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) for a monoprotic acid, we can use the information provided: the concentration of the acid solution (0.0192 M) and the pH of the solution (2.53).<\/p>\n\n\n\n From the pH of the solution, we can calculate the concentration of hydrogen ions ([H^+]) using the following formula:<\/p>\n\n\n\n [ Given that the pH is 2.53, we can rearrange the equation to solve for ([H^+]):<\/p>\n\n\n\n [ So, ([H^+] = 2.95 \\times 10^{-3} \\, \\text{M}).<\/p>\n\n\n\n We will use an ICE (Initial, Change, Equilibrium) table to understand how the dissociation of the monoprotic acid proceeds in water. The dissociation of a monoprotic acid, HA, in water can be represented as:<\/p>\n\n\n\n [ Since each mole of HA dissociates to produce one mole of (H^+) and one mole of (A^-), if x is the amount of HA that dissociates, the change in concentrations will be:<\/p>\n\n\n\n From the pH calculation, we know that ([H^+] = 2.95 \\times 10^{-3} \\, \\text{M}), so (x = 2.95 \\times 10^{-3}\\, \\text{M}).<\/p>\n\n\n\n The expression for the acid dissociation constant (K_a) is:<\/p>\n\n\n\n [ Substitute the equilibrium concentrations into this expression:<\/p>\n\n\n\n [ First, calculate the numerator:<\/p>\n\n\n\n [ Now calculate the denominator:<\/p>\n\n\n\n [ Now, calculate (K_a):<\/p>\n\n\n\n [ The acid dissociation constant (K_a) for the acid is approximately:<\/p>\n\n\n\n [ The acid dissociation constant (K_a) is a measure of the strength of an acid in solution. A larger (K_a) indicates a stronger acid, meaning it dissociates more completely in water. In this case, the value of (K_a = 5.35 \\times 10^{-4}) suggests that the acid is a weak acid, as the dissociation is not complete, and only a small fraction of the acid dissociates to produce (H^+) ions. By using the pH to find the concentration of hydrogen ions and applying it to the equilibrium expression, we can calculate the (K_a), which gives us insight into the acid’s dissociation behavior.<\/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 acid, we can use the information provided: 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-202282","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202282","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=202282"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202282\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202282"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202282"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202282"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Determine the concentration of ([H^+])<\/h3>\n\n\n\n
\\text{pH} = -\\log [H^+]
]<\/p>\n\n\n\n
[H^+] = 10^{-\\text{pH}} = 10^{-2.53} \\approx 2.95 \\times 10^{-3} \\, \\text{M}
]<\/p>\n\n\n\nStep 2: Set up an ICE table<\/h3>\n\n\n\n
\\text{HA} (aq) \\rightleftharpoons \\text{H}^+ (aq) + \\text{A}^- (aq)
]<\/p>\n\n\n\nInitial concentrations:<\/h4>\n\n\n\n
\n
Change in concentrations:<\/h4>\n\n\n\n
\n
Equilibrium concentrations:<\/h4>\n\n\n\n
\n
Step 3: Solve for Ka<\/h3>\n\n\n\n
K_a = \\frac{[H^+][A^-]}{[HA]}
]<\/p>\n\n\n\n
K_a = \\frac{(2.95 \\times 10^{-3})(2.95 \\times 10^{-3})}{0.0192 – 2.95 \\times 10^{-3}}
]<\/p>\n\n\n\n
(2.95 \\times 10^{-3})^2 = 8.70 \\times 10^{-6}
]<\/p>\n\n\n\n
0.0192 – 2.95 \\times 10^{-3} = 0.01625 \\, \\text{M}
]<\/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\nFinal Answer:<\/h3>\n\n\n\n
K_a \\approx 5.35 \\times 10^{-4}
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n