{"id":205374,"date":"2025-03-24T11:36:14","date_gmt":"2025-03-24T11:36:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=205374"},"modified":"2025-03-24T11:36:16","modified_gmt":"2025-03-24T11:36:16","slug":"the-ka-for-benzoic-acid-is-6-5-x-10%e2%88%925","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/24\/the-ka-for-benzoic-acid-is-6-5-x-10%e2%88%925\/","title":{"rendered":"The Ka for benzoic acid is 6.5 \u00d7 10\u22125 ."},"content":{"rendered":"\n

The Ka for benzoic acid is 6.5 \u00d7 10\u22125 . Calculate the pH of a 0.10 M benzoic acid solution.<\/p>\n\n\n\n

The Correct Answer And Explantion is:<\/mark><\/strong><\/p>\n\n\n\n

To calculate the pH of a 0.10 M benzoic acid (C6H5COOHC_6H_5COOHC6\u200bH5\u200bCOOH) solution, we use the acid dissociation constant (KaK_aKa\u200b) and set up an equilibrium expression.<\/p>\n\n\n\n

Step 1: Write the Dissociation Equation<\/h3>\n\n\n\n

Benzoic acid dissociates in water as: C6H5COOH\u21ccC6H5COO\u2212+H+C_6H_5COOH \\rightleftharpoons C_6H_5COO^- + H^+C6\u200bH5\u200bCOOH\u21ccC6\u200bH5\u200bCOO\u2212+H+<\/p>\n\n\n\n

Step 2: Set Up the ICE Table<\/h3>\n\n\n\n

Let the initial concentration of benzoic acid be 0.10 M. Before dissociation:<\/p>\n\n\n\n

    \n
  • [C6H5COOH]=0.10M[C_6H_5COOH] = 0.10 M[C6\u200bH5\u200bCOOH]=0.10M<\/li>\n\n\n\n
  • [C6H5COO\u2212]=0[C_6H_5COO^-] = 0[C6\u200bH5\u200bCOO\u2212]=0<\/li>\n\n\n\n
  • [H+]=0[H^+] = 0[H+]=0<\/li>\n<\/ul>\n\n\n\n

    Let xxx be the amount of benzoic acid that dissociates at equilibrium. Then, the equilibrium concentrations are:<\/p>\n\n\n\n

      \n
    • [C6H5COOH]=0.10\u2212x[C_6H_5COOH] = 0.10 – x[C6\u200bH5\u200bCOOH]=0.10\u2212x<\/li>\n\n\n\n
    • [C6H5COO\u2212]=x[C_6H_5COO^-] = x[C6\u200bH5\u200bCOO\u2212]=x<\/li>\n\n\n\n
    • [H+]=x[H^+] = x[H+]=x<\/li>\n<\/ul>\n\n\n\n

      Step 3: Write the KaK_aKa\u200b Expression<\/h3>\n\n\n\n

      Ka=[C6H5COO\u2212][H+][C6H5COOH]K_a = \\frac{[C_6H_5COO^-][H^+]}{[C_6H_5COOH]}Ka\u200b=[C6\u200bH5\u200bCOOH][C6\u200bH5\u200bCOO\u2212][H+]\u200b<\/p>\n\n\n\n

      Substituting values: 6.5\u00d710\u22125=x20.10\u2212x6.5 \\times 10^{-5} = \\frac{x^2}{0.10 – x}6.5\u00d710\u22125=0.10\u2212xx2\u200b<\/p>\n\n\n\n

      Since KaK_aKa\u200b is small, we approximate 0.10\u2212x\u22480.100.10 – x \\approx 0.100.10\u2212x\u22480.10: 6.5\u00d710\u22125=x20.106.5 \\times 10^{-5} = \\frac{x^2}{0.10}6.5\u00d710\u22125=0.10×2\u200b x2=(6.5\u00d710\u22125)\u00d70.10x^2 = (6.5 \\times 10^{-5}) \\times 0.10×2=(6.5\u00d710\u22125)\u00d70.10 x2=6.5\u00d710\u22126x^2 = 6.5 \\times 10^{-6}x2=6.5\u00d710\u22126 x=6.5\u00d710\u22126=2.55\u00d710\u22123x = \\sqrt{6.5 \\times 10^{-6}} = 2.55 \\times 10^{-3}x=6.5\u00d710\u22126\u200b=2.55\u00d710\u22123<\/p>\n\n\n\n

      Thus, [H+]=2.55\u00d710\u22123[H^+] = 2.55 \\times 10^{-3}[H+]=2.55\u00d710\u22123.<\/p>\n\n\n\n

      Step 4: Calculate pH<\/h3>\n\n\n\n

      pH=\u2212log\u2061[H+]\\text{pH} = -\\log [H^+]pH=\u2212log[H+] pH=\u2212log\u2061(2.55\u00d710\u22123)\\text{pH} = -\\log(2.55 \\times 10^{-3})pH=\u2212log(2.55\u00d710\u22123) pH\u22482.59\\text{pH} \\approx 2.59pH\u22482.59<\/p>\n\n\n\n

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

      The pH of a 0.10 M benzoic acid solution is calculated using the acid dissociation constant (KaK_aKa\u200b). Since benzoic acid is a weak acid, it only partially dissociates in water. We start by writing the dissociation equation and setting up an ICE table to determine the equilibrium concentrations of the species involved.<\/p>\n\n\n\n

      The acid dissociation constant equation is then used to solve for [H+][H^+][H+], assuming that the change in the initial concentration is negligible. This simplifies the quadratic equation to a straightforward square root calculation. By solving for xxx, which represents [H+][H^+][H+], we find it to be 2.55\u00d710\u221232.55 \\times 10^{-3}2.55\u00d710\u22123 M.<\/p>\n\n\n\n

      Using the pH formula, we take the negative logarithm of the hydrogen ion concentration and find the pH to be 2.59<\/strong>. This result makes sense because benzoic acid is a weak acid, and a 0.10 M solution should have a pH lower than 7 but higher than that of a strong acid of the same concentration.<\/p>\n\n\n\n

      In conclusion, the pH of a 0.10 M benzoic acid solution is 2.59<\/strong>, confirming the weak acidic nature of benzoic acid.<\/p>\n\n\n\n

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

      The Ka for benzoic acid is 6.5 \u00d7 10\u22125 . Calculate the pH of a 0.10 M benzoic acid solution. The Correct Answer And Explantion is: To calculate the pH of a 0.10 M benzoic acid (C6H5COOHC_6H_5COOHC6\u200bH5\u200bCOOH) solution, we use the acid dissociation constant (KaK_aKa\u200b) and set up an equilibrium expression. Step 1: Write 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-205374","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/205374","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=205374"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/205374\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=205374"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=205374"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=205374"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}