{"id":183813,"date":"2025-01-17T06:22:31","date_gmt":"2025-01-17T06:22:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183813"},"modified":"2025-01-17T06:22:33","modified_gmt":"2025-01-17T06:22:33","slug":"calculate-the-kb-for-the-base-c2h5nh2-if-the-ka-for-the-corresponding-acid-is-1-8-x-10-11","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/17\/calculate-the-kb-for-the-base-c2h5nh2-if-the-ka-for-the-corresponding-acid-is-1-8-x-10-11\/","title":{"rendered":"Calculate the Kb for the base C2H5NH2 if the Ka for the corresponding acid is 1.8 x 10-11"},"content":{"rendered":"\n

Calculate the Kb for the base C2H5NH2 if the Ka for the corresponding acid is 1.8 x 10-11 Determine the [OH-], [H3O +], and the pH of a solution that has a pOH of 9.50. The pH of 0.10 M formic acid, HCOOH, is 2.39. What is the Ka of this acid? What is the pH of a 0.40 M NH3 solution? Kb = 1.8 x 10-5. Find the pH of a 0.036 M nitrous acid (HNO2) solution. Ka = 4.5 x 10-4 HNO2 + H2O H3O+ + NO2- What is the equilibrium concentration of HNO2 and NO2- in the solution? Calculate the percent ionization for the reaction in question 12. Calculate the pH and the concentration o f all species in solution at equilibrium for a 0.15 M solution of ascorbic acid, or H2C6H6O6. For ascorbic acid, Ka1 = 8.0 10-5 and Ka2 = 1.6 10-12 Calculate the pH of a buffer that was prepared by mixing 25.0 mL of M HAc and 20.0 mL 0.150 M Ac. The Ka for the reaction is 1.8 x 10-5 What is the pH o f the buffer in question 13 after 5.0 mL of 0.075 M HCL is added to it? Calculate the pH at the following points in the titration of 50.0 mL of 0.15 M HC2H3O with 0.15 M NaOH Ka = 1.8 x 10-5 After the addition o f 25.0 mL of 0.15 M NaOH. After the addition of 50.0 mL of 0.15 M NaOH After the addition of 60.0 mL of 0.15 M NaOH.<\/p>\n\n\n\n

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

This is a multi-part question, and I’ll break it down step by step to calculate the required values. Here is the solution and explanation for each part.<\/p>\n\n\n\n

\n\n\n\n

1. Calculate KbK_b for C2H5NH2C_2H_5NH_2:<\/strong><\/h3>\n\n\n\n

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

    \n
  • KaK_a for C2H5NH3+=1.8\u00d710\u221211C_2H_5NH_3^+ = 1.8 \\times 10^{-11}<\/li>\n\n\n\n
  • Kw=1.0\u00d710\u221214K_w = 1.0 \\times 10^{-14}<\/li>\n<\/ul>\n\n\n\n

    Kb=KwKa=1.0\u00d710\u2212141.8\u00d710\u221211=5.56\u00d710\u22124K_b = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{1.8 \\times 10^{-11}} = 5.56 \\times 10^{-4}<\/p>\n\n\n\n

    \n\n\n\n

    2. Determine [OH\u207b], [H\u2083O\u207a], and pH if pOH=9.50pOH = 9.50:<\/strong><\/h3>\n\n\n\n

    [OH\u2212]=10\u2212pOH=10\u22129.50=3.16\u00d710\u221210[OH^-] = 10^{-\\text{pOH}} = 10^{-9.50} = 3.16 \\times 10^{-10} [H3O+]=Kw[OH\u2212]=1.0\u00d710\u2212143.16\u00d710\u221210=3.16\u00d710\u22125[H_3O^+] = \\frac{K_w}{[OH^-]} = \\frac{1.0 \\times 10^{-14}}{3.16 \\times 10^{-10}} = 3.16 \\times 10^{-5} pH=14\u2212pOH=14\u22129.50=4.50\\text{pH} = 14 – \\text{pOH} = 14 – 9.50 = 4.50<\/p>\n\n\n\n

    \n\n\n\n

    3. Calculate KaK_a for formic acid (HCOOH):<\/strong><\/h3>\n\n\n\n

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

      \n
    • [HCOOH]=0.10\u2009M[HCOOH] = 0.10 \\, \\text{M}<\/li>\n\n\n\n
    • pH = 2.39<\/li>\n<\/ul>\n\n\n\n

