{"id":185399,"date":"2025-01-22T13:32:44","date_gmt":"2025-01-22T13:32:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185399"},"modified":"2025-01-22T13:32:47","modified_gmt":"2025-01-22T13:32:47","slug":"write-out-the-chemical-reaction-for-the-dissociation-of-acetic-acid","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/22\/write-out-the-chemical-reaction-for-the-dissociation-of-acetic-acid\/","title":{"rendered":"Write out the chemical reaction for the dissociation of acetic acid"},"content":{"rendered":"\n

Write out the chemical reaction for the dissociation of acetic acid, CH3COOH, in water and expression of its equilibrium constant. b\/ What is the pH of a 5M acetic acid solution given its Ka= 1.7510-5at 25oC. c\/ Given that the Kaof acetic is 1.70<\/em>10-5at 5oC.Calculate the Kaof the acetic acid at 95oC.
Write out the chemical reaction for the dissociation of acetic acid, CH3COOH, in water and expression of its equilibrium constant.<\/p>\n\n\n\n

b\/ What is the pH of a 5M acetic acid solution given its Ka= 1.75*10-5at 25oC.<\/p>\n\n\n\n

c\/ Given that the Kaof acetic is 1.70*10-5at 5oC.Calculate the Kaof the acetic acid at 95oC.<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Part a: Dissociation of Acetic Acid and Expression of Ka<\/strong><\/h3>\n\n\n\n

The dissociation of acetic acid ((\\text{CH}_3\\text{COOH})) in water is represented as:<\/p>\n\n\n\n

[
\\text{CH}_3\\text{COOH} \\, + \\, \\text{H}_2\\text{O} \\, \\rightleftharpoons \\, \\text{CH}_3\\text{COO}^- \\, + \\, \\text{H}_3\\text{O}^+
]<\/p>\n\n\n\n

The equilibrium constant for this reaction ((K_a)) is expressed as:<\/p>\n\n\n\n

[
K_a = \\frac{[\\text{CH}_3\\text{COO}^-][\\text{H}_3\\text{O}^+]}{[\\text{CH}_3\\text{COOH}]}
]<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • ([\\text{CH}_3\\text{COOH}]) = concentration of undissociated acetic acid<\/li>\n\n\n\n
  • ([\\text{CH}_3\\text{COO}^-]) = concentration of acetate ion<\/li>\n\n\n\n
  • ([\\text{H}_3\\text{O}^+]) = concentration of hydronium ion<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Part b: pH of a 5 M Acetic Acid Solution at 25\u00b0C<\/strong><\/h3>\n\n\n\n

    Given:
    [
    K_a = 1.75 \\times 10^{-5}, \\, [\\text{CH}_3\\text{COOH}] = 5 \\, \\text{M}
    ]<\/p>\n\n\n\n

    Using the relationship:
    [
    K_a = \\frac{x^2}{[CH_3COOH]_0 – x}
    ]<\/p>\n\n\n\n

    Where (x) represents ([\\text{H}_3\\text{O}^+]), and assuming (x \\ll 5), simplify to:
    [
    K_a \\approx \\frac{x^2}{5}
    ]<\/p>\n\n\n\n

    Solve for (x):
    [
    x = \\sqrt{K_a \\cdot 5} = \\sqrt{1.75 \\times 10^{-5} \\cdot 5} = \\sqrt{8.75 \\times 10^{-5}} \\approx 9.35 \\times 10^{-3} \\, \\text{M}
    ]<\/p>\n\n\n\n

    [
    \\text{pH} = -\\log_{10} [\\text{H}3\\text{O}^+] = -\\log<\/em>{10} (9.35 \\times 10^{-3}) \\approx 2.03
    ]<\/p>\n\n\n\n

    Thus, the pH is approximately 2.03<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    Part c: Calculate (K_a) at 95\u00b0C Using Van\u2019t Hoff Equation<\/strong><\/h3>\n\n\n\n

    The Van\u2019t Hoff equation relates (K_a) to temperature:
    [
    \\ln\\left(\\frac{K_{a2}}{K_{a1}}\\right) = \\frac{\\Delta H^\\circ}{R} \\left(\\frac{1}{T_1} – \\frac{1}{T_2}\\right)
    ]<\/p>\n\n\n\n

    Assume (\\Delta H^\\circ \\approx 30 \\, \\text{kJ\/mol}) for acetic acid dissociation. Convert temperatures to Kelvin:
    [
    T_1 = 5^\\circ \\text{C} = 278 \\, \\text{K}, \\, T_2 = 95^\\circ \\text{C} = 368 \\, \\text{K}
    ]<\/p>\n\n\n\n

    Plug in:
    [
    \\ln\\left(\\frac{K_{a2}}{1.70 \\times 10^{-5}}\\right) = \\frac{30,000}{8.314} \\left(\\frac{1}{278} – \\frac{1}{368}\\right)
    ]<\/p>\n\n\n\n

    Calculate the right-hand side:
    [
    \\ln\\left(\\frac{K_{a2}}{1.70 \\times 10^{-5}}\\right) \\approx 2.293
    ]<\/p>\n\n\n\n

    [
    \\frac{K_{a2}}{1.70 \\times 10^{-5}} \\approx e^{2.293} \\approx 9.91
    ]<\/p>\n\n\n\n

    [
    K_{a2} \\approx 1.70 \\times 10^{-5} \\cdot 9.91 \\approx 1.69 \\times 10^{-4}
    ]<\/p>\n\n\n\n

    Thus, (K_a) at 95\u00b0C is approximately (1.69 \\times 10^{-4})<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    Explanation<\/strong><\/h3>\n\n\n\n

    Acetic acid dissociates partially, with its equilibrium constant ((K_a)) describing this balance. To calculate pH, we solve for ([\\text{H}_3\\text{O}^+]) using the simplified quadratic equation. The higher (K_a) at 95\u00b0C (compared to 5\u00b0C) reflects increased dissociation due to the endothermic nature of the process, consistent with Le Chatelier’s principle.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Write out the chemical reaction for the dissociation of acetic acid, CH3COOH, in water and expression of its equilibrium constant. b\/ What is the pH of a 5M acetic acid solution given its Ka= 1.7510-5at 25oC. c\/ Given that the Kaof acetic is 1.7010-5at 5oC.Calculate the Kaof the acetic acid at 95oC.Write out the chemical […]<\/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-185399","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185399","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=185399"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185399\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185399"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185399"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185399"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}