{"id":219030,"date":"2025-05-25T03:55:25","date_gmt":"2025-05-25T03:55:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219030"},"modified":"2025-05-25T03:55:27","modified_gmt":"2025-05-25T03:55:27","slug":"enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-31","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/25\/enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-31\/","title":{"rendered":"Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution."},"content":{"rendered":"\n<p>Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.<\/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>Let&#8217;s analyze the problem step-by-step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Concentration of monoprotic acid, [HA]0=0.0192\u2009M[HA]_0 = 0.0192 \\, M<\/li>\n\n\n\n<li>pH of solution = 2.53<\/li>\n\n\n\n<li>Acid is monoprotic (releases 1 H+^+ per molecule)<\/li>\n\n\n\n<li>Find: KaK_a (acid dissociation constant)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate [H+][H^+] from pH<\/h3>\n\n\n\n<p>pH=\u2212log\u2061[H+]\\text{pH} = -\\log [H^+] [H+]=10\u2212pH=10\u22122.53\u22482.95\u00d710\u22123\u2009M[H^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\, M<\/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: Set up the dissociation equilibrium<\/h3>\n\n\n\n<p>The acid HAHA dissociates as: HA\u21ccH++A\u2212HA \\rightleftharpoons H^+ + A^-<\/p>\n\n\n\n<p>Initial concentration of acid = 0.0192 M<br>Let xx = concentration of H+H^+ and A\u2212A^- formed at equilibrium = 2.95\u00d710\u22123\u2009M2.95 \\times 10^{-3} \\, M (from pH)<\/p>\n\n\n\n<p>At equilibrium:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>[HA]=0.0192\u2212x=0.0192\u22120.00295=0.01625\u2009M[HA] = 0.0192 &#8211; x = 0.0192 &#8211; 0.00295 = 0.01625\\, M<\/li>\n\n\n\n<li>[H+]=x=0.00295\u2009M[H^+] = x = 0.00295\\, M<\/li>\n\n\n\n<li>[A\u2212]=x=0.00295\u2009M[A^-] = x = 0.00295\\, M<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Write expression for KaK_a<\/h3>\n\n\n\n<p>Ka=[H+][A\u2212][HA]=x\u00d7x0.0192\u2212x=x20.0192\u2212xK_a = \\frac{[H^+][A^-]}{[HA]} = \\frac{x \\times x}{0.0192 &#8211; x} = \\frac{x^2}{0.0192 &#8211; x}<\/p>\n\n\n\n<p>Plug in x=0.00295x = 0.00295: Ka=(0.00295)20.0192\u22120.00295=8.7\u00d710\u221260.01625\u22485.35\u00d710\u22124K_a = \\frac{(0.00295)^2}{0.0192 &#8211; 0.00295} = \\frac{8.7 \\times 10^{-6}}{0.01625} \\approx 5.35 \\times 10^{-4}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>Ka\u22485.35\u00d710\u22124\\boxed{K_a \\approx 5.35 \\times 10^{-4}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Explanation <\/h2>\n\n\n\n<p>The acid dissociation constant, KaK_a, measures the strength of an acid in solution by indicating the degree to which the acid dissociates into its ions. For a monoprotic acid, this involves the equilibrium between the undissociated acid HAHA and the ions H+H^+ and A\u2212A^-.<\/p>\n\n\n\n<p>Given the molar concentration of the acid and the pH of the solution, we can first calculate the concentration of hydrogen ions, [H+][H^+], using the pH formula: [H+]=10\u2212pH[H^+] = 10^{-\\text{pH}}<\/p>\n\n\n\n<p>This gives the concentration of H+H^+ in the solution, which equals the concentration of A\u2212A^- produced, assuming the acid dissociates completely into one proton and one conjugate base ion.<\/p>\n\n\n\n<p>Next, using the initial concentration of the acid and the equilibrium concentration of ions, we can calculate the remaining concentration of undissociated acid at equilibrium. This is essential because KaK_a depends on the ratio of concentrations of dissociated and undissociated species.<\/p>\n\n\n\n<p>Substituting these values into the equilibrium expression: Ka=[H+][A\u2212][HA]K_a = \\frac{[H^+][A^-]}{[HA]}<\/p>\n\n\n\n<p>allows us to find the acid dissociation constant. In this problem, the relatively low KaK_a value of 5.35\u00d710\u221245.35 \\times 10^{-4} indicates a weak acid since the acid only partially dissociates in water.<\/p>\n\n\n\n<p>This method of calculating KaK_a is standard in acid-base chemistry and is crucial for understanding the strength and behavior of acids in aqueous solutions. It combines concepts of molarity, pH, and chemical equilibrium to provide quantitative insight into acid 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\/image-418.png\" alt=\"\" class=\"wp-image-219031\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid. The Correct Answer and Explanation is: Let&#8217;s analyze the problem step-by-step. Given: Step 1: Calculate [H+][H^+] from pH pH=\u2212log\u2061[H+]\\text{pH} = -\\log [H^+] [H+]=10\u2212pH=10\u22122.53\u22482.95\u00d710\u22123\u2009M[H^+] = 10^{-\\text{pH}} = [&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-219030","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219030","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=219030"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219030\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219030"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219030"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219030"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}