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 Let’s solve the problem carefully first:<\/p>\n\n\n\n Given:<\/strong><\/p>\n\n\n\n Step 1: Find ([H^+]) from pH.<\/strong> Step 2: Set up the Ka expression.<\/strong><\/p>\n\n\n\n For a monoprotic acid, the dissociation is: Since (HA) dissociates to form equal amounts of (H^+) and (A^-), then: Remaining ([HA]) is: Step 3: Plug into Ka expression:<\/strong> \u2705 Final Answer:<\/strong> This problem asks us to calculate the acid dissociation constant, Ka, for a weak monoprotic acid given the solution concentration and pH. The first step is to recall that pH and hydrogen ion concentration ([H^+]) are related by the formula:<\/p>\n\n\n\n [ Substituting the given pH of 2.53 into the formula, we find:<\/p>\n\n\n\n [ Since the acid is monoprotic (it donates only one proton per molecule), the concentration of the conjugate base ([A^-]) formed is equal to the ([H^+]) concentration.<\/p>\n\n\n\n Next, we calculate how much of the original acid remains undissociated. Initially, the acid concentration is 0.0192 M. After dissociation, some acid turns into (H^+) and (A^-), so the remaining concentration is:<\/p>\n\n\n\n [ The Ka expression for a weak acid is:<\/p>\n\n\n\n [ Substituting the values:<\/p>\n\n\n\n [ First, square (2.95 \\times 10^{-3}), giving (8.70 \\times 10^{-6}), and then divide by 0.01625 to find:<\/p>\n\n\n\n [ This value indicates the strength of the acid: since Ka is relatively small (much less than 1), it confirms the acid is weak and does not dissociate completely. Knowing Ka is crucial for understanding the acid\u2019s behavior in chemical reactions, including buffer solutions and titrations.<\/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 : Let’s solve the problem carefully first: Given: Step 1: Find ([H^+]) from pH.[[H^+] = 10^{-\\text{pH}} = 10^{-2.53} = 2.95 […]<\/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-208531","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208531","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=208531"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208531\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=208531"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=208531"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=208531"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n\n
[
[H^+] = 10^{-\\text{pH}} = 10^{-2.53} = 2.95 \\times 10^{-3} \\, \\text{M}
]<\/p>\n\n\n\n
[
HA \\rightleftharpoons H^+ + A^-
]
Ka is:
[
Ka = \\frac{[H^+][A^-]}{[HA]}
]<\/p>\n\n\n\n
[
[A^-] = [H^+] = 2.95 \\times 10^{-3} \\, \\text{M}
]<\/p>\n\n\n\n
[
[HA]_{\\text{remaining}} = 0.0192 – 2.95 \\times 10^{-3} = 0.01625 \\, \\text{M}
]<\/p>\n\n\n\n
[
Ka = \\frac{(2.95 \\times 10^{-3})^2}{0.01625}
]
[
Ka = \\frac{8.70 \\times 10^{-6}}{0.01625}
]
[
Ka = 5.36 \\times 10^{-4}
]<\/p>\n\n\n\n
\n\n\n\n
[
\\boxed{5.36 \\times 10^{-4}}
]<\/p>\n\n\n\n
\n\n\n\nFull 300-Word Explanation:<\/h3>\n\n\n\n
[H^+] = 10^{-\\text{pH}}
]<\/p>\n\n\n\n
[H^+] = 10^{-2.53} = 2.95 \\times 10^{-3} \\, \\text{M}
]<\/p>\n\n\n\n
[HA] = 0.0192 – 2.95 \\times 10^{-3} = 0.01625 \\, \\text{M}
]<\/p>\n\n\n\n
Ka = \\frac{[H^+][A^-]}{[HA]}
]<\/p>\n\n\n\n
Ka = \\frac{(2.95 \\times 10^{-3})^2}{0.01625}
]<\/p>\n\n\n\n
Ka = 5.36 \\times 10^{-4}
]<\/p>\n\n\n\n