{"id":248203,"date":"2025-07-08T08:44:14","date_gmt":"2025-07-08T08:44:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248203"},"modified":"2025-07-08T08:44:16","modified_gmt":"2025-07-08T08:44:16","slug":"carbonic-acid-h2co3-is-a-diprotic-acid-with-ka14-3x10%e2%88%927-ka25-6x10%e2%88%9211","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/carbonic-acid-h2co3-is-a-diprotic-acid-with-ka14-3x10%e2%88%927-ka25-6x10%e2%88%9211\/","title":{"rendered":"Carbonic acid (H2CO3) is a diprotic acid with\u00a0Ka1=4.3\u00d710\u22127;\u00a0Ka2=5.6\u00d710\u221211."},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\">) Carbonic acid (H2CO3) is a diprotic acid with&nbsp;Ka1=4.3\u00d710\u22127;&nbsp;Ka2=5.6\u00d710\u221211.<\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-295.png\" alt=\"\" class=\"wp-image-248204\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here is the correct solution to the problem presented in the image.<\/p>\n\n\n\n<p><strong>A) Write the reactions for the stepwise dissociation of carbonic acid.<\/strong><br>The handwritten reactions are correct. Carbonic acid (H\u2082CO\u2083) is a diprotic acid, meaning it can donate two protons in a stepwise manner.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>First dissociation:<\/strong><br>H\u2082CO\u2083(aq) + H\u2082O(l) \u21cc H\u2083O\u207a(aq) + HCO\u2083\u207b(aq)<br>This step is associated with the first acid dissociation constant, Ka\u2081 = 4.3 x 10\u207b\u2077.<\/li>\n\n\n\n<li><strong>Second dissociation:<\/strong><br>HCO\u2083\u207b(aq) + H\u2082O(l) \u21cc H\u2083O\u207a(aq) + CO\u2083\u00b2\u207b(aq)<br>This step is associated with the second acid dissociation constant, Ka\u2082 = 5.6 x 10\u207b\u00b9\u00b9.<\/li>\n<\/ul>\n\n\n\n<p><strong>B) Determine the pH of a 0.0053 M solution of carbonic acid.<\/strong><br>The correct pH is&nbsp;<strong>4.32<\/strong>.<\/p>\n\n\n\n<p>The handwritten calculation is incorrect. The calculation&nbsp;-log(4.3 x 10\u207b\u2077) = 6.37&nbsp;determines the pKa\u2081 of carbonic acid, not the pH of the solution. The pH depends on both the acid&#8217;s strength (Ka) and its initial concentration.<\/p>\n\n\n\n<p>Here is the correct method to determine the pH:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Analyze the Dissociation Steps:<\/strong>\u00a0To find the pH, we need the concentration of hydronium ions, [H\u2083O\u207a], at equilibrium. Since carbonic acid is diprotic, both dissociation reactions produce H\u2083O\u207a. However, because Ka\u2081 (4.3 x 10\u207b\u2077) is much larger than Ka\u2082 (5.6 x 10\u207b\u00b9\u00b9), the first dissociation is the dominant source of H\u2083O\u207a ions. The contribution from the second dissociation is negligible and can be ignored for the initial pH calculation.<\/li>\n\n\n\n<li><strong>Use an ICE Table for the First Dissociation:<\/strong>\u00a0We set up an ICE (Initial, Change, Equilibrium) table for the first dissociation to find the equilibrium concentrations. Let &#8216;x&#8217; be the change in concentration.ReactionH\u2082CO\u2083H\u2083O\u207aHCO\u2083\u207b<strong>Initial (I)<\/strong>0.0053 M~0 M0 M<strong>Change (C)<\/strong>-x+x+x<strong>Equilibrium (E)<\/strong>0.0053 &#8211; xxx<\/li>\n\n\n\n<li><strong>Use the Ka\u2081 Expression:<\/strong>\u00a0Now, we write the expression for Ka\u2081 and substitute the equilibrium values.Ka\u2081 = [H\u2083O\u207a][HCO\u2083\u207b] \/ [H\u2082CO\u2083]<br>4.3 x 10\u207b\u2077 = (x)(x) \/ (0.0053 &#8211; x)<\/li>\n\n\n\n<li><strong>Simplify and Solve for x:<\/strong>\u00a0Because Ka\u2081 is very small compared to the initial concentration, we can assume that &#8216;x&#8217; is negligible compared to 0.0053 M (i.e., 0.0053 &#8211; x \u2248 0.0053). This assumption is valid if the initial concentration divided by Ka\u2081 is greater than 100. Here, 0.0053 \/ (4.3 x 10\u207b\u2077) is approximately 12,325, so the approximation is justified.4.3 x 10\u207b\u2077 \u2248 x\u00b2 \/ 0.0053<br>x\u00b2 = (4.3 x 10\u207b\u2077) * (0.0053)<br>x\u00b2 = 2.279 x 10\u207b\u2079<br>x = \u221a(2.279 x 10\u207b\u2079)<br>x \u2248 4.77 x 10\u207b\u2075 M<\/li>\n\n\n\n<li><strong>Calculate the pH:<\/strong>\u00a0The value of x represents the equilibrium concentration of H\u2083O\u207a.[H\u2083O\u207a] = 4.77 x 10\u207b\u2075 M<br>pH = -log[H\u2083O\u207a]<br>pH = -log(4.77 x 10\u207b\u2075)<br>pH \u2248 4.32<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-175.jpeg\" alt=\"\" class=\"wp-image-248205\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>) Carbonic acid (H2CO3) is a diprotic acid with&nbsp;Ka1=4.3\u00d710\u22127;&nbsp;Ka2=5.6\u00d710\u221211. The Correct Answer and Explanation is: Here is the correct solution to the problem presented in the image. A) Write the reactions for the stepwise dissociation of carbonic acid.The handwritten reactions are correct. Carbonic acid (H\u2082CO\u2083) is a diprotic acid, meaning it can donate two protons [&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-248203","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248203","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=248203"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248203\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248203"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248203"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248203"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}