{"id":183811,"date":"2025-01-17T06:20:36","date_gmt":"2025-01-17T06:20:36","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183811"},"modified":"2025-01-17T06:20:38","modified_gmt":"2025-01-17T06:20:38","slug":"prove-that-ka-kb-kw-and-pka-pkb-14-using-the-equations-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/17\/prove-that-ka-kb-kw-and-pka-pkb-14-using-the-equations-below\/","title":{"rendered":"Prove that: Ka * Kb = Kw and pKa + pKb = 14 Using the equations below"},"content":{"rendered":"\n

Prove that: Ka * Kb = Kw and pKa + pKb = 14 Using the equations below. ?? = [HA][OH-] b[A AT1 2H20 -H30+ + OH-<\/p>\n\n\n\n

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

To prove Ka\u22c5Kb=KwK_a \\cdot K_b = K_w and pKa+pKb=14pK_a + pK_b = 14, let’s use the following information and equations:<\/p>\n\n\n\n

    \n
  1. Definition of KaK_a and KbK_b:<\/strong>\n
      \n
    • Ka=[H+][A\u2212][HA]K_a = \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]}, where [HA][\\text{HA}] is the concentration of the weak acid, [A\u2212][\\text{A}^-] is its conjugate base, and [H+][\\text{H}^+] is the hydronium ion concentration.<\/li>\n\n\n\n
    • Kb=[OH\u2212][HB+][B]K_b = \\frac{[\\text{OH}^-][\\text{HB}^+]}{[\\text{B}]}, where [B][\\text{B}] is the concentration of the weak base, [HB+][\\text{HB}^+] is its conjugate acid, and [OH\u2212][\\text{OH}^-] is the hydroxide ion concentration.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Water autoionization constant:<\/strong>\n
        \n
      • Kw=[H+][OH\u2212]K_w = [\\text{H}^+][\\text{OH}^-] at 25\u00b0C, and pKw=\u2212log\u2061Kw=14\\text{p}K_w = -\\log K_w = 14.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
        \n\n\n\n

        Derivation of Ka\u22c5Kb=KwK_a \\cdot K_b = K_w<\/h3>\n\n\n\n
          \n
        1. For a weak acid HA\\text{HA}, the conjugate base A\u2212\\text{A}^- reacts with water: A\u2212+H2O\u2194HA+OH\u2212\\text{A}^- + \\text{H}_2\\text{O} \\leftrightarrow \\text{HA} + \\text{OH}^- Here, Kb=[OH\u2212][HA][A\u2212]K_b = \\frac{[\\text{OH}^-][\\text{HA}]}{[\\text{A}^-]}.<\/li>\n\n\n\n
        2. Multiplying the expressions for KaK_a and KbK_b: Ka\u22c5Kb=([H+][A\u2212][HA])\u22c5([OH\u2212][HA][A\u2212])K_a \\cdot K_b = \\left( \\frac{[\\text{H}^+][\\text{A}^-]}{[\\text{HA}]} \\right) \\cdot \\left( \\frac{[\\text{OH}^-][\\text{HA}]}{[\\text{A}^-]} \\right) Simplify: Ka\u22c5Kb=[H+][OH\u2212]=KwK_a \\cdot K_b = [\\text{H}^+][\\text{OH}^-] = K_w<\/li>\n<\/ol>\n\n\n\n
          \n\n\n\n

          Derivation of pKa+pKb=14pK_a + pK_b = 14<\/h3>\n\n\n\n
            \n
          1. By definition, pKa=\u2212log\u2061KapK_a = -\\log K_a and pKb=\u2212log\u2061KbpK_b = -\\log K_b.<\/li>\n\n\n\n
          2. Take the negative logarithm of both sides of Ka\u22c5Kb=KwK_a \\cdot K_b = K_w: \u2212log\u2061(Ka\u22c5Kb)=\u2212log\u2061Kw-\\log(K_a \\cdot K_b) = -\\log K_w Using log properties: \u2212log\u2061Ka\u2212log\u2061Kb=\u2212log\u2061Kw-\\log K_a – \\log K_b = -\\log K_w Substituting pKw=14\\text{p}K_w = 14: pKa+pKb=14pK_a + pK_b = 14<\/li>\n<\/ol>\n\n\n\n
            \n\n\n\n

            Explanation (300 Words)<\/h3>\n\n\n\n

            The relationships Ka\u22c5Kb=KwK_a \\cdot K_b = K_w and pKa+pKb=14pK_a + pK_b = 14 stem from the equilibrium principles of weak acids and bases and water’s autoionization constant. A weak acid (HA\\text{HA}) dissociates in water to form [H+][\\text{H}^+] and its conjugate base (A\u2212\\text{A}^-), described by KaK_a. Similarly, the conjugate base (A\u2212\\text{A}^-) reacts with water to produce [OH\u2212[\\text{OH}^- and HA\\text{HA}, described by KbK_b.<\/p>\n\n\n\n

            The product Ka\u22c5KbK_a \\cdot K_b links the dissociation of an acid and its conjugate base through the relationship Kw=[H+][OH\u2212]K_w = [\\text{H}^+][\\text{OH}^-], a constant value at a given temperature (25\u00b0C, Kw=1\u00d710\u221214K_w = 1 \\times 10^{-14}). The logarithmic expressions, pKa=\u2212log\u2061KapK_a = -\\log K_a and pKb=\u2212log\u2061KbpK_b = -\\log K_b, reflect the inverse relationship between an acid’s strength and its conjugate base’s strength. Their sum always equals 1414, corresponding to the water ionization constant (pKw=14\\text{p}K_w = 14).<\/p>\n\n\n\n

            This mathematical proof underpins acid-base equilibrium concepts, demonstrating how KaK_a, KbK_b, and KwK_w are interrelated. It emphasizes that strong acids have weak conjugate bases and vice versa, maintaining balance in aqueous systems.<\/p>\n","protected":false},"excerpt":{"rendered":"

            Prove that: Ka * Kb = Kw and pKa + pKb = 14 Using the equations below. ?? = [HA][OH-] b[A AT1 2H20 -H30+ + OH- The Correct Answer and Explanation is : To prove Ka\u22c5Kb=KwK_a \\cdot K_b = K_w and pKa+pKb=14pK_a + pK_b = 14, let’s use the following information and equations: Derivation of […]<\/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-183811","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183811","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=183811"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183811\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183811"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183811"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183811"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}