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 ) for the monoprotic acid, we follow these steps:<\/p>\n\n\n\n Let the monoprotic acid be represented as ( HA ): The given pH of the solution is 2.53<\/strong>. We calculate ( [H^+] ) using: Using a calculator: Since the concentration of ( H^+ ) at equilibrium is ( 2.95 \\times 10^{-3} ) M, we set ( x = 2.95 \\times 10^{-3} ):<\/p>\n\n\n\n [ [ [ [ This problem involves calculating the acid dissociation constant ( K_a ) for a weak monoprotic acid from its pH and initial concentration. We first use the pH to find ( [H^+] ), then construct an ICE table to determine equilibrium concentrations. Finally, we substitute the values into the equilibrium expression to solve for ( K_a ). Since ( [H^+] ) is relatively small compared to the initial acid concentration, we assume minimal dissociation. The result, ( K_a = 5.36 \\times 10^{-4} ), confirms that this is a weak acid, as weak acids typically have ( K_a ) values less than 1.<\/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 ) for 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-203343","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203343","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=203343"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203343\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203343"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203343"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203343"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Define the Dissociation Reaction<\/strong><\/h3>\n\n\n\n
[
HA \\rightleftharpoons H^+ + A^-
]
The equilibrium expression for the dissociation constant ( K_a ) is:
[
K_a = \\frac{[H^+][A^-]}{[HA]}
]<\/p>\n\n\n\nStep 2: Determine ( [H^+] ) from pH<\/strong><\/h3>\n\n\n\n
[
[H^+] = 10^{-\\text{pH}} = 10^{-2.53}
]<\/p>\n\n\n\n
[
[H^+] = 2.95 \\times 10^{-3} \\text{ M}
]<\/p>\n\n\n\nStep 3: Set Up an ICE Table<\/strong><\/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^+ )<\/td> 0<\/td> ( +x )<\/td> ( x )<\/td><\/tr> ( A^- )<\/td> 0<\/td> ( +x )<\/td> ( x )<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
[HA]_{\\text{eq}} = 0.0192 – 2.95 \\times 10^{-3} = 0.01625 \\text{ M}
]<\/p>\n\n\n\nStep 4: Calculate ( K_a )<\/strong><\/h3>\n\n\n\n
K_a = \\frac{[H^+][A^-]}{[HA]} = \\frac{(2.95 \\times 10^{-3})^2}{0.01625}
]<\/p>\n\n\n\n
K_a = \\frac{8.70 \\times 10^{-6}}{0.01625} = 5.36 \\times 10^{-4}
]<\/p>\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\n
K_a = 5.36 \\times 10^{-4}
]<\/p>\n\n\n\nExplanation:<\/strong><\/h3>\n\n\n\n