate acid-base pairs in a chemical equation for a proton-transfer reaction<\/li>\n<li>Explaining the origin of the pH scale and its relationship to [H<sub>3<\/sub>O<sup>1+<\/sup> ] concentration<\/li>\n<li>Writing chemical equations showing how a given oxide or hydroxide exhibits amphoteric behavior<\/li>\n<li>Calculating [H<sub>3<\/sub>O<sup>1+<\/sup> ] concentration from [OH<sup>1\u2013<\/sup>] concentration and vice\u00a0versa using K<sub>w<\/sub> = [H<sub>3<\/sub>O<sup>1+<\/sup>][OH<sup>1\u2013<\/sup>]<\/li>\n<li>Comparing the relative strengths of acids or bases in terms of their dissociation, defining strong and weak\u00a0acids, and being able to cite examples of each<\/li>\n<\/ul>\n<section>\n<h3>Equilibrium<\/h3>\n<p>When working with weak acids, weak bases, or <abbr title=\"The combination of a weak acid or a weak base with its conjugate salt\">buffer<\/abbr> systems we must\u00a0take into account the fact that many species exist in the reaction vessel. Take for example the case of\u00a0calculating the pH of a 0.100 M acetic acid solution whose K<sub>a<\/sub> (acid dissociation constant) is\u00a01.8\u00d710<sup>-5<\/sup>.<\/p>\n<p>Since acetic acid is a weak acid, only part of it dissociates so this becomes an equilibrium problem instead of a\u00a0simple stoichiometry problem (like it would if it were a strong acid).<\/p>\n<p>The reaction is:<\/p>\n<p class=\"center\" style=\"text-align: center;\">HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub> (aq) + H<sub>2<\/sub>O (l) \u2194 H<sub>3<\/sub>O<sup>1+<\/sup>(aq) + C<sub>2<\/sub>H<sub>3<\/sub>O <sub>2<\/sub><sup>1\u2013<\/sup>(aq)<\/p>\n<p>We begin by writing the equilibrium expression:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_003.gif\" width=\"140\" height=\"46\" \/><\/center>In this case:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_004.gif\" width=\"186\" height=\"48\" \/><\/center>Next we need to look at the conditions during the reaction:<\/p>\n<table class=\"gas_law_table\">\n<thead>\n<tr>\n<th><\/th>\n<th>HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><\/th>\n<th>H<sub>3<\/sub>O<sup>1+<\/sup><\/th>\n<th>C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>1-<\/sup><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td valign=\"top\"><strong>Initial <\/strong><\/td>\n<td valign=\"top\">0.100 M<\/td>\n<td valign=\"top\">0<\/td>\n<td valign=\"top\">0<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\"><strong>Change <\/strong>&lt;<\/td>\n<td valign=\"top\">&#8211;<em>x<\/em><\/td>\n<td valign=\"top\">+<em>x<\/em><\/td>\n<td valign=\"top\">+<em>x<\/em><\/td>\n<\/tr>\n<tr>\n<td valign=\"top\"><strong>Equilibrium <\/strong><\/td>\n<td valign=\"top\">0.100 M &#8211; <em>x<\/em><\/td>\n<td valign=\"top\"><em>x<\/em><\/td>\n<td valign=\"top\"><em>x<\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting into the equilibrium expression we get:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_005.gif\" width=\"226\" height=\"44\" \/><\/center>Assuming the effect of <em>x<\/em> on 0.100 M is negligible because of the large difference (power difference &gt;\u00a010<sup>-3<\/sup> units) in concentration and K<sub>a<\/sub>, solving for <em>x<\/em> we get:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_006.gif\" width=\"182\" height=\"50\" \/><\/center>Now we can solve for pH using the equation: pH = -log[H<sub>3<\/sub>O<sup>1+<\/sup>] and find the pH to be\u00a02.87.<\/p>\n<p>Let\u2019s look at another example.<\/p>\n<h4>Sample Problem<\/h4>\n<p>How would we calculate the pH of a 2.00 M ammonia solution with a K<sub>a<\/sub>= 5.71\u00d710<sup>-10<\/sup>\u00a0at 25\u00b0C?<\/p>\n<p>The reaction is NH<sub>3<\/sub><em>aq<\/em>) + H<sub>2<\/sub>O(<em>l<\/em>) \u2194 NH<sub>4<\/sub><sup>1+<\/sup>(<em>aq<\/em>)\u00a0+ OH<sup>1\u2013<\/sup>(<em>aq<\/em>)<\/p>\n<p>Using the equilibrium expression:<\/p>\n<p class=\"center\" style=\"text-align: center;\">K<sub>w<\/sub> = K<sub>a<\/sub>K<sub>b<\/sub> = 1.0\u00d710<sup>\u201314<\/sup><br \/>\nK<sub>b<\/sub> (base dissociation constant) = <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_009.gif\" width=\"30\" height=\"46\" \/>.<\/p>\n<p>Substituting in values we get: K<sub>b<\/sub> = <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_010.gif\" width=\"180\" height=\"44\" \/><\/p>\n<table class=\"gas_law_table\">\n<thead>\n<tr>\n<th><\/th>\n<th>NH<sub>3<\/sub><\/th>\n<th>NH<sub>4<\/sub><sup>1+<\/sup><\/th>\n<th>OH<sup>1-<\/sup><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><strong>Initial<\/strong><\/td>\n<td>2.00 M<\/td>\n<td>0<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td><strong>Change<\/strong><\/td>\n<td>&#8211;<em>x<\/em><\/td>\n<td>+<em>x<\/em><\/td>\n<td>+<em>x<\/em><\/td>\n<\/tr>\n<tr>\n<td><strong>Equilibrium<\/strong><\/td>\n<td>2.00 &#8211; <em>x<\/em><\/td>\n<td><em>x <\/em><\/td>\n<td><em>x <\/em><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Substituting into the equilibrium expression we get:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_011.gif\" width=\"213\" height=\"44\" \/><\/center>Assuming the effect of <em>x<\/em> on 2.00 M is negligible because of the large difference (power difference &gt;\u00a010<sup>-3<\/sup> units) in concentration and K<sub>b<\/sub>, solving for <em>x<\/em> we get:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_012.gif\" width=\"165\" height=\"50\" \/><\/center>Now we can solve for pOH using the equation: pOH = -log[OH<sup>1-<\/sup> ] and find the pOH to be 2.22.