      [H3O+]=10\u2212pH=10\u22122.39=4.07\u00d710\u22123[H_3O^+] = 10^{-\\text{pH}} = 10^{-2.39} = 4.07 \\times 10^{-3}<\/p>\n\n\n\n

      Assume x=[H3O+]=[HCOO\u2212]x = [H_3O^+] = [HCOO^-] at equilibrium: Ka=x2[HCOOH]\u2212x\u2248(4.07\u00d710\u22123)20.10=1.66\u00d710\u22124K_a = \\frac{x^2}{[HCOOH] – x} \\approx \\frac{(4.07 \\times 10^{-3})^2}{0.10} = 1.66 \\times 10^{-4}<\/p>\n\n\n\n

      \n\n\n\n

      4. Calculate pH of 0.40 M NH3NH_3:<\/strong><\/h3>\n\n\n\n

      Given Kb=1.8\u00d710\u22125K_b = 1.8 \\times 10^{-5}: [OH\u2212]=Kb\u22c5[NH3]=(1.8\u00d710\u22125)(0.40)=2.68\u00d710\u22123[OH^-] = \\sqrt{K_b \\cdot [NH_3]} = \\sqrt{(1.8 \\times 10^{-5})(0.40)} = 2.68 \\times 10^{-3} pOH=\u2212log\u2061[OH\u2212]=\u2212log\u2061(2.68\u00d710\u22123)=2.57\\text{pOH} = -\\log[OH^-] = -\\log(2.68 \\times 10^{-3}) = 2.57 pH=14\u2212pOH=14\u22122.57=11.43\\text{pH} = 14 – \\text{pOH} = 14 – 2.57 = 11.43<\/p>\n\n\n\n

      \n\n\n\n

      5. pH of 0.036 M HNO2HNO_2:<\/strong><\/h3>\n\n\n\n

      Given Ka=4.5\u00d710\u22124K_a = 4.5 \\times 10^{-4}: [H3O+]=Ka\u22c5[HNO2]=(4.5\u00d710\u22124)(0.036)=4.02\u00d710\u22123[H_3O^+] = \\sqrt{K_a \\cdot [HNO_2]} = \\sqrt{(4.5 \\times 10^{-4})(0.036)} = 4.02 \\times 10^{-3} pH=\u2212log\u2061[H3O+]=\u2212log\u2061(4.02\u00d710\u22123)=2.40\\text{pH} = -\\log[H_3O^+] = -\\log(4.02 \\times 10^{-3}) = 2.40<\/p>\n\n\n\n

      Equilibrium Concentrations:<\/h4>\n\n\n\n
        \n
      • [HNO2]=0.036\u2212[H3O+]\u22480.036\u22120.00402=0.03198\u2009M[HNO_2] = 0.036 – [H_3O^+] \\approx 0.036 – 0.00402 = 0.03198 \\, \\text{M}<\/li>\n\n\n\n
      • [NO2\u2212]=[H3O+]=4.02\u00d710\u22123\u2009M[NO_2^-] = [H_3O^+] = 4.02 \\times 10^{-3} \\, \\text{M}<\/li>\n<\/ul>\n\n\n\n

        Percent Ionization:<\/h4>\n\n\n\n

        %Ionization=[H3O+][HNO2]initial\u00d7100=4.02\u00d710\u221230.036\u00d7100=11.17%\\% \\text{Ionization} = \\frac{[H_3O^+]}{[HNO_2]_{\\text{initial}}} \\times 100 = \\frac{4.02 \\times 10^{-3}}{0.036} \\times 100 = 11.17\\%<\/p>\n\n\n\n

        \n\n\n\n

        6. pH and species concentrations for 0.15 M ascorbic acid (H2C6H6O6H_2C_6H_6O_6):<\/strong><\/h3>\n\n\n\n

        First dissociation:<\/h4>\n\n\n\n

        [H3O+]=Ka1\u22c5[H2C6H6O6]=(8.0\u00d710\u22125)(0.15)=3.46\u00d710\u22123[H_3O^+] = \\sqrt{K_{a1} \\cdot [H_2C_6H_6O_6]} = \\sqrt{(8.0 \\times 10^{-5})(0.15)} = 3.46 \\times 10^{-3} pH=\u2212log\u2061[H3O+]=2.46\\text{pH} = -\\log[H_3O^+] = 2.46<\/p>\n\n\n\n

