The weak acid hypochlorous acid (HClO) is formed in bleach solutions. If the pH of a 0.12 M solution of
HClO is 4.19, what is the value of the Ka of this weak acid?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the value of the acid dissociation constant (Ka) of hypochlorous acid (HClO), we need to work through the problem step-by-step using the information provided: a 0.12 M solution of HClO with a pH of 4.19.<\/p>\n\n\n\n Hypochlorous acid (HClO) dissociates in water as follows:<\/p>\n\n\n\n [ We are given that the pH of the solution is 4.19. The pH is related to the concentration of hydrogen ions (([H^+])) by the following equation:<\/p>\n\n\n\n [ From this equation, we can calculate the concentration of ( [\\text{H}^+] ) (hydrogen ion concentration).<\/p>\n\n\n\n [ We can now set up an ICE (Initial, Change, Equilibrium) table to represent the dissociation of HClO:<\/p>\n\n\n\n Where ( x ) is the change in concentration due to dissociation. At equilibrium, we know that ( [\\text{H}^+] = 6.46 \\times 10^{-5} \\, \\text{M} ). Therefore, ( x = 6.46 \\times 10^{-5} \\, \\text{M} ).<\/p>\n\n\n\n The acid dissociation constant (Ka) for HClO is given by:<\/p>\n\n\n\n [ Substitute the equilibrium concentrations into the expression:<\/p>\n\n\n\n [ Since ( 6.46 \\times 10^{-5} ) is much smaller than 0.12, we can approximate ( 0.12 – 6.46 \\times 10^{-5} \\approx 0.12 ). Thus, the Ka expression simplifies to:<\/p>\n\n\n\n [ Now, calculate ( K_a ):<\/p>\n\n\n\n [ The value of the acid dissociation constant ( K_a ) for hypochlorous acid (HClO) is approximately 3.48 \u00d7 10\u207b\u2078<\/strong>.<\/p>\n\n\n\n This result shows that HClO is a weak acid because its dissociation constant ( K_a ) is quite small. A small ( K_a ) means that only a small fraction of HClO dissociates into hydrogen ions and hypochlorite ions in solution, which is characteristic of weak acids. The lower the ( K_a ), the weaker the acid, as it indicates that the equilibrium lies more toward the undissociated form (HClO).<\/p>\n","protected":false},"excerpt":{"rendered":" The weak acid hypochlorous acid (HClO) is formed in bleach solutions. If the pH of a 0.12 M solution ofHClO is 4.19, what is the value of the Ka of this weak acid? The correct answer and explanation is : To find the value of the acid dissociation constant (Ka) of hypochlorous acid (HClO), we […]<\/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-204708","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204708","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=204708"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204708\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204708"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204708"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204708"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Understanding the Dissociation of HClO<\/h3>\n\n\n\n
\\text{HClO} \\rightleftharpoons \\text{H}^+ + \\text{ClO}^-
]<\/p>\n\n\n\n
\\text{pH} = -\\log[\\text{H}^+]
]<\/p>\n\n\n\n
[\\text{H}^+] = 10^{-\\text{pH}} = 10^{-4.19} \\approx 6.46 \\times 10^{-5} \\, \\text{M}
]<\/p>\n\n\n\nStep 2: Set Up the Equilibrium Expression<\/h3>\n\n\n\n
Species<\/th> Initial (M)<\/th> Change (M)<\/th> Equilibrium (M)<\/th><\/tr><\/thead> HClO<\/td> 0.12<\/td> -x<\/td> 0.12 – x<\/td><\/tr> H(^+)<\/td> 0<\/td> +x<\/td> x<\/td><\/tr> ClO(^-)<\/td> 0<\/td> +x<\/td> x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Step 3: Use the Ka Expression<\/h3>\n\n\n\n
K_a = \\frac{[\\text{H}^+][\\text{ClO}^-]}{[\\text{HClO}]}
]<\/p>\n\n\n\n
K_a = \\frac{(6.46 \\times 10^{-5})(6.46 \\times 10^{-5})}{0.12 – 6.46 \\times 10^{-5}}
]<\/p>\n\n\n\n
K_a \\approx \\frac{(6.46 \\times 10^{-5})^2}{0.12}
]<\/p>\n\n\n\n
K_a \\approx \\frac{4.18 \\times 10^{-9}}{0.12} = 3.48 \\times 10^{-8}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n