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 find the acid dissociation constant (( K_a )) for the monoprotic acid, we follow these steps:<\/p>\n\n\n\n Since pH is given, we can find the hydronium ion concentration using:<\/p>\n\n\n\n [ [ For the dissociation of the monoprotic acid ( HA ):<\/p>\n\n\n\n [ Since ( [H^+] = x = 2.95 \\times 10^{-3} ), we substitute this into the ICE table values.<\/p>\n\n\n\n [ [ [ [ The ( K_a ) value for the acid is ( 5.36 \\times 10^{-4} )<\/strong>.<\/p>\n\n\n\n This result indicates the acid is weak because its ( K_a ) value is significantly less than 1. The method follows the assumption that ( x ) (the amount of dissociation) is small compared to the initial concentration, which is valid here.<\/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 find 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-204474","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204474","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=204474"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204474\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204474"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204474"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204474"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Define the given values<\/h3>\n\n\n\n
\n
Step 2: Determine ( [H^+] )<\/h3>\n\n\n\n
[H^+] = 10^{-\\text{pH}}
]<\/p>\n\n\n\n
[H^+] = 10^{-2.53} = 2.95 \\times 10^{-3} \\text{ M}
]<\/p>\n\n\n\nStep 3: Set up an ICE table<\/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: Apply the ( K_a ) expression<\/h3>\n\n\n\n
K_a = \\frac{[H^+][A^-]}{[HA]}
]<\/p>\n\n\n\n
K_a = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 – 2.95 \\times 10^{-3}}
]<\/p>\n\n\n\n
K_a = \\frac{(8.70 \\times 10^{-6})}{0.01625}
]<\/p>\n\n\n\n
K_a = 5.36 \\times 10^{-4}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n