{"id":216694,"date":"2025-05-20T07:58:37","date_gmt":"2025-05-20T07:58:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=216694"},"modified":"2025-05-20T07:58:39","modified_gmt":"2025-05-20T07:58:39","slug":"enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-26","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/20\/enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-26\/","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-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s analyze the problem carefully and calculate the acid dissociation constant KaK_a for the monoprotic acid.<\/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>Molarity of the acid solution, [HA]0=0.0192\u2009M[HA]_0 = 0.0192 \\, M<\/li>\n\n\n\n<li>pH of the solution = 2.53<\/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: Understand what is being asked<\/h3>\n\n\n\n<p>We want to find the acid dissociation constant KaK_a for the acid HAHA. This constant describes the equilibrium between the acid and its ions in water: HA\u21ccH++A\u2212HA \\rightleftharpoons H^+ + A^-<\/p>\n\n\n\n<p>Where HAHA is the monoprotic acid, H+H^+ is the hydrogen ion, and A\u2212A^- is the conjugate base.<\/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: Calculate the concentration of H+H^+<\/h3>\n\n\n\n<p>pH is defined as: pH=\u2212log\u2061[H+]pH = -\\log[H^+]<\/p>\n\n\n\n<p>So, [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<p>This means the concentration of H+H^+ ions in solution is approximately 2.95\u00d710\u22123\u2009M2.95 \\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 3: Set up the ICE table for the dissociation<\/h3>\n\n\n\n<p>Let:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial concentration of acid = 0.0192\u2009M0.0192 \\, M<\/li>\n\n\n\n<li>Change in concentration = \u2212x-x for HAHA, and +x+x for H+H^+ and A\u2212A^-<\/li>\n<\/ul>\n\n\n\n<p>At equilibrium: [HA]=0.0192\u2212x\u22480.0192\u22122.95\u00d710\u22123[HA] = 0.0192 &#8211; x \\approx 0.0192 &#8211; 2.95 \\times 10^{-3} [H+]=x=2.95\u00d710\u22123\u2009M[H^+] = x = 2.95 \\times 10^{-3} \\, M [A\u2212]=x=2.95\u00d710\u22123\u2009M[A^-] = x = 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 4: Write the expression for KaK_a<\/h3>\n\n\n\n<p>Ka=[H+][A\u2212][HA]=x\u22c5x0.0192\u2212x=x20.0192\u2212xK_a = \\frac{[H^+][A^-]}{[HA]} = \\frac{x \\cdot x}{0.0192 &#8211; x} = \\frac{x^2}{0.0192 &#8211; x}<\/p>\n\n\n\n<p>Substitute x=2.95\u00d710\u22123x = 2.95 \\times 10^{-3}: Ka=(2.95\u00d710\u22123)20.0192\u22122.95\u00d710\u22123=8.70\u00d710\u221260.01625\u22485.35\u00d710\u22124K_a = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 &#8211; 2.95 \\times 10^{-3}} = \\frac{8.70 \\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\"><strong>Final answer:<\/strong><\/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<h3 class=\"wp-block-heading\">Explanation (300 words):<\/h3>\n\n\n\n<p>The problem involves calculating the acid dissociation constant KaK_a for a monoprotic acid given the solution\u2019s molarity and its pH. A monoprotic acid is one that donates a single proton (H+)(H^+) when it dissociates in water. The KaK_a quantifies the strength of this acid by measuring the extent to which it dissociates into ions.<\/p>\n\n\n\n<p>We begin by converting the pH value to the hydrogen ion concentration. pH is defined as the negative logarithm of the hydrogen ion concentration, so by raising 10 to the negative power of the pH, we find the concentration of H+H^+ ions in the solution. This is crucial because the amount of H+H^+ ions produced is directly linked to how much of the acid dissociates.<\/p>\n\n\n\n<p>Using the initial molarity of the acid and the equilibrium concentration of hydrogen ions, we set up an ICE (Initial, Change, Equilibrium) table to express the dissociation equilibrium. The change in concentration of the acid is equal to the amount dissociated, which equals the concentration of H+H^+ ions formed. Since the acid is monoprotic, the concentration of the conjugate base A\u2212A^- formed is the same as that of H+H^+.<\/p>\n\n\n\n<p>The KaK_a expression comes from the law of mass action: it\u2019s the ratio of the products\u2019 concentrations (hydrogen ion and conjugate base) to the reactant (undissociated acid). Substituting the known values yields the KaK_a, which is a measure of acid strength \u2014 smaller values of KaK_a correspond to weaker acids.<\/p>\n\n\n\n<p>In this problem, the KaK_a calculated as approximately 5.35\u00d710\u221245.35 \\times 10^{-4} indicates a weak acid that does not fully dissociate in solution but produces enough H+H^+ ions to make the pH 2.53. This method is common in chemistry to characterize acids and understand their behavior in aqueous solutions.<\/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: Let&#8217;s analyze the problem carefully and calculate the acid dissociation constant KaK_a for the monoprotic acid. Given: Step 1: Understand [&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-216694","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216694","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=216694"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216694\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=216694"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=216694"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=216694"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}