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 (( K_a )) for the monoprotic acid, we will use the following relationship:<\/p>\n\n\n\n [ From the given pH of the solution, we can calculate the concentration of hydrogen ions. The pH is related to the concentration of hydrogen ions by the equation:<\/p>\n\n\n\n [ Given that the pH is 2.53, we can solve for ([H^+]):<\/p>\n\n\n\n [ Let\u2019s assume the acid is ( HA ), which dissociates as follows:<\/p>\n\n\n\n [ Let the initial concentration of ( HA ) be 0.0192 M, and at equilibrium, the concentration of ( H^+ ) will be ( 2.96 \\times 10^{-3} ) M (as we calculated above).<\/p>\n\n\n\n Let\u2019s define the changes in concentration during dissociation. Initially:<\/p>\n\n\n\n At equilibrium:<\/p>\n\n\n\n Where ( x ) is the amount of acid dissociated. Since we know ([H^+] = 2.96 \\times 10^{-3} \\, \\text{M}), this means ( x = 2.96 \\times 10^{-3} \\, \\text{M} ).<\/p>\n\n\n\n Thus, the equilibrium concentration of ( HA ) is:<\/p>\n\n\n\n [ The acid dissociation constant, ( K_a ), is given by:<\/p>\n\n\n\n [ Substitute the values we have:<\/p>\n\n\n\n [ The acid dissociation constant ( K_a ) for the acid is ( 5.4 \\times 10^{-4} ).<\/p>\n\n\n\n The process involves calculating the concentration of ( H^+ ) from the pH, then setting up an equilibrium expression to account for the dissociation of the acid. The value of ( K_a ) is determined using the equilibrium concentrations of the dissociated and undissociated species. This approach is common when determining the strength of an acid in solution, as ( K_a ) reflects the extent to which the acid 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 acid dissociation constant (( K_a )) for the monoprotic acid, we will use the following relationship: […]<\/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-203010","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203010","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=203010"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203010\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203010"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203010"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203010"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\text{pH} = -\\log[H^+]
]<\/p>\n\n\n\nStep 1: Determine the concentration of hydrogen ions ([H^+])<\/h3>\n\n\n\n
\\text{pH} = -\\log[H^+]
]<\/p>\n\n\n\n
2.53 = -\\log[H^+]
]
[
[H^+] = 10^{-2.53} = 2.96 \\times 10^{-3} \\, \\text{M}
]<\/p>\n\n\n\nStep 2: Set up the equilibrium expression for the dissociation of the acid<\/h3>\n\n\n\n
HA \\rightleftharpoons H^+ + A^-
]<\/p>\n\n\n\n\n
\n
[HA] = 0.0192 – 2.96 \\times 10^{-3} = 0.01624 \\, \\text{M}
]<\/p>\n\n\n\nStep 3: Write the expression for ( K_a )<\/h3>\n\n\n\n
K_a = \\frac{[H^+][A^-]}{[HA]}
]<\/p>\n\n\n\n
K_a = \\frac{(2.96 \\times 10^{-3})(2.96 \\times 10^{-3})}{0.01624}
]
[
K_a = \\frac{8.76 \\times 10^{-6}}{0.01624} = 5.4 \\times 10^{-4}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n