{"id":193392,"date":"2025-02-19T11:15:33","date_gmt":"2025-02-19T11:15:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=193392"},"modified":"2025-02-19T11:15:36","modified_gmt":"2025-02-19T11:15:36","slug":"calculating-percent-dissociation-of-a-weak-acid-a-student-prepares-a-0-49-m-aqueous-solution-of-4-chlorobutanoic-acid-chcicoh","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/19\/calculating-percent-dissociation-of-a-weak-acid-a-student-prepares-a-0-49-m-aqueous-solution-of-4-chlorobutanoic-acid-chcicoh\/","title":{"rendered":"Calculating Percent Dissociation Of A Weak Acid A Student Prepares A 0.49 M Aqueous Solution Of 4-Chlorobutanoic Acid (C,H,CICO,H)"},"content":{"rendered":"\n<p>Calculating Percent Dissociation Of A Weak Acid A Student Prepares A 0.49 M Aqueous Solution Of 4-Chlorobutanoic Acid (C,H,CICO,H). Calculate The Fraction Of 4-Chlorobutanoic Acid That Is In The Dissociated Form In His Solution. Express Your Answer As A Percentage. You Will Probably Find Some Useful Data In The ALEKS Data Resource. Round Your Answer To 2 digits<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To determine the fraction of 4-chlorobutanoic acid dissociated in a 0.49 M aqueous solution, we need to calculate its percent dissociation. This involves understanding the acid dissociation constant (Ka) and applying it to the dissociation equilibrium.<\/p>\n\n\n\n<p><strong>Step 1: Understanding the Acid Dissociation Constant (Ka)<\/strong><\/p>\n\n\n\n<p>The dissociation of 4-chlorobutanoic acid (C\u2084H\u2087ClO\u2082) in water can be represented as:<\/p>\n\n\n\n<p>[ \\text{C\u2084H\u2087ClO\u2082} \\rightleftharpoons \\text{C\u2084H\u2086ClO\u2082}\u207b + \\text{H}\u207a ]<\/p>\n\n\n\n<p>The strength of this acid is quantified by its acid dissociation constant (Ka). A lower pKa value indicates a stronger acid, as it dissociates more readily. For 4-chlorobutanoic acid, the pKa is approximately 4.52. \ue200cite\ue202turn0search7\ue201<\/p>\n\n\n\n<p><strong>Step 2: Calculating the Percent Dissociation<\/strong><\/p>\n\n\n\n<p>The percent dissociation is calculated using the formula:<\/p>\n\n\n\n<p>[ \\text{Percent Dissociation} = \\left( \\frac{[\\text{H}\u207a]}{[\\text{HA}]} \\right) \\times 100 ]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>[H\u207a] is the concentration of hydrogen ions (H\u207a) in the solution.<\/li>\n\n\n\n<li>[HA] is the initial concentration of the acid.<\/li>\n<\/ul>\n\n\n\n<p>Given that the pKa is 4.52, we can calculate the Ka:<\/p>\n\n\n\n<p>[ \\text{Ka} = 10^{-\\text{pKa}} = 10^{-4.52} \\approx 3.02 \\times 10^{-5} ]<\/p>\n\n\n\n<p>For a weak acid, the dissociation is minimal, so we can assume that the concentration of dissociated acid ([H\u207a]) is much smaller than the initial concentration of the acid. Therefore, we can approximate the percent dissociation as:<\/p>\n\n\n\n<p>[ \\text{Percent Dissociation} \\approx \\left( \\frac{[\\text{H}\u207a]}{[\\text{HA}]} \\right) \\times 100 ]<\/p>\n\n\n\n<p>Substituting the given concentration:<\/p>\n\n\n\n<p>[ \\text{Percent Dissociation} \\approx \\left( \\frac{0.49}{0.49} \\right) \\times 100 = 100\\% ]<\/p>\n\n\n\n<p>Therefore, the fraction of 4-chlorobutanoic acid that is dissociated in the solution is approximately 100%.<\/p>\n\n\n\n<p><strong>Step 3: Conclusion<\/strong><\/p>\n\n\n\n<p>In a 0.49 M aqueous solution of 4-chlorobutanoic acid, the acid dissociates to a significant extent, resulting in a high concentration of hydrogen ions and a low pH. This indicates that the acid is relatively strong in this concentration.<\/p>\n\n\n\n<p><em>Note: The calculation assumes ideal behavior and does not account for activity coefficients or other real-world factors that might slightly alter the dissociation in a concentrated solution.<\/em><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculating Percent Dissociation Of A Weak Acid A Student Prepares A 0.49 M Aqueous Solution Of 4-Chlorobutanoic Acid (C,H,CICO,H). Calculate The Fraction Of 4-Chlorobutanoic Acid That Is In The Dissociated Form In His Solution. Express Your Answer As A Percentage. You Will Probably Find Some Useful Data In The ALEKS Data Resource. Round Your Answer [&hellip;]<\/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-193392","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193392","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=193392"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193392\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193392"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193392"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193392"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}