Hypochlorous acid (HOCl) is a weak acid with pKa = 7.54. Suppose a 45.00 mL sample of 0.240 M HOCl solution is titrated with a 0.250 M standard KOH solution. What is the pH at the equivalence point?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n At the equivalence point of a titration between a weak acid like hypochlorous acid (HOCl) and a strong base like potassium hydroxide (KOH), the resulting solution contains the conjugate base of the weak acid\u2014in this case, the hypochlorite ion (OCl\u207b). This conjugate base undergoes hydrolysis, reacting with water to produce hydroxide ions (OH\u207b), which makes the solution basic. Consequently, the pH at the equivalence point will be greater than 7.<\/p>\n\n\n\n Calculating the pH at the Equivalence Point:<\/strong><\/p>\n\n\n\n Therefore, the pH at the equivalence point is approximately 10.77<\/strong>, indicating a basic solution due to the hydrolysis of the hypochlorite ion.<\/p>\n","protected":false},"excerpt":{"rendered":" Hypochlorous acid (HOCl) is a weak acid with pKa = 7.54. Suppose a 45.00 mL sample of 0.240 M HOCl solution is titrated with a 0.250 M standard KOH solution. What is the pH at the equivalence point? The correct answer and explanation is : At the equivalence point of a titration between a weak […]<\/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-202379","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202379","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=202379"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202379\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202379"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202379"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202379"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
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[ \\text{OCl}^- + \\text{H}_2\\text{O} \\rightleftharpoons \\text{HOCl} + \\text{OH}^- ]<\/li>\n\n\n\n
[ K_b = \\frac{K_w}{K_a} ]
Given that the pKa of HOCl is 7.54,
[ K_a = 10^{-7.54} \u2248 3.61 \\times 10^{-8} ]
Therefore,
[ K_b = \\frac{1.0 \\times 10^{-14}}{3.61 \\times 10^{-8}} \u2248 2.77 \\times 10^{-7} ]<\/li>\n\n\n\n
[ K_b = \\frac{[\\text{HOCl}][\\text{OH}^-]}{[\\text{OCl}^-]} ]
Assuming x is the concentration of OH\u207b produced:
[ 2.77 \\times 10^{-7} = \\frac{x^2}{0.1225} ]
Solving for x:
[ x^2 = 2.77 \\times 10^{-7} \\times 0.1225 = 3.40 \\times 10^{-8} ]
[ x = \\sqrt{3.40 \\times 10^{-8}} \u2248 5.83 \\times 10^{-4} \\, \\text{M} ]<\/li>\n\n\n\n
[ \\text{pOH} = -\\log(5.83 \\times 10^{-4}) \u2248 3.23 ]<\/li>\n\n\n\n
[ \\text{pH} = 14 – \\text{pOH} = 14 – 3.23 = 10.77 ]<\/li>\n<\/ul>\n\n\n\n