{"id":202377,"date":"2025-03-17T12:41:24","date_gmt":"2025-03-17T12:41:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=202377"},"modified":"2025-03-17T12:41:26","modified_gmt":"2025-03-17T12:41:26","slug":"the-pka-of-hypochlorous-acid-is-7-530-a-53-0-ml-solution-of-0-100-m-sodium-hypochlorite-naocl-is-titrated-with-0-303-m-hcl","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/17\/the-pka-of-hypochlorous-acid-is-7-530-a-53-0-ml-solution-of-0-100-m-sodium-hypochlorite-naocl-is-titrated-with-0-303-m-hcl\/","title":{"rendered":"The pKa of hypochlorous acid is 7.530. A 53.0 mL solution of 0.100 M sodium hypochlorite (NaOCl) is titrated with 0.303 M HCl"},"content":{"rendered":"\n

The pKa of hypochlorous acid is 7.530. A 53.0 mL solution of 0.100 M sodium hypochlorite (NaOCl) is titrated with 0.303 M HCl. Calculate the pH of the solution<\/p>\n\n\n\n

a) after the addition of 6.47 mL of 0.303 M HCl.<\/p>\n\n\n\n

b) after the addition of 18.4 mL of 0.303 M HCl.<\/p>\n\n\n\n

c) at the equivalence point with 0.303 M HCl.<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

In this titration, sodium hypochlorite (NaOCl), the sodium salt of hypochlorous acid (HOCl), is neutralized by hydrochloric acid (HCl). The reaction between NaOCl and HCl forms HOCl and NaCl:<\/p>\n\n\n\n

[ \\text{NaOCl} + \\text{HCl} \\rightarrow \\text{HOCl} + \\text{NaCl} ]<\/p>\n\n\n\n

The pKa of HOCl is given as 7.530. We can apply the Henderson-Hasselbalch equation to calculate the pH at various stages of the titration:<\/p>\n\n\n\n

[ \\text{pH} = \\text{pKa} + \\log \\left( \\frac{[\\text{OCl}^-]}{[\\text{HOCl}]} \\right) ]<\/p>\n\n\n\n

a) After adding 6.47 mL of 0.303 M HCl:<\/strong><\/p>\n\n\n\n

    \n
  1. Initial moles of NaOCl:<\/strong> [ \\text{moles of NaOCl} = 0.100\\, \\text{M} \\times 0.053\\, \\text{L} = 0.00530\\, \\text{mol} ]<\/li>\n\n\n\n
  2. Moles of HCl added:<\/strong> [ \\text{moles of HCl} = 0.303\\, \\text{M} \\times 0.00647\\, \\text{L} = 0.00196\\, \\text{mol} ]<\/li>\n\n\n\n
  3. Moles of NaOCl remaining:<\/strong> [ \\text{remaining NaOCl} = 0.00530\\, \\text{mol} – 0.00196\\, \\text{mol} = 0.00334\\, \\text{mol} ]<\/li>\n\n\n\n
  4. Moles of HOCl formed:<\/strong> [ \\text{HOCl formed} = 0.00196\\, \\text{mol} ]<\/li>\n\n\n\n
  5. Total volume of the solution:<\/strong> [ \\text{total volume} = 53.0\\, \\text{mL} + 6.47\\, \\text{mL} = 59.47\\, \\text{mL} = 0.05947\\, \\text{L} ]<\/li>\n\n\n\n
  6. Concentrations of HOCl and OCl\u207b:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • [ [\\text{HOCl}] = \\frac{0.00196\\, \\text{mol}}{0.05947\\, \\text{L}} = 0.0329\\, \\text{M} ]<\/li>\n\n\n\n
    • [ [\\text{OCl}^-] = \\frac{0.00334\\, \\text{mol}}{0.05947\\, \\text{L}} = 0.0561\\, \\text{M} ]<\/li>\n<\/ul>\n\n\n\n
        \n
      1. pH calculation:<\/strong> [ \\text{pH} = 7.530 + \\log \\left( \\frac{0.0561}{0.0329} \\right) = 7.530 + \\log (1.705) \\approx 7.530 + 0.232 = 7.762 ]<\/li>\n<\/ol>\n\n\n\n

        b) After adding 18.4 mL of 0.303 M HCl:<\/strong><\/p>\n\n\n\n

          \n
        1. Moles of HCl added:<\/strong> [ \\text{moles of HCl} = 0.303\\, \\text{M} \\times 0.0184\\, \\text{L} = 0.00558\\, \\text{mol} ]<\/li>\n\n\n\n
        2. Moles of NaOCl remaining:<\/strong> [ \\text{remaining NaOCl} = 0.00530\\, \\text{mol} – 0.00558\\, \\text{mol} = -0.00028\\, \\text{mol} ] A negative value indicates that all NaOCl has reacted, and excess HCl remains.<\/li>\n\n\n\n
        3. Excess moles of HCl:<\/strong> [ \\text{excess HCl} = 0.00558\\, \\text{mol} – 0.00530\\, \\text{mol} = 0.00028\\, \\text{mol} ]<\/li>\n\n\n\n
        4. Total volume of the solution:<\/strong> [ \\text{total volume} = 53.0\\, \\text{mL} + 18.4\\, \\text{mL} = 71.4\\, \\text{mL} = 0.0714\\, \\text{L} ]<\/li>\n\n\n\n
        5. Concentration of excess HCl:<\/strong> [ [\\text{HCl}] = \\frac{0.00028\\, \\text{mol}}{0.0714\\, \\text{L}} = 0.00392\\, \\text{M} ]<\/li>\n\n\n\n
        6. pH calculation:<\/strong> [ \\text{pH} = -\\log (0.00392) \\approx 2.407 ]<\/li>\n<\/ol>\n\n\n\n

          c) At the equivalence point with 0.303 M HCl:<\/strong><\/p>\n\n\n\n

            \n
          1. Volume of HCl required to reach equivalence:<\/strong> [ \\text{volume of HCl} = \\frac{0.00530\\, \\text{mol}}{0.303\\, \\text{M}} = 0.0175\\, \\text{L} = 17.5\\, \\text{mL} ]<\/li>\n\n\n\n
          2. Concentration of HOCl at equivalence:<\/strong> [ [\\text{HOCl}] = \\frac{0.00530\\, \\text{mol}}{0.053\\, \\text{L} + 0.0175\\, \\text{L}} = \\frac{0.00530\\, \\text{mol}}{0.0705\\, \\text{L}} \\approx 0.0751\\, \\text{M} ]<\/li>\n\n\n\n
          3. pH calculation using the formula for a weak acid:<\/strong> [ \\text{pH} = \\frac{1}{2} \\left( \\text{pKa} – \\log [\\text{HOCl}] \\right) = \\frac{1}{2} \\left( 7.530 – \\log (0.0751)<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

            The pKa of hypochlorous acid is 7.530. A 53.0 mL solution of 0.100 M sodium hypochlorite (NaOCl) is titrated with 0.303 M HCl. Calculate the pH of the solution a) after the addition of 6.47 mL of 0.303 M HCl. b) after the addition of 18.4 mL of 0.303 M HCl. c) at the equivalence […]<\/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-202377","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202377","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=202377"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202377\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202377"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202377"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202377"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}