{"id":28976,"date":"2025-06-20T18:31:14","date_gmt":"2025-06-20T18:31:14","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=28976"},"modified":"2025-06-20T18:31:16","modified_gmt":"2025-06-20T18:31:16","slug":"what-is-the-ph-of-a-0-10-m-solution-of-sodium-cyanide-nacn","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/what-is-the-ph-of-a-0-10-m-solution-of-sodium-cyanide-nacn\/","title":{"rendered":"What is the pH of a 0.10 M solution of sodium cyanide (NaCN"},"content":{"rendered":"\n

What is the pH of a 0.10 M solution of sodium cyanide (NaCN)? The pKa of hydrocyanic acid (HCN) is 9.21. A. 4.37 B. 5.43 C. 9.62 D. 11.1 I already know the answer but I was hoping someone could explain why?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Correct Answer: D. 11.1<\/strong><\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n

Sodium cyanide (NaCN) is a salt of a weak acid (HCN)<\/strong> and a strong base (NaOH). When NaCN dissolves in water, it dissociates completely into Na\u207a and CN\u207b ions. The Na\u207a ion does not affect the pH, but the CN\u207b ion can react with water<\/strong> in a process called hydrolysis<\/strong>:CN\u2212+H2O\u21ccHCN+OH\u2212\\text{CN}^- + \\text{H}_2\\text{O} \\rightleftharpoons \\text{HCN} + \\text{OH}^-CN\u2212+H2\u200bO\u21ccHCN+OH\u2212<\/p>\n\n\n\n

This reaction produces hydroxide ions (OH\u207b), which makes the solution basic<\/strong>. To find the pH, we follow these steps:<\/p>\n\n\n\n

\n\n\n\n

Step 1: Find the Kb<\/strong> for CN\u207b<\/h3>\n\n\n\n

We are given the pKa of HCN = 9.21<\/strong>. Use the relationship:pKa+pKb=14\u21d2pKb=14\u22129.21=4.79\\text{pKa} + \\text{pKb} = 14 \\Rightarrow \\text{pKb} = 14 – 9.21 = 4.79pKa+pKb=14\u21d2pKb=14\u22129.21=4.79<\/p>\n\n\n\n

Now convert pKb to Kb:Kb=10\u22124.79\u22481.62\u00d710\u22125\\text{Kb} = 10^{-4.79} \\approx 1.62 \\times 10^{-5}Kb=10\u22124.79\u22481.62\u00d710\u22125<\/p>\n\n\n\n

\n\n\n\n

Step 2: Use the ICE table<\/strong> and set up the expression:<\/h3>\n\n\n\n

Let CN\u207b = 0.10 M initially.Kb=[OH\u2212]2[CN\u2212]=x20.10\\text{Kb} = \\frac{[OH^-]^2}{[CN^-]} = \\frac{x^2}{0.10}Kb=[CN\u2212][OH\u2212]2\u200b=0.10×2\u200b1.62\u00d710\u22125=x20.10\u21d2x2=1.62\u00d710\u22126\u21d2x=[OH\u2212]=1.62\u00d710\u22126\u22481.27\u00d710\u221231.62 \\times 10^{-5} = \\frac{x^2}{0.10} \\Rightarrow x^2 = 1.62 \\times 10^{-6} \\Rightarrow x = [OH^-] = \\sqrt{1.62 \\times 10^{-6}} \\approx 1.27 \\times 10^{-3}1.62\u00d710\u22125=0.10×2\u200b\u21d2x2=1.62\u00d710\u22126\u21d2x=[OH\u2212]=1.62\u00d710\u22126\u200b\u22481.27\u00d710\u22123<\/p>\n\n\n\n

\n\n\n\n

Step 3: Find the pOH<\/strong> and then the pH<\/strong>:<\/h3>\n\n\n\n

pOH=\u2212log\u2061(1.27\u00d710\u22123)\u22482.90\\text{pOH} = -\\log(1.27 \\times 10^{-3}) \\approx 2.90pOH=\u2212log(1.27\u00d710\u22123)\u22482.90pH=14\u22122.90=11.1\\text{pH} = 14 – 2.90 = 11.1pH=14\u22122.90=11.1<\/p>\n\n\n\n

\n\n\n\n

Conclusion:<\/h3>\n\n\n\n

Since CN\u207b produces OH\u207b through hydrolysis, the solution becomes basic, and the pH rises above 7. With calculations complete, the pH of a 0.10 M NaCN solution is 11.1<\/strong>, which matches choice D<\/strong>.<\/p>\n\n\n\n

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

What is the pH of a 0.10 M solution of sodium cyanide (NaCN)? The pKa of hydrocyanic acid (HCN) is 9.21. A. 4.37 B. 5.43 C. 9.62 D. 11.1 I already know the answer but I was hoping someone could explain why? The Correct Answer and Explanation is: Correct Answer: D. 11.1 Explanation: Sodium cyanide […]<\/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-28976","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28976","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=28976"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28976\/revisions"}],"predecessor-version":[{"id":28978,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/28976\/revisions\/28978"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=28976"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=28976"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=28976"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}