{"id":221587,"date":"2025-05-29T20:05:44","date_gmt":"2025-05-29T20:05:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221587"},"modified":"2025-05-29T20:05:46","modified_gmt":"2025-05-29T20:05:46","slug":"a-diprotic-acid-h2a-has-ka1-3-4-x-10-4-and-ka2-6-7-x-10-9","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/a-diprotic-acid-h2a-has-ka1-3-4-x-10-4-and-ka2-6-7-x-10-9\/","title":{"rendered":"A diprotic acid, H2A, has Ka1 = 3.4 x 10-4 and Ka2 = 6.7 x 10-9."},"content":{"rendered":"\n<p>A diprotic acid, H2A, has Ka1 = 3.4 x 10-4 and Ka2 = 6.7 x 10-9. What is the pH of a 0.36 M solution of H2A?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To find the <strong>pH of a 0.36 M solution of the diprotic acid H\u2082A<\/strong>, we need to account for the <strong>stepwise dissociation<\/strong> of the acid and the relative strength of each dissociation step:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Stepwise dissociation:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>H\u2082A \u21cc H\u207a + HA\u207b<\/strong>\u2003\u2003\u2003\u2003Ka\u2081 = 3.4 \u00d7 10\u207b\u2074<\/li>\n\n\n\n<li><strong>HA\u207b \u21cc H\u207a + A\u00b2\u207b<\/strong>\u2003\u2003\u2003\u2003Ka\u2082 = 6.7 \u00d7 10\u207b\u2079<\/li>\n<\/ol>\n\n\n\n<p>Since <strong>Ka\u2081 \u226b Ka\u2082<\/strong>, the <strong>first dissociation<\/strong> contributes the <strong>majority of H\u207a ions<\/strong>, while the second contributes very little and can be initially neglected in calculating the pH.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: First dissociation \u2013 ICE Table<\/h3>\n\n\n\n<p>Let the initial concentration of H\u2082A be <strong>0.36 M<\/strong>. Let <strong>x<\/strong> be the amount that dissociates:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Species<\/th><th>Initial (M)<\/th><th>Change (M)<\/th><th>Equilibrium (M)<\/th><\/tr><\/thead><tbody><tr><td>H\u2082A<\/td><td>0.36<\/td><td>-x<\/td><td>0.36 &#8211; x<\/td><\/tr><tr><td>H\u207a<\/td><td>0<\/td><td>+x<\/td><td>x<\/td><\/tr><tr><td>HA\u207b<\/td><td>0<\/td><td>+x<\/td><td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Ka\u2081 = x\u00b2 \/ (0.36 &#8211; x) \u2248 x\u00b2 \/ 0.36 (since x \u226a 0.36)<\/p>\n\n\n\n<p><strong>3.4 \u00d7 10\u207b\u2074 = x\u00b2 \/ 0.36<\/strong><br>\u27f9 <strong>x\u00b2 = (3.4 \u00d7 10\u207b\u2074)(0.36) = 1.224 \u00d7 10\u207b\u2074<\/strong><br>\u27f9 <strong>x \u2248 \u221a(1.224 \u00d7 10\u207b\u2074) \u2248 0.0111 M<\/strong><\/p>\n\n\n\n<p>This x is the concentration of H\u207a contributed by the first dissociation.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Check second dissociation<\/h3>\n\n\n\n<p>The second dissociation (Ka\u2082 = 6.7 \u00d7 10\u207b\u2079) is very weak. The amount of additional H\u207a is negligible compared to 0.0111 M, so we can safely ignore it for pH calculation.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate pH<\/h3>\n\n\n\n<p>pH=\u2212log\u2061[H+]=\u2212log\u2061(0.0111)\u22481.95\\text{pH} = -\\log[H\u207a] = -\\log(0.0111) \\approx 1.95<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer: <strong>pH \u2248 1.95<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>A diprotic acid like H\u2082A donates two protons in solution, but not all at once\u2014the dissociation occurs in two steps, each with its own acid dissociation constant (Ka). In this case, Ka\u2081 = 3.4 \u00d7 10\u207b\u2074 and Ka\u2082 = 6.7 \u00d7 10\u207b\u2079. These values tell us that the first proton is much more easily donated than the second. The strength of Ka\u2081 implies a moderate acid, whereas Ka\u2082 indicates a very weak acid.<\/p>\n\n\n\n<p>To calculate pH, we assess how much H\u207a is produced in the first dissociation step. Setting up an ICE table (Initial, Change, Equilibrium) helps us model the concentrations as the acid dissociates. By solving the expression for Ka\u2081, we find that the concentration of H\u207a at equilibrium is about 0.0111 M.<\/p>\n\n\n\n<p>The second step does generate more H\u207a ions, but the extremely small Ka\u2082 means that the contribution is negligible in comparison to the first dissociation. Mathematically, it would contribute less than 0.00001 M of H\u207a, a drop in the ocean compared to 0.0111 M.<\/p>\n\n\n\n<p>Finally, we apply the pH formula: pH = \u2013log[H\u207a]. Using our calculated [H\u207a] from the first step, the pH comes out to be approximately 1.95. This result aligns with expectations for a moderately strong diprotic acid where the first dissociation dominates the solution\u2019s acidity.<\/p>\n\n\n\n<p>In summary, the key to solving diprotic acid problems lies in comparing Ka values. If the first Ka is significantly larger than the second, it&#8217;s safe to approximate pH based on only the first dissociation.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner6-145.jpeg\" alt=\"\" class=\"wp-image-221588\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A diprotic acid, H2A, has Ka1 = 3.4 x 10-4 and Ka2 = 6.7 x 10-9. What is the pH of a 0.36 M solution of H2A? The Correct Answer and Explanation is: To find the pH of a 0.36 M solution of the diprotic acid H\u2082A, we need to account for the stepwise dissociation [&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-221587","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221587","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=221587"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221587\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221587"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221587"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221587"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}