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 (Ka)<\/strong> for the monoprotic acid, we begin by interpreting the information given:<\/p>\n\n\n\n We use the formula:<\/p>\n\n\n\n $$ This is the concentration of dissociated H\u207a ions<\/strong>, which equals the concentration of the conjugate base $[A^-]$, because the acid is monoprotic.<\/p>\n\n\n\n So, at equilibrium:<\/p>\n\n\n\n $$ $$ $$ $$ The acid dissociation constant, Ka, is a measure of the strength of an acid in solution. It represents the equilibrium constant for the dissociation of an acid (HA) into hydrogen ions (H\u207a) and its conjugate base (A\u207b). In this problem, we are given the concentration of a monoprotic acid solution and its pH, and we are asked to determine Ka.<\/p>\n\n\n\n We start by using the pH value to determine the concentration of hydrogen ions. pH is the negative logarithm of the hydrogen ion concentration, so using the inverse logarithmic function gives us [H\u207a]. For a monoprotic acid, the concentration of [H\u207a] is equal to the concentration of the conjugate base [A\u207b] because each molecule of acid that dissociates produces one H\u207a and one A\u207b.<\/p>\n\n\n\n Next, we consider how much of the acid has dissociated. We subtract the amount dissociated from the initial acid concentration to get the equilibrium concentration of the undissociated acid [HA].<\/p>\n\n\n\n We then use the Ka expression, which is derived from the law of mass action. Plugging in the equilibrium concentrations, we calculate Ka. This value gives us insight into the acid’s strength. A larger Ka means a stronger acid (more dissociation), while a smaller Ka indicates a weaker acid (less dissociation).<\/p>\n\n\n\n In this case, the Ka value of approximately 5.35 \u00d7 10\u207b\u2074 suggests the acid is weak but partially dissociates in solution.<\/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 (Ka) for the monoprotic acid, we begin by interpreting the information given: Step […]<\/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-211819","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211819","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=211819"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211819\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=211819"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=211819"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=211819"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
\n\n\n\nStep 1: Calculate $H\u207a$ from the pH<\/h3>\n\n\n\n
[H^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\text{ M}
$$<\/p>\n\n\n\n
\n\n\n\nStep 2: Use an ICE Table<\/h3>\n\n\n\n
Species<\/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\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
[HA] = 0.0192 – 0.00295 = 0.01625 \\text{ M}
$$<\/p>\n\n\n\n
\n\n\n\nStep 3: Use the Ka Expression<\/h3>\n\n\n\n
K_a = \\frac{[H^+][A^-]}{[HA]} = \\frac{(2.95 \\times 10^{-3})^2}{0.01625}
$$<\/p>\n\n\n\n
K_a = \\frac{8.70 \\times 10^{-6}}{0.01625} \\approx 5.35 \\times 10^{-4}
$$<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\\boxed{K_a = 5.35 \\times 10^{-4}}
$$<\/p>\n\n\n\n
\n\n\n\nExplanation (300 Words):<\/h3>\n\n\n\n