{"id":31849,"date":"2025-06-22T08:59:13","date_gmt":"2025-06-22T08:59:13","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=31849"},"modified":"2025-06-22T08:59:19","modified_gmt":"2025-06-22T08:59:19","slug":"the-pka-of-hf-is-3-17","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-pka-of-hf-is-3-17\/","title":{"rendered":"The pKa of HF is 3.17"},"content":{"rendered":"\n<p>The pKa of HF is 3.17. a. Calculate the pH of a 1.00 L solution that is 1.00 M HF and 1.50 M NaF. b. What is the pH of this solution after addition of 40.0 ml of 10.0 M HCl?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s solve this step-by-step.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>pKapK_a of HF = 3.17<\/li>\n\n\n\n<li>Initial solution: 1.00 L of 1.00 M HF and 1.50 M NaF<\/li>\n\n\n\n<li>Volume of HCl added = 40.0 mL (0.0400 L) of 10.0 M HCl<\/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\">Part (a): Calculate the pH of the buffer solution<\/h3>\n\n\n\n<p>This is a classic buffer problem containing a weak acid (HF) and its conjugate base (F\u207b from NaF). The pH can be found using the Henderson-Hasselbalch equation: pH=pKa+log\u2061([A\u2212][HA])pH = pK_a + \\log\\left(\\frac{[\\text{A}^-]}{[\\text{HA}]}\\right)<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>[HA][\\text{HA}] = concentration of HF = 1.00 M<\/li>\n\n\n\n<li>[A\u2212][\\text{A}^-] = concentration of F\u207b from NaF = 1.50 M<\/li>\n<\/ul>\n\n\n\n<p>Substitute values: pH=3.17+log\u2061(1.501.00)pH = 3.17 + \\log\\left(\\frac{1.50}{1.00}\\right)<\/p>\n\n\n\n<p>Calculate the logarithm: log\u2061(1.50)=0.176\\log(1.50) = 0.176<\/p>\n\n\n\n<p>Therefore: pH=3.17+0.176=3.35pH = 3.17 + 0.176 = 3.35<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Part (b): Calculate the pH after addition of 40.0 mL of 10.0 M HCl<\/h3>\n\n\n\n<p>First, calculate moles of species before HCl addition:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles HF initially = 1.00\u2009mol\/L\u00d71.00\u2009L=1.00\u2009mol1.00 \\, \\text{mol\/L} \\times 1.00 \\, \\text{L} = 1.00 \\, \\text{mol}<\/li>\n\n\n\n<li>Moles F\u207b initially = 1.50\u2009mol\/L\u00d71.00\u2009L=1.50\u2009mol1.50 \\, \\text{mol\/L} \\times 1.00 \\, \\text{L} = 1.50 \\, \\text{mol}<\/li>\n<\/ul>\n\n\n\n<p>Moles of HCl added: 0.0400\u2009L\u00d710.0\u2009mol\/L=0.400\u2009mol&nbsp;HCl0.0400 \\, \\text{L} \\times 10.0 \\, \\text{mol\/L} = 0.400 \\, \\text{mol HCl}<\/p>\n\n\n\n<p>HCl is a strong acid and will react with the base (F\u207b) to form HF: F\u2212+H+\u2192HF\\text{F}^- + \\text{H}^+ \\rightarrow \\text{HF}<\/p>\n\n\n\n<p>Calculate new moles after reaction:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>New moles F\u207b = 1.50\u22120.400=1.10\u2009mol1.50 &#8211; 0.400 = 1.10 \\, \\text{mol}<\/li>\n\n\n\n<li>New moles HF = 1.00+0.400=1.40\u2009mol1.00 + 0.400 = 1.40 \\, \\text{mol}<\/li>\n<\/ul>\n\n\n\n<p>The total volume after adding HCl is: 1.00\u2009L+0.0400\u2009L=1.040\u2009L1.00 \\, \\text{L} + 0.0400 \\, \\text{L} = 1.040 \\, \\text{L}<\/p>\n\n\n\n<p>Calculate new concentrations: [F\u2212]=1.101.040=1.058\u2009M[\\text{F}^-] = \\frac{1.10}{1.040} = 1.058 \\, M [HF]=1.401.040=1.346\u2009M[\\text{HF}] = \\frac{1.40}{1.040} = 1.346 \\, M<\/p>\n\n\n\n<p>Use Henderson-Hasselbalch equation again: pH=3.17+log\u2061(1.0581.346)pH = 3.17 + \\log\\left(\\frac{1.058}{1.346}\\right)<\/p>\n\n\n\n<p>Calculate the ratio and logarithm: 1.0581.346=0.786\\frac{1.058}{1.346} = 0.786 log\u2061(0.786)=\u22120.104\\log(0.786) = -0.104<\/p>\n\n\n\n<p>Thus: pH=3.17\u22120.104=3.07pH = 3.17 &#8211; 0.104 = 3.07<\/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>Initial pH<\/strong> = 3.35<\/li>\n\n\n\n<li><strong>pH after HCl addition<\/strong> = 3.07<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The initial solution is a buffer consisting of a weak acid (HF) and its conjugate base (F\u207b). The Henderson-Hasselbalch equation relates the pH to the ratio of conjugate base to acid. Adding HCl introduces strong acid, which reacts with the base F\u207b, reducing its amount and increasing the HF concentration. This shifts the ratio in favor of the acid, lowering the pH. The buffer resists large pH changes, so although a significant amount of HCl is added, the pH changes only moderately from 3.35 to 3.07. The slight volume increase is accounted for in concentration calculations, which slightly affects the final pH. This example shows the effectiveness of buffers in maintaining relatively stable pH when small amounts of strong acid are added.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The pKa of HF is 3.17. a. Calculate the pH of a 1.00 L solution that is 1.00 M HF and 1.50 M NaF. b. What is the pH of this solution after addition of 40.0 ml of 10.0 M HCl? The correct answer and explanation is: Let&#8217;s solve this step-by-step. Given: Part (a): Calculate [&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-31849","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31849","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=31849"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31849\/revisions"}],"predecessor-version":[{"id":31850,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31849\/revisions\/31850"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=31849"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=31849"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=31849"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}