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 determine the acid dissociation constant (( K_a )) of the monoprotic acid, we follow these steps:<\/p>\n\n\n\n The hydrogen ion concentration can be determined from the pH formula: For the dissociation of a monoprotic acid ( HA ):<\/p>\n\n\n\n [ Since ( [H^+] = x = 2.95 \\times 10^{-3} ) M, we substitute this into the equilibrium expression.<\/p>\n\n\n\n [ [ [ The acid dissociation constant (( K_a )) quantifies the strength of a weak acid in solution. A higher ( K_a ) value indicates stronger acid dissociation, meaning more ( H^+ ) ions are produced.<\/p>\n\n\n\n To calculate ( K_a ), we start by determining the hydrogen ion concentration using the given pH. Since ( pH = -\\log [H^+] ), we take the inverse log (antilog) to find ( [H^+] ). The result, ( 2.95 \\times 10^{-3} ) M, represents the equilibrium concentration of both ( H^+ ) and ( A^- ) because they are produced in a 1:1 ratio from the acid.<\/p>\n\n\n\n Next, we construct an ICE table (Initial, Change, Equilibrium) to track how the acid dissociates. Initially, the acid concentration is 0.0192 M, and since some dissociates into ( H^+ ) and ( A^- ), we subtract ( x ) (which equals ( [H^+] )) from the initial concentration of ( HA ). Using these equilibrium values, we apply the expression for ( K_a ), substitute the known values, and solve for ( K_a ).<\/p>\n\n\n\n The final answer, ( 5.34 \\times 10^{-4} ), confirms that the acid is weak (since ( K_a ) values for strong acids are typically much larger). This method is crucial in chemistry for determining acid strength, buffer capacities, and pH behavior in various solutions.<\/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 determine the acid dissociation constant (( K_a )) of the monoprotic acid, we follow these steps: 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-203581","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203581","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=203581"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203581\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203581"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203581"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203581"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Define Known Values<\/strong><\/h3>\n\n\n\n
\n
Step 2: Calculate the ( [H^+] ) Concentration<\/strong><\/h3>\n\n\n\n
[
[H^+] = 10^{-\\text{pH}}
]
[
[H^+] = 10^{-2.53}
]
[
[H^+] = 2.95 \\times 10^{-3} \\text{ M}
]<\/p>\n\n\n\nStep 3: Set Up the ICE Table<\/strong><\/h3>\n\n\n\n
HA \\rightleftharpoons H^+ + A^-
]<\/p>\n\n\n\nSpecies<\/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^+ )<\/td> 0<\/td> +x<\/td> x<\/td><\/tr> ( A^- )<\/td> 0<\/td> +x<\/td> x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Step 4: Write the ( K_a ) Expression<\/strong><\/h3>\n\n\n\n
K_a = \\frac{[H^+][A^-]}{[HA]}
]
[
K_a = \\frac{(2.95 \\times 10^{-3}) (2.95 \\times 10^{-3})}{0.0192 – 2.95 \\times 10^{-3}}
]<\/p>\n\n\n\nStep 5: Solve for ( K_a )<\/strong><\/h3>\n\n\n\n
K_a = \\frac{(8.70 \\times 10^{-6})}{0.0163}
]
[
K_a = 5.34 \\times 10^{-4}
]<\/p>\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\n
K_a = \\mathbf{5.34 \\times 10^{-4}}
]<\/p>\n\n\n\nExplanation (300 Words)<\/strong><\/h3>\n\n\n\n