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
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To calculate the Ka<\/strong> (acid dissociation constant) for a monoprotic weak acid<\/strong>, we can use the following approach:<\/p>\n\n\n\n $$ $$ Let the weak monoprotic acid be HA<\/strong>, which dissociates as:<\/p>\n\n\n\n $$ Initial concentrations:<\/p>\n\n\n\n Change:<\/p>\n\n\n\n At equilibrium:<\/p>\n\n\n\n $$ $$ $$ The acid dissociation constant, Ka<\/strong>, is a quantitative measure of the strength of a weak acid in solution. It expresses how much the acid dissociates into hydrogen ions (H\u207a) and conjugate base (A\u207b) in water. For a monoprotic acid<\/strong>, which releases only one proton (H\u207a) per molecule, the dissociation reaction is straightforward:<\/p>\n\n\n\n $$ We’re given the initial concentration of the acid (0.0192 M) and the pH (2.53). Since pH is a measure of hydrogen ion concentration, we first convert pH to [H\u207a] using the inverse log:<\/p>\n\n\n\n $$ Because each mole of HA that dissociates gives one mole of H\u207a, the concentration of A\u207b is also 2.95 \u00d7 10\u207b\u00b3 M at equilibrium. The remaining concentration of the undissociated acid HA is the initial minus the amount that dissociated:<\/p>\n\n\n\n $$ Now, we substitute these equilibrium concentrations into the Ka expression<\/strong>:<\/p>\n\n\n\n $$ Solving this gives a Ka of approximately 5.35 \u00d7 10\u207b\u2074<\/strong>.<\/p>\n\n\n\n This Ka value indicates a weak acid<\/strong> because it is significantly less than 1, meaning only partial ionization occurs in solution. This calculation is crucial in understanding acid strength, buffering capacity, and equilibrium behavior in biological and chemical systems.<\/p>\n","protected":false},"excerpt":{"rendered":" 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 calculate the Ka (acid dissociation constant) for a monoprotic weak acid, we can use the following approach: Step 1: […]<\/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-217349","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217349","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=217349"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217349\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217349"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217349"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217349"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Understand the given information<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 2: Use pH to find [H\u207a] (also written as [H\u2083O\u207a])<\/strong><\/h3>\n\n\n\n
\\text{pH} = -\\log[H\u207a]
$$<\/p>\n\n\n\n
[H\u207a] = 10^{-2.53} \u2248 2.95 \\times 10^{-3} \\, \\text{M}
$$<\/p>\n\n\n\n
\n\n\n\nStep 3: Set up an ICE table for the dissociation<\/strong><\/h3>\n\n\n\n
\\text{HA} \\rightleftharpoons \\text{H}^+ + \\text{A}^-
$$<\/p>\n\n\n\n\n
\n
\n
\n\n\n\nStep 4: Use the Ka expression<\/strong><\/h3>\n\n\n\n
Ka = \\frac{[H^+][A^-]}{[HA]}
$$<\/p>\n\n\n\n
Ka = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 – 2.95 \\times 10^{-3}}
$$<\/p>\n\n\n\n
Ka = \\frac{8.70 \\times 10^{-6}}{0.01625} \u2248 5.35 \\times 10^{-4}
$$<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer: Ka \u2248 5.35 \u00d7 10\u207b\u2074<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation (300+ words)<\/strong><\/h3>\n\n\n\n
HA \\rightleftharpoons H^+ + A^-
$$<\/p>\n\n\n\n
[H\u207a] = 10^{-pH} = 10^{-2.53} \u2248 2.95 \\times 10^{-3} \\, \\text{M}
$$<\/p>\n\n\n\n
[HA] = 0.0192 – 2.95 \\times 10^{-3} = 0.01625 \\, \\text{M}
$$<\/p>\n\n\n\n
Ka = \\frac{[H^+][A^-]}{[HA]} = \\frac{(2.95 \\times 10^{-3})^2}{0.01625}
$$<\/p>\n\n\n\n