{"id":47349,"date":"2025-07-02T11:43:04","date_gmt":"2025-07-02T11:43:04","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=47349"},"modified":"2025-07-02T11:43:06","modified_gmt":"2025-07-02T11:43:06","slug":"calculate-the-percent-dissociation-of-acetic-acid","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculate-the-percent-dissociation-of-acetic-acid\/","title":{"rendered":"Calculate the percent dissociation of acetic acid"},"content":{"rendered":"\n

Calculate the percent dissociation of acetic acid (Ka=1.8×10^-5) in: A) 1.00 M HC2H3O2 B) 0.100 M C2H3O2H C) For solutions of any weak acid HA, H+ decreases as HA decreases. But the percent dissociation increases as HA decreases. Explain why this is the case.<\/p>\n\n\n\n

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

To calculate the percent dissociation of acetic acid (HC\u2082H\u2083O\u2082), we use the acid dissociation constant (Ka), which is given as 1.8 \u00d7 10\u207b\u2075. We also need to apply an ICE table (Initial, Change, Equilibrium) for each concentration of acetic acid.<\/p>\n\n\n\n

A) For 1.00 M HC\u2082H\u2083O\u2082:<\/h3>\n\n\n\n
    \n
  1. Write the dissociation equation<\/strong>:
    HC\u2082H\u2083O\u2082 \u21cc H\u207a + C\u2082H\u2083O\u2082\u207b<\/li>\n\n\n\n
  2. Set up the ICE table<\/strong>:<\/li>\n<\/ol>\n\n\n\n
    <\/th>HC\u2082H\u2083O\u2082<\/th>H\u207a<\/th>C\u2082H\u2083O\u2082\u207b<\/th><\/tr><\/thead>
    Initial<\/td>1.00 M<\/td>0<\/td>0<\/td><\/tr>
    Change<\/td>-x<\/td>+x<\/td>+x<\/td><\/tr>
    Equilibrium<\/td>1.00 – x<\/td>x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
      \n
    1. Write the expression for Ka<\/strong>:<\/li>\n<\/ol>\n\n\n\n

      Ka=[H+][C2H3O2\u2212][HC2H3O2]Ka = \\frac{[H\u207a][C\u2082H\u2083O\u2082\u207b]}{[HC\u2082H\u2083O\u2082]}Ka=[HC2\u200bH3\u200bO2\u200b][H+][C2\u200bH3\u200bO2\u2212\u200b]\u200b<\/p>\n\n\n\n

      Substitute the equilibrium concentrations:1.8\u00d710\u22125=x21.00\u2212x1.8 \\times 10^{-5} = \\frac{x^2}{1.00 – x}1.8\u00d710\u22125=1.00\u2212xx2\u200b<\/p>\n\n\n\n

      Since KaKaKa is small, we can assume that xxx will be much smaller than 1.00, so 1.00\u2212x\u22481.001.00 – x \\approx 1.001.00\u2212x\u22481.00. Therefore:1.8\u00d710\u22125=x21.001.8 \\times 10^{-5} = \\frac{x^2}{1.00}1.8\u00d710\u22125=1.00×2\u200b<\/p>\n\n\n\n

      Solve for xxx:x2=1.8\u00d710\u22125x^2 = 1.8 \\times 10^{-5}x2=1.8\u00d710\u22125x=1.8\u00d710\u22125\u22484.24\u00d710\u22123\u2009Mx = \\sqrt{1.8 \\times 10^{-5}} \\approx 4.24 \\times 10^{-3} \\, \\text{M}x=1.8\u00d710\u22125\u200b\u22484.24\u00d710\u22123M<\/p>\n\n\n\n

      The percent dissociation is given by:Percent dissociation=x[HC2H3O2]\u00d7100=4.24\u00d710\u221231.00\u00d7100\u22480.424%\\text{Percent dissociation} = \\frac{x}{[HC\u2082H\u2083O\u2082]} \\times 100 = \\frac{4.24 \\times 10^{-3}}{1.00} \\times 100 \\approx 0.424\\%Percent dissociation=[HC2\u200bH3\u200bO2\u200b]x\u200b\u00d7100=1.004.24\u00d710\u22123\u200b\u00d7100\u22480.424%<\/p>\n\n\n\n

