{"id":220187,"date":"2025-05-27T19:45:45","date_gmt":"2025-05-27T19:45:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220187"},"modified":"2025-05-27T19:45:47","modified_gmt":"2025-05-27T19:45:47","slug":"calculate-the-ph-of-a-solution-made-by-mixing-50-00-ml-of-0-100-m-nacn-with-a-4-20-ml-of-0-438-m-hclo4-b-11-82-ml-of-0-438-m","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/27\/calculate-the-ph-of-a-solution-made-by-mixing-50-00-ml-of-0-100-m-nacn-with-a-4-20-ml-of-0-438-m-hclo4-b-11-82-ml-of-0-438-m\/","title":{"rendered":"Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN with (a) 4.20 mL of 0.438 M HClO4 (b) 11.82 mL of 0.438 M"},"content":{"rendered":"\n<p>. Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN with (a) 4.20 mL of 0.438 M HClO4 (b) 11.82 mL of 0.438 M HClO4 (c) What is the pH at the equivalence point with 0.438 M HClO4?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">T he Correct Answer and Explanation is :<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the pH of the solution after mixing NaCN (a weak base) and HClO\u2084 (a strong acid), we consider the acid-base reaction:<\/p>\n\n\n\n<p><strong>CN\u207b + H\u2083O\u207a \u2192 HCN + H\u2082O<\/strong><\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of NaCN = 50.00 mL = 0.05000 L<\/li>\n\n\n\n<li>Molarity of NaCN = 0.100 M<\/li>\n\n\n\n<li>Moles of CN\u207b = 0.100 mol\/L \u00d7 0.05000 L = <strong>0.005000 mol<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">(a) Add 4.20 mL of 0.438 M HClO\u2084<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume HClO\u2084 = 4.20 mL = 0.00420 L<\/li>\n\n\n\n<li>Moles HClO\u2084 = 0.438 mol\/L \u00d7 0.00420 L = <strong>0.001840 mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Since CN\u207b is in excess, this is a <strong>buffer<\/strong> solution (both CN\u207b and HCN present):<\/p>\n\n\n\n<p>Moles CN\u207b after reaction = 0.005000 &#8211; 0.001840 = <strong>0.003160 mol<\/strong><br>Moles HCN formed = <strong>0.001840 mol<\/strong><\/p>\n\n\n\n<p>Use the Henderson-Hasselbalch equation:<\/p>\n\n\n\n<p><strong>pKa of HCN = 9.21<\/strong> (Ka = 6.2 \u00d7 10\u207b\u00b9\u2070) pH=pKa+log\u2061([base][acid])=9.21+log\u2061(0.0031600.001840)\\text{pH} = \\text{pKa} + \\log\\left(\\frac{[\\text{base}]}{[\\text{acid}]}\\right) = 9.21 + \\log\\left(\\frac{0.003160}{0.001840}\\right) pH\u22489.21+log\u2061(1.717)\u22489.21+0.235=9.45\\text{pH} \u2248 9.21 + \\log(1.717) \u2248 9.21 + 0.235 = \\boxed{9.45}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(b) Add 11.82 mL of 0.438 M HClO\u2084<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume HClO\u2084 = 11.82 mL = 0.01182 L<\/li>\n\n\n\n<li>Moles HClO\u2084 = 0.438 mol\/L \u00d7 0.01182 L = <strong>0.005176 mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>All CN\u207b reacts and HClO\u2084 is in excess:<\/p>\n\n\n\n<p>Moles excess HClO\u2084 = 0.005176 &#8211; 0.005000 = <strong>0.000176 mol<\/strong><\/p>\n\n\n\n<p>Total volume = 50.00 mL + 11.82 mL = 61.82 mL = 0.06182 L<br>[H\u2083O\u207a] = 0.000176 mol \/ 0.06182 L \u2248 2.847 \u00d7 10\u207b\u00b3 M pH=\u2212log\u2061(2.847\u00d710\u22123)\u22482.55\\text{pH} = -\\log(2.847 \\times 10^{-3}) \\approx \\boxed{2.55}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(c) Equivalence point with 0.438 M HClO\u2084<\/h3>\n\n\n\n<p>Moles CN\u207b = 0.005000 mol<br>To reach equivalence, moles HClO\u2084 needed = 0.005000 mol<br>Volume HClO\u2084 = 0.005000 mol \/ 0.438 mol\/L = <strong>0.01142 L = 11.42 mL<\/strong><\/p>\n\n\n\n<p>At equivalence, all CN\u207b is converted to <strong>HCN<\/strong> (a weak acid). Now calculate pH from HCN dissociation: Ka=6.2\u00d710\u221210,total&nbsp;volume=61.42&nbsp;mL=0.06142&nbsp;L\\text{Ka} = 6.2 \\times 10^{-10},\\quad \\text{total volume} = 61.42 \\text{ mL} = 0.06142 \\text{ L} [HCN]=0.0050000.06142=0.0814&nbsp;M[\\text{HCN}] = \\frac{0.005000}{0.06142} = 0.0814\\ \\text{M}<\/p>\n\n\n\n<p>Set up ICE table and solve: Ka=x20.0814=6.2\u00d710\u221210\u21d2x=[H\u2083O\u207a]=6.2\u00d710\u221210\u00d70.0814=7.1\u00d710\u22126\\text{Ka} = \\frac{x^2}{0.0814} = 6.2 \\times 10^{-10} \\Rightarrow x = [\\text{H\u2083O\u207a}] = \\sqrt{6.2 \\times 10^{-10} \\times 0.0814} = 7.1 \\times 10^{-6} pH=\u2212log\u2061(7.1\u00d710\u22126)\u22485.15\\text{pH} = -\\log(7.1 \\times 10^{-6}) \u2248 \\boxed{5.15}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>(a) pH = 9.45<\/strong> (buffer solution)<\/li>\n\n\n\n<li><strong>(b) pH = 2.55<\/strong> (excess strong acid)<\/li>\n\n\n\n<li><strong>(c) pH = 5.15<\/strong> (weak acid at equivalence)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>This problem involves a weak base (NaCN, providing CN\u207b) reacting with a strong acid (HClO\u2084). The key idea is to determine the pH based on the mole ratio of base to acid and understand the solution&#8217;s nature (buffer, excess acid, or equivalence).<\/p>\n\n\n\n<p>In <strong>(a)<\/strong>, less acid is added than base present. The strong acid partially neutralizes CN\u207b, forming HCN. Since both the weak acid (HCN) and its conjugate base (CN\u207b) coexist, the result is a <strong>buffer solution<\/strong>. We apply the <strong>Henderson-Hasselbalch equation<\/strong> to find pH, which reflects the buffer&#8217;s resistance to pH change.<\/p>\n\n\n\n<p>In <strong>(b)<\/strong>, more HClO\u2084 is added than the available CN\u207b. All CN\u207b is neutralized, and the excess HClO\u2084 determines the solution\u2019s pH. Since it&#8217;s a strong acid, the pH is directly calculated from the hydronium ion concentration using <strong>pH = -log[H\u2083O\u207a]<\/strong>. The solution is now <strong>acidic<\/strong>, with no buffering capacity.<\/p>\n\n\n\n<p>In <strong>(c)<\/strong>, the amount of acid exactly neutralizes the base, forming HCN only. HCN, being a <strong>weak acid<\/strong>, partially ionizes, producing some H\u2083O\u207a ions. The pH is calculated using the <strong>Ka of HCN<\/strong> and solving the equilibrium expression from an <strong>ICE table<\/strong>. This results in a pH typical of a weak acid\u2014lower than neutral but not as low as strong acid solutions.<\/p>\n\n\n\n<p>Understanding these scenarios is fundamental in acid-base chemistry, particularly in <strong>titration curves<\/strong> and <strong>buffer systems<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner4-47.jpeg\" alt=\"\" class=\"wp-image-220188\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>. Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN with (a) 4.20 mL of 0.438 M HClO4 (b) 11.82 mL of 0.438 M HClO4 (c) What is the pH at the equivalence point with 0.438 M HClO4? T he Correct Answer and Explanation is : To calculate the [&hellip;]<\/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-220187","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220187","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=220187"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220187\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220187"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220187"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220187"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}