{"id":231198,"date":"2025-06-10T15:37:51","date_gmt":"2025-06-10T15:37:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231198"},"modified":"2025-06-10T15:37:53","modified_gmt":"2025-06-10T15:37:53","slug":"the-substance-benzoic-acid-c6h5cooh-is-a-weak-acid-ka-6-3a-10-5","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/10\/the-substance-benzoic-acid-c6h5cooh-is-a-weak-acid-ka-6-3a-10-5\/","title":{"rendered":"The substance benzoic acid (C6H5COOH) is a weak acid (Ka = 6.3\u00c3\u201410^-5)."},"content":{"rendered":"\n

The substance benzoic acid (C6H5COOH) is a weak acid (Ka = 6.3\u00c3\u201410^-5). What is the pH of a 5.09\u00c3\u201410^-2 M aqueous solution of potassium benzoate, KC6H5COO<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Answer:<\/strong> The pH of a <\/p>\n\n\n\n

5.09\u00d710\u221225.09\u00d710\u22122<\/code><\/pre>\n\n\n\n
 M aqueous solution of potassium benzoate is 8.45<\/strong>.<\/p>\n\n\n\n
Explanation<\/h3>\n\n\n\n
This problem requires the determination of the pH of a salt solution. The salt in question, potassium benzoate (<\/p>\n\n\n\n
KC6H5COOK<\/em>C<\/em>6\u200bH<\/em>5\u200bCOO<\/em><\/code><\/pre>\n\n\n\n
), is formed from a strong base (potassium hydroxide, KOH) and a weak acid (benzoic acid, <\/p>\n\n\n\n
C6H5COOHC<\/em>6\u200bH<\/em>5\u200bCOO<\/em>H<\/em><\/code><\/pre>\n\n\n\n
). When dissolved in water, the salt dissociates completely into its constituent ions:<\/p>\n\n\n\n
KC6H5COO(aq)\u2192K+(aq)+C6H5COO\u2212(aq)K<\/em>C<\/em>6\u200bH<\/em>5\u200bCOO<\/em>(a<\/em>q<\/em>)\u2192K<\/em>+(a<\/em>q<\/em>)+C<\/em>6\u200bH<\/em>5\u200bCO<\/em>O<\/em>\u2212(a<\/em>q<\/em>)<\/code><\/pre>\n\n\n\n
The potassium ion (<\/p>\n\n\n\n
K+K<\/em>+<\/code><\/pre>\n\n\n\n
) is the conjugate acid of a strong base and is thus a spectator ion that does not react with water. However, the benzoate ion (<\/p>\n\n\n\n
C6H5COO\u2212C<\/em>6\u200bH<\/em>5\u200bCO<\/em>O<\/em>\u2212<\/code><\/pre>\n\n\n\n
) is the conjugate base of a weak acid. It will react with water in a hydrolysis reaction, accepting a proton and producing hydroxide ions (<\/p>\n\n\n\n
OH\u2212O<\/em>H<\/em>\u2212<\/code><\/pre>\n\n\n\n
), making the solution basic.<\/p>\n\n\n\n
C6H5COO\u2212(aq)+H2O(l)\u21ccC6H5COOH(aq)+OH\u2212(aq)C<\/em>6\u200bH<\/em>5\u200bCO<\/em>O<\/em>\u2212(a<\/em>q<\/em>)+H<\/em>2\u200bO<\/em>(l<\/em>)\u21ccC<\/em>6\u200bH<\/em>5\u200bCOO<\/em>H<\/em>(a<\/em>q<\/em>)+O<\/em>H<\/em>\u2212(a<\/em>q<\/em>)<\/code><\/pre>\n\n\n\n
To calculate the concentration of <\/p>\n\n\n\n
OH\u2212O<\/em>H<\/em>\u2212<\/code><\/pre>\n\n\n\n
 at equilibrium, the base dissociation constant (<\/p>\n\n\n\n
KbK<\/em>b<\/em>\u200b<\/code><\/pre>\n\n\n\n
) for the benzoate ion is needed. It can be calculated from the acid dissociation constant (<\/p>\n\n\n\n
KaK<\/em>a<\/em>\u200b<\/code><\/pre>\n\n\n\n
) of its conjugate acid, benzoic acid, using the ion-product constant for water, <\/p>\n\n\n\n
Kw=1.0\u00d710\u221214K<\/em>w<\/em>\u200b=1.0\u00d710\u221214<\/code><\/pre>\n\n\n\n
.<\/p>\n\n\n\n
Kb=KwKa=1.0\u00d710\u2212146.3\u00d710\u22125=1.59\u00d710\u221210K<\/em>b<\/em>\u200b=K<\/em>a<\/em>\u200bK<\/em>w<\/em>\u200b\u200b=6.3\u00d710\u221251.0\u00d710\u221214\u200b=1.59\u00d710\u221210<\/code><\/pre>\n\n\n\n
