{"id":204721,"date":"2025-03-22T13:45:13","date_gmt":"2025-03-22T13:45:13","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=204721"},"modified":"2025-03-22T13:45:15","modified_gmt":"2025-03-22T13:45:15","slug":"find-the-ph-of-0-2-m-a-weak-acid-ha-with-ka-8-80-x-10-6","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/22\/find-the-ph-of-0-2-m-a-weak-acid-ha-with-ka-8-80-x-10-6\/","title":{"rendered":"Find the pH of 0.2 M a weak acid (HA) with Ka = 8.80 x 10-6"},"content":{"rendered":"\n

Find the pH of 0.2 M a weak acid (HA) with Ka = 8.80 x 10-6.<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To find the pH of a 0.2 M solution of a weak acid (HA) with a given acid dissociation constant (Ka), we can use the following steps:<\/p>\n\n\n\n

Step 1: Write the dissociation equation of the acid.<\/h3>\n\n\n\n

For a weak acid HA dissociating in water:
[ \\text{HA (aq)} \\rightleftharpoons \\text{H}^+ (aq) + \\text{A}^- (aq) ]<\/p>\n\n\n\n

Step 2: Set up the expression for the equilibrium constant (Ka).<\/h3>\n\n\n\n

The expression for ( K_a ) is:
[
K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]}
]
where:<\/p>\n\n\n\n

    \n
  • ( [\\text{H}^+] ) is the concentration of hydrogen ions,<\/li>\n\n\n\n
  • ( [\\text{A}^-] ) is the concentration of the conjugate base,<\/li>\n\n\n\n
  • ( [\\text{HA}] ) is the concentration of the weak acid at equilibrium.<\/li>\n<\/ul>\n\n\n\n

    Step 3: Use the ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations.<\/h3>\n\n\n\n

    We start with 0.2 M of HA, and initially, no ( \\text{H}^+ ) or ( \\text{A}^- ) ions are present.<\/p>\n\n\n\n

    Species<\/th>Initial Concentration<\/th>Change in Concentration<\/th>Equilibrium Concentration<\/th><\/tr><\/thead>
    HA<\/td>0.2 M<\/td>-x<\/td>0.2 – x<\/td><\/tr>
    H\u207a<\/td>0 M<\/td>+x<\/td>x<\/td><\/tr>
    A\u207b<\/td>0 M<\/td>+x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    At equilibrium, the concentration of hydrogen ions ( [\\text{H}^+] = x ), and the concentration of HA is ( [\\text{HA}] = 0.2 – x ).<\/p>\n\n\n\n

    Step 4: Substitute the equilibrium concentrations into the expression for Ka.<\/h3>\n\n\n\n

    [
    K_a = \\frac{x \\cdot x}{0.2 – x}
    ]
    Given that ( K_a = 8.80 \\times 10^{-6} ), we can substitute this value:
    [
    8.80 \\times 10^{-6} = \\frac{x^2}{0.2 – x}
    ]<\/p>\n\n\n\n

    Step 5: Make an assumption for small ( x ).<\/h3>\n\n\n\n

    Since the acid dissociates very weakly (due to the small ( K_a )), we can assume that ( x ) will be very small compared to 0.2 M, so we approximate ( 0.2 – x \\approx 0.2 ). This simplifies the equation to:
    [
    8.80 \\times 10^{-6} = \\frac{x^2}{0.2}
    ]<\/p>\n\n\n\n

    Step 6: Solve for ( x ), the concentration of H\u207a.<\/h3>\n\n\n\n

    [
    x^2 = (8.80 \\times 10^{-6}) \\times 0.2
    ]
    [
    x^2 = 1.76 \\times 10^{-6}
    ]
    [
    x = \\sqrt{1.76 \\times 10^{-6}} = 1.33 \\times 10^{-3} \\, \\text{M}
    ]<\/p>\n\n\n\n

    Step 7: Calculate the pH.<\/h3>\n\n\n\n

    The pH is given by:
    [
    \\text{pH} = -\\log [\\text{H}^+]
    ]
    Substitute the value of ( x ) (the concentration of H\u207a):
    [
    \\text{pH} = -\\log (1.33 \\times 10^{-3}) = 2.876
    ]<\/p>\n\n\n\n

    Final Answer:<\/h3>\n\n\n\n

    The pH of the 0.2 M weak acid solution is approximately 2.88<\/strong>.<\/p>\n\n\n\n

    Explanation:<\/h3>\n\n\n\n

    To find the pH of a weak acid solution, we used the acid dissociation constant ( K_a ) and an equilibrium approach. The dissociation of the acid is minimal, so we made the assumption that ( x ) is very small, which simplifies the calculations. The result shows that the solution is acidic, with a pH of 2.88, indicating that the acid does not fully dissociate in water.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Find the pH of 0.2 M a weak acid (HA) with Ka = 8.80 x 10-6. The correct answer and explanation is : To find the pH of a 0.2 M solution of a weak acid (HA) with a given acid dissociation constant (Ka), we can use the following steps: Step 1: Write the dissociation […]<\/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-204721","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204721","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=204721"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204721\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204721"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204721"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204721"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}