{"id":187460,"date":"2025-02-05T03:34:31","date_gmt":"2025-02-05T03:34:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187460"},"modified":"2025-02-05T03:34:33","modified_gmt":"2025-02-05T03:34:33","slug":"using-the-kas-for-hc2h3o2-1-8x10-5-and-hco3-5-6x10-11","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/05\/using-the-kas-for-hc2h3o2-1-8x10-5-and-hco3-5-6x10-11\/","title":{"rendered":"Using the Ka\u2019s for HC2H3O2 (1.8×10-5) and HCO3- (5.6×10-11)"},"content":{"rendered":"\n
    \n
  1. Using the Ka\u2019s for HC2H3O2 (1.8×10-5) and HCO3- (5.6×10-11) :<\/li>\n<\/ol>\n\n\n\n

    a) Calculate the Kb for C2H3O2-<\/p>\n\n\n\n

    b)Calculate the Kb for CO32-<\/p>\n\n\n\n

      \n
    1. In order to prepare 50.0 mL of 0.200 M HCl you will addmL of 1.00 M HCI tomL of water.<\/li>\n<\/ol>\n\n\n\n

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

      1. Calculating Kb for C\u2082H\u2083O\u2082\u207b and CO\u2083\u00b2\u207b<\/strong><\/h3>\n\n\n\n

      To find the base dissociation constant (Kb<\/strong>) for the conjugate bases C\u2082H\u2083O\u2082\u207b<\/strong> (acetate) and CO\u2083\u00b2\u207b<\/strong> (carbonate), we use the relationship between Ka and Kb:<\/p>\n\n\n\n

      [
      K_w = K_a \\times K_b
      ]<\/p>\n\n\n\n

      where Kw<\/strong> (the ionization constant for water) is 1.0 \u00d7 10\u207b\u00b9\u2074<\/strong> at 25\u00b0C.<\/p>\n\n\n\n

      (a) Calculating Kb for C\u2082H\u2083O\u2082\u207b<\/strong><\/h4>\n\n\n\n

      Given:<\/p>\n\n\n\n

        \n
      • Ka<\/strong> for HC\u2082H\u2083O\u2082<\/strong> (acetic acid) = 1.8 \u00d7 10\u207b\u2075<\/strong><\/li>\n\n\n\n
      • Kw<\/strong> = 1.0 \u00d7 10\u207b\u00b9\u2074<\/strong><\/li>\n<\/ul>\n\n\n\n

        [
        K_b = \\frac{K_w}{K_a}
        ]<\/p>\n\n\n\n

        [
        K_b = \\frac{1.0 \\times 10^{-14}}{1.8 \\times 10^{-5}}
        ]<\/p>\n\n\n\n

        [
        K_b = 5.56 \\times 10^{-10}
        ]<\/p>\n\n\n\n

        (b) Calculating Kb for CO\u2083\u00b2\u207b<\/strong><\/h4>\n\n\n\n

        Given:<\/p>\n\n\n\n

          \n
        • Ka<\/strong> for HCO\u2083\u207b<\/strong> (bicarbonate) = 5.6 \u00d7 10\u207b\u00b9\u00b9<\/strong><\/li>\n\n\n\n
        • Kw<\/strong> = 1.0 \u00d7 10\u207b\u00b9\u2074<\/strong><\/li>\n<\/ul>\n\n\n\n

          [
          K_b = \\frac{K_w}{K_a}
          ]<\/p>\n\n\n\n

          [
          K_b = \\frac{1.0 \\times 10^{-14}}{5.6 \\times 10^{-11}}
          ]<\/p>\n\n\n\n

          [
          K_b = 1.79 \\times 10^{-4}
          ]<\/p>\n\n\n\n

          \n\n\n\n

          2. Preparing 50.0 mL of 0.200 M HCl from 1.00 M HCl<\/strong><\/h3>\n\n\n\n

          To dilute a stock solution, we use the dilution formula<\/strong>:<\/p>\n\n\n\n

          [
          M_1V_1 = M_2V_2
          ]<\/p>\n\n\n\n

          where:<\/p>\n\n\n\n

            \n
          • M\u2081<\/strong> = initial concentration (1.00 M),<\/li>\n\n\n\n
          • V\u2081<\/strong> = volume of stock solution needed (mL),<\/li>\n\n\n\n
          • M\u2082<\/strong> = final concentration (0.200 M),<\/li>\n\n\n\n
          • V\u2082<\/strong> = final total volume (50.0 mL).<\/li>\n<\/ul>\n\n\n\n

            Solving for V\u2081<\/strong>:<\/p>\n\n\n\n

            [
            V_1 = \\frac{M_2 \\times V_2}{M_1}
            ]<\/p>\n\n\n\n

            [
            V_1 = \\frac{(0.200 \\text{ M}) (50.0 \\text{ mL})}{1.00 \\text{ M}}
            ]<\/p>\n\n\n\n

            [
            V_1 = 10.0 \\text{ mL}
            ]<\/p>\n\n\n\n

            Thus, you need 10.0 mL<\/strong> of 1.00 M HCl<\/strong>.<\/p>\n\n\n\n

            To get a total volume of 50.0 mL<\/strong>, the amount of water needed is:<\/p>\n\n\n\n

            [
            V_{\\text{water}} = V_2 – V_1
            ]<\/p>\n\n\n\n

            [
            V_{\\text{water}} = 50.0 \\text{ mL} – 10.0 \\text{ mL}
            ]<\/p>\n\n\n\n

            [
            V_{\\text{water}} = 40.0 \\text{ mL}
            ]<\/p>\n\n\n\n

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

            Dilution involves reducing the concentration of a solution by adding more solvent (water). Since M\u2081V\u2081 = M\u2082V\u2082<\/strong> holds true, this ensures the number of moles of solute remains constant before and after dilution. Here, 10.0 mL<\/strong> of 1.00 M HCl<\/strong> contains the same number of moles as 50.0 mL<\/strong> of 0.200 M HCl<\/strong> but spread over a larger volume.<\/p>\n\n\n\n

            To prepare this solution correctly:<\/p>\n\n\n\n

              \n
            1. Measure 10.0 mL<\/strong> of 1.00 M HCl<\/strong> using a pipette.<\/li>\n\n\n\n
            2. Transfer<\/strong> it to a 50.0 mL<\/strong> volumetric flask or a graduated cylinder.<\/li>\n\n\n\n
            3. Add distilled water<\/strong> until the total volume reaches 50.0 mL<\/strong>.<\/li>\n\n\n\n
            4. Mix well<\/strong> to ensure uniformity.<\/li>\n<\/ol>\n\n\n\n

              It is important to add acid to water<\/strong> and not the other way around, as adding water to concentrated acid can cause heat release<\/strong>, leading to splashing or even dangerous reactions.<\/p>\n","protected":false},"excerpt":{"rendered":"

              a) Calculate the Kb for C2H3O2- b)Calculate the Kb for CO32- The Correct Answer and Explanation is : 1. Calculating Kb for C\u2082H\u2083O\u2082\u207b and CO\u2083\u00b2\u207b To find the base dissociation constant (Kb) for the conjugate bases C\u2082H\u2083O\u2082\u207b (acetate) and CO\u2083\u00b2\u207b (carbonate), we use the relationship between Ka and Kb: [K_w = K_a \\times K_b] where […]<\/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-187460","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187460","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=187460"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187460\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187460"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187460"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187460"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}