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 :<\/strong><\/mark><\/p>\n\n\n\n To calculate the acid dissociation constant (Ka) for the monoprotic acid, we need to use the information provided: the concentration of the acid (0.0192 M) and the pH of the solution (2.53). We can follow a series of steps to find the value of Ka.<\/p>\n\n\n\n The monoprotic acid (HA) dissociates in water as follows:<\/p>\n\n\n\n $$ The acid dissociation constant (Ka) is given by:<\/p>\n\n\n\n $$ The pH of the solution is 2.53. The pH is related to the concentration of hydrogen ions ([H\u207a]) by the equation:<\/p>\n\n\n\n $$ Rearranging to find [H\u207a]:<\/p>\n\n\n\n $$ We can now set up an ICE (Initial, Change, Equilibrium) table to track the concentrations of the acid, hydrogen ion, and conjugate base at equilibrium.<\/p>\n\n\n\n At equilibrium, the concentration of hydrogen ions [H\u207a] is $x$, which we have already calculated as 2.95 \u00d7 10\u207b\u00b3 M. Therefore:<\/p>\n\n\n\n $$ $$ $$ Now we can calculate the Ka using the equilibrium concentrations:<\/p>\n\n\n\n $$ $$ The acid dissociation constant, $K_a$, for the acid is approximately 5.35 \u00d7 10\u207b\u2074<\/strong>.<\/p>\n\n\n\n To summarize, the pH of the solution gives us the concentration of hydrogen ions, and we can use the ICE table to track how the concentrations of acid and its dissociation products change. With the equilibrium concentrations of HA, H\u207a, and A\u207b, we apply the formula for the acid dissociation constant (Ka) to calculate the value. This process reveals that the acid has a moderate strength since the Ka value is neither very high (strong acid) nor very low (weak acid).<\/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 acid dissociation constant (Ka) for the monoprotic acid, we need to use the information provided: 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-212213","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212213","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=212213"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212213\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=212213"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=212213"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=212213"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Understanding the Dissociation Process<\/h3>\n\n\n\n
\\text{HA} \\rightleftharpoons \\text{H}^+ + \\text{A}^-
$$<\/p>\n\n\n\n
K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]}
$$<\/p>\n\n\n\nStep 2: Calculate [H\u207a] from pH<\/h3>\n\n\n\n
\\text{pH} = -\\log[\\text{H}^+]
$$<\/p>\n\n\n\n
[\\text{H}^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\, \\text{M}
$$<\/p>\n\n\n\nStep 3: Set up the ICE Table<\/h3>\n\n\n\n
Species<\/th> Initial (M)<\/th> Change (M)<\/th> Equilibrium (M)<\/th><\/tr><\/thead> HA<\/td> 0.0192<\/td> -x<\/td> 0.0192 – x<\/td><\/tr> H\u207a<\/td> 0<\/td> +x<\/td> x<\/td><\/tr> A\u207b<\/td> 0<\/td> +x<\/td> x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
[\\text{HA}]_{\\text{eq}} = 0.0192 – x \\approx 0.0192 – 2.95 \\times 10^{-3} = 0.01625 \\, \\text{M}
$$<\/p>\n\n\n\n
[\\text{H}^+] = 2.95 \\times 10^{-3} \\, \\text{M}
$$<\/p>\n\n\n\n
[\\text{A}^-] = 2.95 \\times 10^{-3} \\, \\text{M}
$$<\/p>\n\n\n\nStep 4: Calculate Ka<\/h3>\n\n\n\n
K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]} = \\frac{(2.95 \\times 10^{-3})(2.95 \\times 10^{-3})}{0.01625}
$$<\/p>\n\n\n\n
K_a = \\frac{8.70 \\times 10^{-6}}{0.01625} \\approx 5.35 \\times 10^{-4}
$$<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n