      B) For 0.100 M HC\u2082H\u2083O\u2082:<\/h3>\n\n\n\n
        \n
      1. Set up the ICE table<\/strong>:<\/li>\n<\/ol>\n\n\n\n
        <\/th>HC\u2082H\u2083O\u2082<\/th>H\u207a<\/th>C\u2082H\u2083O\u2082\u207b<\/th><\/tr><\/thead>
        Initial<\/td>0.100 M<\/td>0<\/td>0<\/td><\/tr>
        Change<\/td>-x<\/td>+x<\/td>+x<\/td><\/tr>
        Equilibrium<\/td>0.100 – x<\/td>x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
          \n
        1. Write the expression for Ka<\/strong>:<\/li>\n<\/ol>\n\n\n\n

          Ka=x20.100\u2212xKa = \\frac{x^2}{0.100 – x}Ka=0.100\u2212xx2\u200b<\/p>\n\n\n\n

          Assume xxx is small compared to 0.100, so 0.100\u2212x\u22480.1000.100 – x \\approx 0.1000.100\u2212x\u22480.100:1.8\u00d710\u22125=x20.1001.8 \\times 10^{-5} = \\frac{x^2}{0.100}1.8\u00d710\u22125=0.100×2\u200b<\/p>\n\n\n\n

          Solve for xxx:x2=1.8\u00d710\u22125\u00d70.100=1.8\u00d710\u22126x^2 = 1.8 \\times 10^{-5} \\times 0.100 = 1.8 \\times 10^{-6}x2=1.8\u00d710\u22125\u00d70.100=1.8\u00d710\u22126x=1.8\u00d710\u22126\u22481.34\u00d710\u22123\u2009Mx = \\sqrt{1.8 \\times 10^{-6}} \\approx 1.34 \\times 10^{-3} \\, \\text{M}x=1.8\u00d710\u22126\u200b\u22481.34\u00d710\u22123M<\/p>\n\n\n\n

          The percent dissociation is:Percent dissociation=x[HC2H3O2]\u00d7100=1.34\u00d710\u221230.100\u00d7100=1.34%\\text{Percent dissociation} = \\frac{x}{[HC\u2082H\u2083O\u2082]} \\times 100 = \\frac{1.34 \\times 10^{-3}}{0.100} \\times 100 = 1.34\\%Percent dissociation=[HC2\u200bH3\u200bO2\u200b]x\u200b\u00d7100=0.1001.34\u00d710\u22123\u200b\u00d7100=1.34%<\/p>\n\n\n\n

          C) Explanation of Why Percent Dissociation Increases as HA Decreases:<\/h3>\n\n\n\n

          As the concentration of a weak acid HA decreases, its dissociation becomes more significant in terms of the total concentration of acid. In a solution of weak acid, the dissociation is governed by an equilibrium where the concentration of dissociated ions is very small compared to the undissociated acid. At higher concentrations of HA, the dissociation is suppressed because the equilibrium favors the undissociated acid.<\/p>\n\n\n\n

          However, as the concentration of HA decreases (such as in more dilute solutions), the concentration of undissociated acid decreases, so the relative amount of dissociation increases. In other words, with less HA present, a larger proportion of the acid dissociates to maintain the equilibrium, leading to a higher percent dissociation. This explains why percent dissociation increases as the concentration of the acid decreases.<\/p>\n\n\n\n

          \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

          Calculate the percent dissociation of acetic acid (Ka=1.8×10^-5) in: A) 1.00 M HC2H3O2 B) 0.100 M C2H3O2H C) For solutions of any weak acid HA, H+ decreases as HA decreases. But the percent dissociation increases as HA decreases. Explain why this is the case. The Correct Answer and Explanation is: To calculate the percent dissociation […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-47349","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47349","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=47349"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47349\/revisions"}],"predecessor-version":[{"id":47358,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47349\/revisions\/47358"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=47349"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=47349"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=47349"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}