<\/p>\n<p class=\"center\" style=\"text-align: center;\">pH + pOH = 14<br \/>\npH = 14 \u2013 2.22<br \/>\npH = 11.78<\/p>\n<h3>Titrations<\/h3>\n<p>A titration is the experimental procedure used to determine the concentration of an unknown acid or base. The\u00a0necessary chemical components for a titration include a <abbr title=\"A solution of known concentration \">standard\u00a0solution<\/abbr>, a solution\u00a0of unknown concentration, and an <abbr title=\"A dye that changes color when the pH level changes\">indicator<\/abbr>.<\/p>\n<p>A titration curve is a way to visually represent the change in pH during a titration. Titrations and titration\u00a0curves tell you many things about a reaction. During a titration, the indicator changes color at the <abbr title=\"The time in a titration when the indicator changes color. The endpoint may or may not occur at the equivalence point.\">endpoint<\/abbr> of the titration. If the best indicator is chosen, the endpoint and the <abbr title=\"The stoichiometric end of a titration, which is the time when there are equal amounts of acid and base \">equivalence\u00a0point<\/abbr> (when there\u00a0are equal amounts of acid and base) occur at nearly the same time.<\/p>\n<p>Common indicators used in acid\/base titrations include phenolphthalein (clear to pink pH = 8-10) and bromothymol\u00a0blue (yellow to blue pH = 6-8). The equivalence point can be visually determined from the titration curve. It is\u00a0the area where the slope of the line approaches infinity (or is nearly vertical).<\/p>\n<table>\n<tbody>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/relationacidsbasessalts4.strong_monoprotic.gif\" width=\"250\" height=\"252\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/relationacidsbasessalts4.weak_monoprotic.gif\" width=\"250\" height=\"250\" \/><\/td>\n<\/tr>\n<tr>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/relationacidsbasessalts4.strong_diprotic.gif\" width=\"250\" height=\"250\" \/><\/td>\n<td><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/relationacidsbasessalts4.weak_diprotic.gif\" width=\"250\" height=\"250\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p class=\"figcaption\">Four titration curves. A monoprotic acid is an acid with only one hydrogen ion to donate. A diprotic acid has two hydrogen ions to donate.<\/p>\n<h3>Buffers and the Henderson-Hasselbalch Equation<\/h3>\n<p>Combinations of weak acids or weak bases with a salt containing a corresponding ion form buffers. Buffers are\u00a0solutions which resist changes in pH when hydrogen or hydroxide ions are added. The amount of acid or base a\u00a0buffer can neutralize before the pH of the system changes to any appreciable degree is the system\u2019s buffer\u00a0capacity. The larger the concentration of the buffer components, the larger the buffer capacity.<\/p>\n<p>The Henderson-Hasselbalch equation can be used to solve for the pH of a buffer system.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_013.gif\" width=\"164\" height=\"46\" align=\"absmiddle\" \/> or <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_014.gif\" width=\"180\" height=\"46\" align=\"absmiddle\" \/><\/p>\n<p class=\"center\">pK<sub>a<\/sub> and pK<sub>b<\/sub> are \u2212log( K<sub>a<\/sub>) and \u2212log( K<sub>b<\/sub>) respectively<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>What is the pH of a solution that contains 3.00 moles of ammonia and 2.50 moles of ammonium chloride in 1.00\u00a0liter of solution? The K<sub>b<\/sub> of ammonia is 1.81 \u00d7 10<sup>-5<\/sup>.<\/p>\n<ol>\n<li>4.66<\/li>\n<li>4.82<\/li>\n<li>9.17<\/li>\n<li>9.34<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is D. Using <img loading=\"lazy\" decoding=\"async\" class=\"no_margin\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_014.gif\" width=\"180\" height=\"46\" align=\"absmiddle\" \/> , we substitute in the given\u00a0information:<\/p>\n<p class=\"center q-reveal\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/gaslaw\/s6_016.gif\" width=\"296\" height=\"114\" \/><br \/>\nSolve for pH using pH + pOH = 14<br \/>\npH = 14 \u2212 pOH<br \/>\npH = 14 \u22124.66314 = 9.34<\/p>\n<\/section>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/the-development-and-theories-of-acids-and-bases\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/gas-laws-and-solutions\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/review-5\">Next Lesson\u00a0\u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/section>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson \u27a1 Relationships Between Acids, Bases, and Salts Objective In this lesson we will examine the relationships between acids, bases, and salts and equilibrium in both experimental situations and common calculations. Previously we covered&#8230; Comparing and contrasting the Arrhenius, Br\u00f8nsted-Lowry, and Lewis definitions of acids and bases Identifying the conjugate acid-base [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-523","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/523","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=523"}],"version-history":[{"count":11,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/523\/revisions"}],"predecessor-version":[{"id":957,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/523\/revisions\/957"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=523"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}