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 Ka<\/strong> (acid dissociation constant) of a monoprotic acid<\/strong> given its concentration and pH, we follow these steps:<\/p>\n\n\n\n We use the pH equation:<\/p>\n\n\n\n [ A monoprotic acid<\/strong> HA dissociates like this:<\/p>\n\n\n\n [ [ [ [ The Ka<\/strong> value, or acid dissociation constant, quantifies the strength of a weak acid. It measures how much the acid dissociates into ions in solution. A monoprotic acid<\/strong> is an acid that donates only one proton (H\u207a)<\/strong> per molecule.<\/p>\n\n\n\n In this problem, we are given the initial concentration of the acid (0.0192 M) and the resulting pH (2.53). Since pH tells us the concentration of hydrogen ions in solution, we can convert it to [H\u207a] using the formula:<\/p>\n\n\n\n [ Once we know the concentration of H\u207a ions, we can assume this is also the concentration of A\u207b (the conjugate base), since the acid releases one H\u207a for every A\u207b it forms.<\/p>\n\n\n\n We then subtract the amount of HA that dissociated (which is x = [H\u207a]) from the initial HA concentration to find the equilibrium concentration of HA.<\/p>\n\n\n\n With all equilibrium concentrations known, we substitute into the Ka expression<\/strong>:<\/p>\n\n\n\n [ This calculation gives us a Ka value of 5.35 \u00d7 10\u207b\u2074<\/strong>, which tells us this is a weak acid (since strong acids have very high Ka values, and weak acids have small ones).<\/p>\n\n\n\n Understanding Ka helps chemists predict how acids behave in reactions, determine pH, and calculate buffer capacities, all of which are essential in both lab and real-life applications like medicine and environmental science.<\/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 Ka (acid dissociation constant) of a monoprotic acid given its concentration and pH, we follow these […]<\/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-207492","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207492","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=207492"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207492\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207492"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207492"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207492"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n\u2705 Step 1: Find the concentration of H\u207a (or H\u2083O\u207a) ions from pH<\/strong><\/h3>\n\n\n\n
[\\text{H}^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\, \\text{M}
]<\/p>\n\n\n\n
\n\n\n\n\u2705 Step 2: Set up an ICE table<\/strong><\/h3>\n\n\n\n
\\text{HA} \\rightleftharpoons \\text{H}^+ + \\text{A}^-
]<\/p>\n\n\n\nSpecies<\/th> Initial (M)<\/th> Change (M)<\/th> Equilibrium (M)<\/th><\/tr><\/thead> HA<\/td> 0.0192<\/td> \u2013x<\/td> 0.0192 \u2013 x<\/td><\/tr> H\u207a<\/td> 0<\/td> +x<\/td> x = 2.95 \u00d7 10\u207b\u00b3<\/td><\/tr> A\u207b<\/td> 0<\/td> +x<\/td> x = 2.95 \u00d7 10\u207b\u00b3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n\u2705 Step 3: Use the Ka expression<\/strong><\/h3>\n\n\n\n
K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]} = \\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} \\approx 5.35 \\times 10^{-4}
]<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\n
\\boxed{K_a = 5.35 \\times 10^{-4}}
]<\/p>\n\n\n\n
\n\n\n\n\ud83e\udde0 Explanation (300 words)<\/strong><\/h3>\n\n\n\n
[\\text{H}^+] = 10^{-\\text{pH}}
]<\/p>\n\n\n\n
K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]}
]<\/p>\n\n\n\n