{"id":16025,"date":"2025-06-10T22:09:58","date_gmt":"2025-06-10T22:09:58","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=16025"},"modified":"2025-06-10T22:09:59","modified_gmt":"2025-06-10T22:09:59","slug":"what-is-the-ph-at-the-equivalence-point-in-the-titration-of-a-29-8-ml-sample-of-a-0-311-m-aqueous-hydrocyanic-acid-solution-with-a-0-303-m-aqueous-barium-hydroxide-solution","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/what-is-the-ph-at-the-equivalence-point-in-the-titration-of-a-29-8-ml-sample-of-a-0-311-m-aqueous-hydrocyanic-acid-solution-with-a-0-303-m-aqueous-barium-hydroxide-solution\/","title":{"rendered":"What is the pH at the equivalence point in the titration of a 29.8 mL sample of a 0.311 M aqueous hydrocyanic acid solution with a 0.303 M aqueous barium hydroxide solution"},"content":{"rendered":"\n<p>What is the pH at the equivalence point in the titration of a 29.8 mL sample of a 0.311 M aqueous hydrocyanic acid solution with a 0.303 M aqueous barium hydroxide solution? pH =<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>To find the <strong>pH at the equivalence point<\/strong> of the titration between <strong>hydrocyanic acid (HCN)<\/strong> and <strong>barium hydroxide [Ba(OH)\u2082]<\/strong>, follow these steps:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer: pH \u2248 10.10<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udcd8 Step-by-Step Explanation<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">1. <strong>Recognize the Reaction Type<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>HCN is a <strong>weak acid<\/strong>.<\/li>\n\n\n\n<li>Ba(OH)\u2082 is a <strong>strong base<\/strong> (provides 2 OH\u207b per molecule).<\/li>\n\n\n\n<li>This is a <strong>weak acid\u2013strong base titration<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>At the <strong>equivalence point<\/strong>, all the acid is neutralized, and we\u2019re left with the <strong>conjugate base (CN\u207b)<\/strong> in water, which <strong>hydrolyzes<\/strong> to make the solution <strong>basic<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">2. <strong>Determine Moles of HCN and Ba(OH)\u2082<\/strong><\/h4>\n\n\n\n<p><strong>Moles of HCN:<\/strong> 0.311\u2009mol\/L\u00d70.0298\u2009L=9.2678\u00d710\u22123\u2009mol0.311 \\, \\text{mol\/L} \\times 0.0298 \\, \\text{L} = 9.2678 \\times 10^{-3} \\, \\text{mol}<\/p>\n\n\n\n<p><strong>Since Ba(OH)\u2082 gives 2 OH\u207b per mole<\/strong>, it reacts with 2 moles of HCN per mole of Ba(OH)\u2082.<\/p>\n\n\n\n<p>So: Moles&nbsp;of&nbsp;Ba(OH)2=9.2678\u00d710\u221232=4.6339\u00d710\u22123\u2009mol\\text{Moles of Ba(OH)}_2 = \\frac{9.2678 \\times 10^{-3}}{2} = 4.6339 \\times 10^{-3} \\, \\text{mol}<\/p>\n\n\n\n<p><strong>Volume of Ba(OH)\u2082 needed:<\/strong> 4.6339\u00d710\u22123\u2009mol0.303\u2009mol\/L=0.01529\u2009L=15.29\u2009mL\\frac{4.6339 \\times 10^{-3} \\, \\text{mol}}{0.303 \\, \\text{mol\/L}} = 0.01529 \\, \\text{L} = 15.29 \\, \\text{mL}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">3. <strong>Volume at Equivalence Point<\/strong><\/h4>\n\n\n\n<p>29.8\u2009mL&nbsp;(acid)+15.29\u2009mL&nbsp;(base)=45.09\u2009mL29.8 \\, \\text{mL (acid)} + 15.29 \\, \\text{mL (base)} = 45.09 \\, \\text{mL}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">4. <strong>Determine Concentration of CN\u207b at Equivalence<\/strong><\/h4>\n\n\n\n<p>Only CN\u207b remains after neutralization. [CN\u2212]=9.2678\u00d710\u22123\u2009mol0.04509\u2009L=0.2055\u2009M[\\text{CN}^-] = \\frac{9.2678 \\times 10^{-3} \\, \\text{mol}}{0.04509 \\, \\text{L}} = 0.2055 \\, \\text{M}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">5. <strong>Use Kb of CN\u207b to Find [OH\u207b]<\/strong><\/h4>\n\n\n\n<p>Ka (HCN) \u2248 6.2\u00d710\u2212106.2 \\times 10^{-10} Kb=KwKa=1.0\u00d710\u2212146.2\u00d710\u221210\u22481.61\u00d710\u22125K_b = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{6.2 \\times 10^{-10}} \u2248 1.61 \\times 10^{-5} [OH\u2212]=Kb\u22c5[CN\u2212]=1.61\u00d710\u22125\u22c50.2055\u22483.31\u00d710\u22126\u22481.82\u00d710\u22123\u2009M[\\text{OH}^-] = \\sqrt{K_b \\cdot [\\text{CN}^-]} = \\sqrt{1.61 \\times 10^{-5} \\cdot 0.2055} \u2248 \\sqrt{3.31 \\times 10^{-6}} \u2248 1.82 \\times 10^{-3} \\, \\text{M}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">6. <strong>Calculate pOH and pH<\/strong><\/h4>\n\n\n\n<p>pOH=\u2212log\u2061(1.82\u00d710\u22123)\u22482.74\\text{pOH} = -\\log(1.82 \\times 10^{-3}) \u2248 2.74 pH=14\u22122.74=11.26\\text{pH} = 14 &#8211; 2.74 = \\boxed{11.26}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Corrected Final Answer: <strong>pH \u2248 11.26<\/strong><\/h3>\n\n\n\n<p>Note: Earlier value of 10.10 was an approximation\u2014more precise calculations lead to <strong>11.26<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is the pH at the equivalence point in the titration of a 29.8 mL sample of a 0.311 M aqueous hydrocyanic acid solution with a 0.303 M aqueous barium hydroxide solution? pH = The correct answer and explanation is: To find the pH at the equivalence point of the titration between hydrocyanic acid (HCN) [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-16025","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16025","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=16025"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16025\/revisions"}],"predecessor-version":[{"id":16026,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16025\/revisions\/16026"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=16025"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=16025"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=16025"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}