{"id":206652,"date":"2025-06-16T09:47:47","date_gmt":"2025-06-16T09:47:47","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=206652"},"modified":"2025-06-16T09:48:17","modified_gmt":"2025-06-16T09:48:17","slug":"enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-32","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/16\/enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-32\/","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><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 find the acid dissociation constant KaK_aKa\u200b of a <strong>monoprotic acid<\/strong>, we use the information provided:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Concentration of acid solution, [HA] = 0.0192 M<\/strong><\/li>\n\n\n\n<li><strong>pH = 2.53<\/strong><\/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\"><strong>Step 1: Find the [H+][H^+][H+] concentration<\/strong><\/h3>\n\n\n\n<p>Use the pH equation: pH=\u2212log\u2061[H+]\\text{pH} = -\\log[H^+]pH=\u2212log[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} \\, \\text{M}[H+]=10\u2212pH=10\u22122.53\u22482.95\u00d710\u22123M<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Set up the ICE table for a monoprotic acid<\/strong><\/h3>\n\n\n\n<p>HA\u21ccH++A\u2212\\text{HA} \\rightleftharpoons \\text{H}^+ + \\text{A}^-HA\u21ccH++A\u2212<\/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>HA<\/td><td>0.0192<\/td><td>\u2212x-x\u2212x<\/td><td>0.0192\u2212x0.0192 &#8211; x0.0192\u2212x<\/td><\/tr><tr><td>H\u207a<\/td><td>0<\/td><td>+x+x+x<\/td><td>x=[H+]=2.95\u00d710\u22123x = [H^+] = 2.95 \\times 10^{-3}x=[H+]=2.95\u00d710\u22123<\/td><\/tr><tr><td>A\u207b<\/td><td>0<\/td><td>+x+x+x<\/td><td>x=2.95\u00d710\u22123x = 2.95 \\times 10^{-3}x=2.95\u00d710\u22123<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Plug into the Ka expression<\/strong><\/h3>\n\n\n\n<p>Ka=[H+][A\u2212][HA]=x2[HA]\u2212xK_a = \\frac{[H^+][A^-]}{[HA]} = \\frac{x^2}{[HA] &#8211; x}Ka\u200b=[HA][H+][A\u2212]\u200b=[HA]\u2212xx2\u200b Ka=(2.95\u00d710\u22123)20.0192\u22122.95\u00d710\u22123=8.70\u00d710\u221260.01625K_a = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 &#8211; 2.95 \\times 10^{-3}} = \\frac{8.70 \\times 10^{-6}}{0.01625}Ka\u200b=0.0192\u22122.95\u00d710\u22123(2.95\u00d710\u22123)2\u200b=0.016258.70\u00d710\u22126\u200b Ka\u22485.35\u00d710\u22124K_a \\approx 5.35 \\times 10^{-4}Ka\u200b\u22485.35\u00d710\u22124<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>Ka=5.35\u00d710\u22124\\boxed{K_a = 5.35 \\times 10^{-4}}Ka\u200b=5.35\u00d710\u22124\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udca1 <strong>Explanation (300 words)<\/strong><\/h3>\n\n\n\n<p>In acid-base chemistry, the acid dissociation constant KaK_aKa\u200b measures the strength of a weak acid in water. For a <strong>monoprotic acid<\/strong>, which donates only one proton (H\u207a) per molecule, the dissociation follows: HA\u21ccH++A\u2212\\text{HA} \\rightleftharpoons \\text{H}^+ + \\text{A}^-HA\u21ccH++A\u2212<\/p>\n\n\n\n<p>The extent to which this acid dissociates determines the concentration of hydrogen ions and, hence, the <strong>pH<\/strong> of the solution. In this problem, we were given the <strong>pH (2.53)<\/strong> and the <strong>initial concentration of the acid (0.0192 M)<\/strong>, but not directly the KaK_aKa\u200b.<\/p>\n\n\n\n<p>To find it, we first calculated the hydrogen ion concentration using the inverse log of the pH. This yielded [H+]=2.95\u00d710\u22123\u2009M[H^+] = 2.95 \\times 10^{-3} \\, \\text{M}[H+]=2.95\u00d710\u22123M. Assuming all of this came from the dissociation of HA, we said that the amount of HA dissociated is also x=[H+]x = [H^+]x=[H+].<\/p>\n\n\n\n<p>Then, using the expression for Ka=[H+][A\u2212][HA]K_a = \\frac{[H^+][A^-]}{[HA]}Ka\u200b=[HA][H+][A\u2212]\u200b, we substituted the values from the ICE table. Since HA loses a proton to form H\u207a and A\u207b in equal amounts, we set both product concentrations equal to xxx. The remaining undissociated HA was the initial amount minus xxx.<\/p>\n\n\n\n<p>Solving the equation gave us Ka=5.35\u00d710\u22124K_a = 5.35 \\times 10^{-4}Ka\u200b=5.35\u00d710\u22124, confirming that the acid is weak (since strong acids fully dissociate and do not have a measurable Ka in dilute aqueous solution).<\/p>\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: To find the acid dissociation constant KaK_aKa\u200b of a monoprotic acid, we use the information provided: Step 1: Find the [&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-206652","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206652","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=206652"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206652\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=206652"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=206652"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=206652"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}