An ICE table can be used to find the equilibrium concentration of <\/p>\n\n\n\n
OH\u2212O<\/em>H<\/em>\u2212<\/code><\/pre>\n\n\n\n
. The initial concentration of the benzoate ion is <\/p>\n\n\n\n
5.09\u00d710\u221225.09\u00d710\u22122<\/code><\/pre>\n\n\n\n
 M.<\/p>\n\n\n\n
<\/td>C6H5COO\u2212C<\/em>6\u200bH<\/em>5\u200bCO<\/em>O<\/em>\u2212<\/code><\/td>C6H5COOHC<\/em>6\u200bH<\/em>5\u200bCOO<\/em>H<\/em><\/code><\/td>OH\u2212O<\/em>H<\/em>\u2212<\/code><\/td><\/tr>
Initial<\/strong><\/td>5.09\u00d710\u221225.09\u00d710\u22122<\/code><\/td>0<\/td>0<\/td><\/tr>
Change<\/strong><\/td>-x<\/td>+x<\/td>+x<\/td><\/tr>
Equilibrium<\/strong><\/td>(5.09\u00d710\u22122\u2212x)(5.09\u00d710\u22122\u2212x<\/em>)<\/code><\/td>x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
The <\/p>\n\n\n\n
KbK<\/em>b<\/em>\u200b<\/code><\/pre>\n\n\n\n
 expression is:<\/p>\n\n\n\n
Kb=[C6H5COOH][OH\u2212][C6H5COO\u2212]=(x)(x)(5.09\u00d710\u22122\u2212x)K<\/em>b<\/em>\u200b=[C<\/em>6\u200bH<\/em>5\u200bCO<\/em>O<\/em>\u2212][C<\/em>6\u200bH<\/em>5\u200bCOO<\/em>H<\/em>][O<\/em>H<\/em>\u2212]\u200b=(5.09\u00d710\u22122\u2212x<\/em>)(x<\/em>)(x<\/em>)\u200b<\/code><\/pre>\n\n\n\n
Since <\/p>\n\n\n\n
KbK<\/em>b<\/em>\u200b<\/code><\/pre>\n\n\n\n
 is very small, we can assume <\/p>\n\n\n\n
xx<\/em><\/code><\/pre>\n\n\n\n
 is negligible compared to the initial concentration, simplifying the equation to:<\/p>\n\n\n\n
1.59\u00d710\u221210\u2248x25.09\u00d710\u221221.59\u00d710\u221210\u22485.09\u00d710\u22122x<\/em>2\u200b<\/code><\/pre>\n\n\n\n
x2=(1.59\u00d710\u221210)(5.09\u00d710\u22122)=8.09\u00d710\u221212x<\/em>2=(1.59\u00d710\u221210)(5.09\u00d710\u22122)=8.09\u00d710\u221212<\/code><\/pre>\n\n\n\n
x=[OH\u2212]=8.09\u00d710\u221212=2.85\u00d710\u22126 Mx<\/em>=[O<\/em>H<\/em>\u2212]=8.09\u00d710\u221212\u200b=2.85\u00d710\u22126 M<\/code><\/pre>\n\n\n\n
Now, the pOH can be calculated, followed by the pH.<\/p>\n\n\n\n
pOH=\u2212log\u2061[OH\u2212]=\u2212log\u2061(2.85\u00d710\u22126)=5.55pO<\/em>H<\/em>=\u2212log[O<\/em>H<\/em>\u2212]=\u2212log(2.85\u00d710\u22126)=5.55<\/code><\/pre>\n\n\n\n
pH=14.00\u2212pOH=14.00\u22125.55=8.45pH<\/em>=14.00\u2212pOH<\/em>=14.00\u22125.55=8.45<\/code><\/pre>\n\n\n\n
\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
The substance benzoic acid (C6H5COOH) is a weak acid (Ka = 6.3\u00c3\u201410^-5). What is the pH of a 5.09\u00c3\u201410^-2 M aqueous solution of potassium benzoate, KC6H5COO The Correct Answer and Explanation is: Answer: The pH of a   M aqueous solution of potassium benzoate is 8.45. Explanation This problem requires the determination of the pH of a salt […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-231198","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231198","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=231198"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231198\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231198"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231198"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231198"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}