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 acid<\/strong>, we use the relationship between pH, concentration, and the expression for Ka:<\/p>\n\n\n\n Use the formula:<\/p>\n\n\n\n $$ $$ From step 1: $x = [H^+] = 2.95 \\times 10^{-3} \\, \\text{M}$<\/p>\n\n\n\n $$ Substitute values:<\/p>\n\n\n\n $$ $$ $$ The problem involves calculating the acid dissociation constant (Ka)<\/strong> for a monoprotic acid<\/strong>, which dissociates in water as follows:<\/p>\n\n\n\n $$ The Ka quantifies how much of the acid dissociates. A stronger acid has a higher Ka because more $H^+$ ions are produced.<\/p>\n\n\n\n We are given the initial concentration<\/strong> of the acid and the pH<\/strong>. First, we convert the pH to $[H^+]$ using the formula:<\/p>\n\n\n\n $$ This gives the amount of acid that has dissociated into $H^+$ (and $A^-$, in equal amounts for a monoprotic acid). We assume $x = [H^+] = 2.95 \\times 10^{-3} \\, \\text{M}$, meaning 0.00295 M of the acid has dissociated.<\/p>\n\n\n\n Now, we calculate how much of the acid remains: $0.0192 – x = 0.01625 \\, \\text{M}$. The Ka formula for this dissociation is:<\/p>\n\n\n\n $$ By plugging in the known values:<\/p>\n\n\n\n $$ This result shows the acid is a weak acid<\/strong> because its Ka is less than 1. If all the acid had dissociated, it would be a strong acid, and the pH would be much lower. This calculation method\u2014using an ICE table and pH to find Ka\u2014is standard in acid-base equilibrium problems and is especially important when the acid is weak and only partially dissociates.<\/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 acid, we use the relationship between pH, concentration, and the […]<\/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-214780","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214780","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=214780"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214780\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=214780"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=214780"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=214780"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Calculate $[H^+]$ from pH<\/strong><\/h3>\n\n\n\n
[H^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\, \\text{M}
$$<\/p>\n\n\n\n
\n\n\n\nStep 2: Set up ICE Table<\/strong><\/h3>\n\n\n\n
\\begin{array}{c|ccc}
& \\text{HA} & \\rightleftharpoons & \\text{H}^+ & + & \\text{A}^- \\
\\hline
\\text{Initial (M)} & 0.0192 & & 0 & & 0 \\
\\text{Change (M)} & -x & & +x & & +x \\
\\text{Equilibrium (M)} & 0.0192 – x & & x & & x \\
\\end{array}
$$<\/p>\n\n\n\n
\n\n\n\nStep 3: Use Ka Expression<\/strong><\/h3>\n\n\n\n
Ka = \\frac{[H^+][A^-]}{[HA]} = \\frac{x^2}{0.0192 – x}
$$<\/p>\n\n\n\n
Ka = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 – 2.95 \\times 10^{-3}} = \\frac{8.70 \\times 10^{-6}}{0.01625}
$$<\/p>\n\n\n\n
Ka \\approx 5.35 \\times 10^{-4}
$$<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\n
\\boxed{5.35 \\times 10^{-4}}
$$<\/p>\n\n\n\n
\n\n\n\nExplanation (300+ words):<\/strong><\/h3>\n\n\n\n
HA \\rightleftharpoons H^+ + A^-
$$<\/p>\n\n\n\n
[H^+] = 10^{-\\text{pH}} = 10^{-2.53}
$$<\/p>\n\n\n\n
Ka = \\frac{[H^+][A^-]}{[HA]} = \\frac{x^2}{[HA]_{\\text{remaining}}}
$$<\/p>\n\n\n\n
Ka = \\frac{(2.95 \\times 10^{-3})^2}{0.01625} = 5.35 \\times 10^{-4}
$$<\/p>\n\n\n\n