        Second dissociation is negligible due to Ka2K_{a2}, so:<\/h4>\n\n\n\n
          \n
        • [H3O+]=3.46\u00d710\u22123\u2009M[H_3O^+] = 3.46 \\times 10^{-3} \\, \\text{M}<\/li>\n\n\n\n
        • [H2C6H6O6]=0.15\u22123.46\u00d710\u22123\u22480.147\u2009M[H_2C_6H_6O_6] = 0.15 – 3.46 \\times 10^{-3} \\approx 0.147 \\, \\text{M}<\/li>\n\n\n\n
        • [HC6H6O6\u2212]=3.46\u00d710\u22123\u2009M[HC_6H_6O_6^-] = 3.46 \\times 10^{-3} \\, \\text{M}<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          7. Buffer pH (HAc and Ac\u207b):<\/strong><\/h3>\n\n\n\n

          Using the Henderson-Hasselbalch equation: pH=pKa+log\u2061([A\u2212][HA])\\text{pH} = \\text{p}K_a + \\log\\left(\\frac{[A^-]}{[HA]}\\right) pH=\u2212log\u2061(1.8\u00d710\u22125)+log\u2061((20.0\u2009mL\u00d70.150\u2009M)(25.0\u2009mL\u00d70.100\u2009M))=4.74+log\u2061(1.2)=4.83\\text{pH} = -\\log(1.8 \\times 10^{-5}) + \\log\\left(\\frac{(20.0 \\, \\text{mL} \\times 0.150 \\, \\text{M})}{(25.0 \\, \\text{mL} \\times 0.100 \\, \\text{M})}\\right) = 4.74 + \\log(1.2) = 4.83<\/p>\n\n\n\n

          \n\n\n\n

          8. Buffer pH after adding HClHCl:<\/strong><\/h3>\n\n\n\n

          Moles of HCl=5.0\u2009mL\u00d70.075\u2009M=0.375\u2009mmol\\text{Moles of } HCl = 5.0 \\, \\text{mL} \\times 0.075 \\, \\text{M} = 0.375 \\, \\text{mmol} New [A\u2212]=3.0\u22120.375=2.625\u2009mmol\\text{New } [A^-] = 3.0 – 0.375 = 2.625 \\, \\text{mmol} New [HA]=2.5+0.375=2.875\u2009mmol\\text{New } [HA] = 2.5 + 0.375 = 2.875 \\, \\text{mmol} pH=4.74+log\u2061(2.6252.875)=4.74\u22120.04=4.70\\text{pH} = 4.74 + \\log\\left(\\frac{2.625}{2.875}\\right) = 4.74 – 0.04 = 4.70<\/p>\n\n\n\n

          \n\n\n\n

          9. Titration pH calculations:<\/strong><\/h3>\n\n\n\n

          (a) After 25.0 mL NaOH:<\/strong> Half-equivalence point: pH=pKa=4.74\\text{pH} = \\text{p}K_a = 4.74<\/p>\n\n\n\n

          (b) After 50.0 mL NaOH:<\/strong> Equivalence point: pH=7+12(pKb\u2212pKa)\\text{pH} = 7 + \\frac{1}{2}\\left(\\text{p}K_b – \\text{p}K_a\\right) Kb=KwKa=5.56\u00d710\u221210K_b = \\frac{K_w}{K_a} = 5.56 \\times 10^{-10} pH\u22488.87\\text{pH} \\approx 8.87<\/p>\n\n\n\n

          (c) After 60.0 mL NaOH:<\/strong> Excess OH\u2212OH^-: [OH\u2212]=10\u2009mmol110\u2009mL=9.09\u00d710\u22124[OH^-] = \\frac{10 \\, \\text{mmol}}{110 \\, \\text{mL}} = 9.09 \\times 10^{-4} pOH=\u2212log\u2061[OH\u2212]=3.04\\text{pOH} = -\\log[OH^-] = 3.04 pH=14\u22123.04=10.96\\text{pH} = 14 – 3.04 = 10.96<\/p>\n","protected":false},"excerpt":{"rendered":"

          Calculate the Kb for the base C2H5NH2 if the Ka for the corresponding acid is 1.8 x 10-11 Determine the [OH-], [H3O +], and the pH of a solution that has a pOH of 9.50. The pH of 0.10 M formic acid, HCOOH, is 2.39. What is the Ka of this acid? What is the […]<\/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-183813","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183813","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=183813"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183813\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183813"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183813"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183813"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}