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><\/p>\n\n\n\n To calculate the acid dissociation constant (Ka)<\/strong> for a monoprotic weak acid<\/strong>, we use the pH<\/strong> to determine the concentration of H\u207a ions<\/strong>, then apply the expression for Ka<\/strong> using the ICE table method<\/strong>.<\/p>\n\n\n\n The pH of the solution is given as 2.53<\/strong>. We use the formula: [H+]=10\u2212pH=10\u22122.53[\\text{H}^+] = 10^{-\\text{pH}} = 10^{-2.53} [H+]\u22482.95\u00d710\u22123 M[\\text{H}^+] \\approx 2.95 \\times 10^{-3} \\text{ M}<\/p>\n\n\n\n Since it\u2019s a monoprotic acid<\/strong> (denoted as HA), the reaction is: HA\u21ccH++A\u2212\\text{HA} \\rightleftharpoons \\text{H}^+ + \\text{A}^-<\/p>\n\n\n\n Initial concentration of HA = 0.0192 M Ka=[H+][A\u2212][HA]K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]}<\/p>\n\n\n\n Substitute known values:<\/p>\n\n\n\n Ka=(2.95\u00d710\u22123)20.01625K_a = \\frac{(2.95 \\times 10^{-3})^2}{0.01625} Ka=8.70\u00d710\u221260.01625K_a = \\frac{8.70 \\times 10^{-6}}{0.01625} Ka\u22485.35\u00d710\u22124K_a \\approx 5.35 \\times 10^{-4}<\/p>\n\n\n\n Ka\u22485.35\u00d710\u22124\\boxed{K_a \\approx 5.35 \\times 10^{-4}}<\/p>\n\n\n\n In acid-base chemistry, the Ka (acid dissociation constant)<\/strong> quantifies the strength of a weak acid. For monoprotic acids<\/strong> (which donate one proton), the dissociation in water is represented by: HA\u21ccH++A\u2212\\text{HA} \\rightleftharpoons \\text{H}^+ + \\text{A}^-<\/p>\n\n\n\n We’re given the initial concentration of the acid solution as 0.0192 M<\/strong> and a pH of 2.53<\/strong>. Since pH is the negative logarithm of the hydrogen ion concentration, we first determine [H\u207a] by reversing the logarithmic function: [H+]=10\u2212pH=10\u22122.53\u22482.95\u00d710\u22123 M[\\text{H}^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\text{ M}<\/p>\n\n\n\n This value also represents the concentration of A\u207b<\/strong>, because for every mole of HA that dissociates, one mole each of H\u207a and A\u207b is produced.<\/p>\n\n\n\n We then use an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations. Initially, HA is at 0.0192 M, and no H\u207a or A\u207b is present. As dissociation occurs, HA decreases by x (which is 2.95 \u00d7 10\u207b\u00b3), and H\u207a and A\u207b each increase by x.<\/p>\n\n\n\n With equilibrium concentrations known, we plug values into the Ka expression: Ka=[H+][A\u2212][HA]=(2.95\u00d710\u22123)20.01625=5.35\u00d710\u22124K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]} = \\frac{(2.95 \\times 10^{-3})^2}{0.01625} = 5.35 \\times 10^{-4}<\/p>\n\n\n\n This value reflects a moderate<\/strong> acid strength: not too weak, but not strong either. It shows a small but significant degree of ionization, which is typical for weak acids.<\/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 a monoprotic weak acid, we use the pH to determine the concentration […]<\/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-211303","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211303","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=211303"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211303\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=211303"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=211303"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=211303"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Use pH to find [H\u207a]<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 2: Write the dissociation equation<\/strong><\/h3>\n\n\n\n
Change in [H\u207a] = +2.95 \u00d7 10\u207b\u00b3 M
Change in [A\u207b] = +2.95 \u00d7 10\u207b\u00b3 M
Change in [HA] = \u20132.95 \u00d7 10\u207b\u00b3 M<\/p>\n\n\n\n
\n\n\n\nStep 3: Set up the Ka expression<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation (300+ words):<\/strong><\/h3>\n\n\n\n