{"id":185622,"date":"2025-01-22T18:51:05","date_gmt":"2025-01-22T18:51:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185622"},"modified":"2025-01-22T18:51:07","modified_gmt":"2025-01-22T18:51:07","slug":"the-pka-of-hf-is-3-17","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/22\/the-pka-of-hf-is-3-17\/","title":{"rendered":"The pKa of HF is 3.17"},"content":{"rendered":"\n<p>a) The pKa of HF is 3.17. 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 50.0 mL of 10.0 M HCl?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part A: pH of a 1.00 M HF and 1.50 M NaF Solution<\/h3>\n\n\n\n<p>The solution contains both a weak acid (HF) and its conjugate base (F\u207b) from NaF. This setup is a <strong>buffer solution<\/strong>, which we can analyze using the <strong>Henderson-Hasselbalch equation<\/strong>:<\/p>\n\n\n\n<p>[<br>\\text{pH} = \\text{pKa} + \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right)<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>pKa of HF = 3.17<\/li>\n\n\n\n<li>[A\u207b] = concentration of the conjugate base (F\u207b) = 1.50 M (from NaF)<\/li>\n\n\n\n<li>[HA] = concentration of the weak acid (HF) = 1.00 M<\/li>\n<\/ul>\n\n\n\n<p>Substituting the values into the equation:<\/p>\n\n\n\n<p>[<br>\\text{pH} = 3.17 + \\log \\left( \\frac{1.50}{1.00} \\right)<br>]<\/p>\n\n\n\n<p>First, calculate the ratio of concentrations:<\/p>\n\n\n\n<p>[<br>\\frac{1.50}{1.00} = 1.50<br>]<\/p>\n\n\n\n<p>Now calculate the logarithm:<\/p>\n\n\n\n<p>[<br>\\log(1.50) \\approx 0.176<br>]<\/p>\n\n\n\n<p>Therefore, the pH is:<\/p>\n\n\n\n<p>[<br>\\text{pH} = 3.17 + 0.176 = 3.35<br>]<\/p>\n\n\n\n<p>Thus, the pH of the solution is <strong>3.35<\/strong>.<\/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: pH After Addition of 50.0 mL of 10.0 M HCl<\/h3>\n\n\n\n<p>After adding HCl, the solution will experience an increase in the concentration of ( \\text{H}^+ ) ions, which will interact with the conjugate base ( \\text{F}^- ) in the buffer. The reaction will be:<\/p>\n\n\n\n<p>[<br>\\text{F}^- + \\text{H}^+ \\rightleftharpoons \\text{HF}<br>]<\/p>\n\n\n\n<p>First, calculate the number of moles of HCl added:<\/p>\n\n\n\n<p>[<br>\\text{moles of HCl} = (10.0 \\, \\text{M}) \\times (50.0 \\, \\text{mL}) = 10.0 \\times 0.050 = 0.50 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p>The initial total volume is 1.00 L of buffer solution + 0.050 L of HCl solution, so the final volume is:<\/p>\n\n\n\n<p>[<br>\\text{final volume} = 1.00 \\, \\text{L} + 0.050 \\, \\text{L} = 1.05 \\, \\text{L}<br>]<\/p>\n\n\n\n<p>Next, calculate the change in the concentrations of F\u207b and HF:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial moles of F\u207b = ( 1.50 \\, \\text{M} \\times 1.00 \\, \\text{L} = 1.50 \\, \\text{mol} )<\/li>\n\n\n\n<li>Initial moles of HF = ( 1.00 \\, \\text{M} \\times 1.00 \\, \\text{L} = 1.00 \\, \\text{mol} )<\/li>\n<\/ul>\n\n\n\n<p>After the addition of HCl, the ( \\text{H}^+ ) ions will react with F\u207b, converting some F\u207b to HF. The moles of F\u207b will decrease, and the moles of HF will increase by the amount of H\u207a added. Since 0.50 mol of HCl is added, this will react with an equivalent amount of F\u207b.<\/p>\n\n\n\n<p>After the reaction:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of F\u207b = ( 1.50 &#8211; 0.50 = 1.00 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of HF = ( 1.00 + 0.50 = 1.50 \\, \\text{mol} )<\/li>\n<\/ul>\n\n\n\n<p>Now, calculate the new concentrations:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>[F\u207b] = ( \\frac{1.00 \\, \\text{mol}}{1.05 \\, \\text{L}} = 0.952 \\, \\text{M} )<\/li>\n\n\n\n<li>[HF] = ( \\frac{1.50 \\, \\text{mol}}{1.05 \\, \\text{L}} = 1.43 \\, \\text{M} )<\/li>\n<\/ul>\n\n\n\n<p>Using the Henderson-Hasselbalch equation again:<\/p>\n\n\n\n<p>[<br>\\text{pH} = 3.17 + \\log \\left( \\frac{0.952}{1.43} \\right)<br>]<\/p>\n\n\n\n<p>First, calculate the ratio:<\/p>\n\n\n\n<p>[<br>\\frac{0.952}{1.43} \\approx 0.665<br>]<\/p>\n\n\n\n<p>Now calculate the logarithm:<\/p>\n\n\n\n<p>[<br>\\log(0.665) \\approx -0.176<br>]<\/p>\n\n\n\n<p>Thus, the pH is:<\/p>\n\n\n\n<p>[<br>\\text{pH} = 3.17 + (-0.176) = 2.99<br>]<\/p>\n\n\n\n<p>Therefore, the pH of the solution after the addition of HCl is <strong>2.99<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Results:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>pH of the buffer solution<\/strong> (1.00 M HF, 1.50 M NaF) = 3.35<\/li>\n\n\n\n<li><strong>pH after addition of 50.0 mL of 10.0 M HCl<\/strong> = 2.99<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>a) The pKa of HF is 3.17. 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 50.0 mL of 10.0 M HCl? The Correct Answer and Explanation is : Part A: pH of a 1.00 M [&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-185622","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185622","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=185622"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185622\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185622"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185622"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185622"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}