What is the Ka of HA if 0.10 M solution has pH = 3.50?<\/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 weak acid ( HA ), we use the given information:<\/p>\n\n\n\n The pH is related to the hydrogen ion concentration by the formula:<\/p>\n\n\n\n [ Solving for ( [H^+] ):<\/p>\n\n\n\n [ [ For the dissociation of the weak acid:<\/p>\n\n\n\n [ From Step 1, we found ( [H^+] = x = 3.16 \\times 10^{-4} ) M.<\/p>\n\n\n\n The acid dissociation constant is given by:<\/p>\n\n\n\n [ Substituting the values:<\/p>\n\n\n\n [ Approximating ( 0.10 – 3.16 \\times 10^{-4} \\approx 0.10 ):<\/p>\n\n\n\n [ [ [ [ The problem involves a weak acid ( HA ) that partially ionizes in solution. The pH provides the hydrogen ion concentration, which we use in the equilibrium expression. By assuming ( x ) is small compared to 0.10 M, we simplify calculations and determine ( K_a ) as ( 1.0 \\times 10^{-6} ), characterizing the acid’s strength.<\/p>\n","protected":false},"excerpt":{"rendered":" What is the Ka of HA if 0.10 M solution has pH = 3.50? The correct answer and explanation is : To determine the acid dissociation constant ( K_a ) of the weak acid ( HA ), we use the given information: Step 1: Determine ( [H^+] ) The pH is related to the hydrogen […]<\/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-204858","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204858","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=204858"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204858\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204858"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204858"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204858"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Step 1: Determine ( [H^+] )<\/strong><\/h3>\n\n\n\n
\\text{pH} = -\\log [H^+]
]<\/p>\n\n\n\n
[H^+] = 10^{-\\text{pH}} = 10^{-3.50}
]<\/p>\n\n\n\n
[H^+] = 3.16 \\times 10^{-4} \\text{ M}
]<\/p>\n\n\n\nStep 2: 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> 0.10<\/td> (-x)<\/td> (0.10 – 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 3: Calculate ( K_a )<\/strong><\/h3>\n\n\n\n
K_a = \\frac{[H^+][A^-]}{[HA]}
]<\/p>\n\n\n\n
K_a = \\frac{(3.16 \\times 10^{-4}) (3.16 \\times 10^{-4})}{0.10 – 3.16 \\times 10^{-4}}
]<\/p>\n\n\n\n
K_a \\approx \\frac{(3.16 \\times 10^{-4})^2}{0.10}
]<\/p>\n\n\n\n
K_a \\approx \\frac{9.98 \\times 10^{-8}}{0.10}
]<\/p>\n\n\n\n
K_a \\approx 1.00 \\times 10^{-6}
]<\/p>\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\n
K_a = 1.0 \\times 10^{-6}
]<\/p>\n\n\n\nExplanation (Summary)<\/strong><\/h3>\n\n\n\n