{"id":197047,"date":"2025-03-06T13:35:41","date_gmt":"2025-03-06T13:35:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197047"},"modified":"2025-03-06T13:35:44","modified_gmt":"2025-03-06T13:35:44","slug":"hac-is-the-abbreviation-for-acetic-acid-ch3cooh","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/06\/hac-is-the-abbreviation-for-acetic-acid-ch3cooh\/","title":{"rendered":"HAc is the abbreviation for acetic acid (CH3COOH)"},"content":{"rendered":"\n<p>HAc is the abbreviation for acetic acid (CH3COOH); it is by random chance that the numerical value of the acid dissociation constant for acetic acid is equal to the base hydrolysis constant of ammonia. 1.00L of buffer is prepared in which [HAc] = 0.35M and [NaAc] = 0.75M. What is the pH of this buffer?<\/p>\n\n\n\n<p>Ka,HAc = 1.8&#215;10^-5 Kb,NH3 = 1.8&#215;10^-5<\/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<p>To determine the pH of the buffer solution consisting of acetic acid (HAc) and sodium acetate (NaAc), we use 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><strong>pKa<\/strong> is the negative logarithm of the acid dissociation constant ( K_a ).<\/li>\n\n\n\n<li><strong>[A\u207b]<\/strong> is the concentration of the conjugate base (sodium acetate, NaAc).<\/li>\n\n\n\n<li><strong>[HA]<\/strong> is the concentration of the weak acid (acetic acid, HAc).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate pKa of Acetic Acid<\/h3>\n\n\n\n<p>[<br>\\text{pKa} = -\\log(1.8 \\times 10^{-5})<br>]<\/p>\n\n\n\n<p>[<br>\\text{pKa} \\approx 4.74<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Apply the Henderson-Hasselbalch Equation<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>[HAc]<\/strong> = 0.35 M<\/li>\n\n\n\n<li><strong>[NaAc]<\/strong> = 0.75 M<\/li>\n<\/ul>\n\n\n\n<p>[<br>\\text{pH} = 4.74 + \\log \\left( \\frac{0.75}{0.35} \\right)<br>]<\/p>\n\n\n\n<p>[<br>\\text{pH} = 4.74 + \\log (2.14)<br>]<\/p>\n\n\n\n<p>[<br>\\text{pH} = 4.74 + 0.33<br>]<\/p>\n\n\n\n<p>[<br>\\text{pH} \\approx 5.07<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>The <strong>buffer solution<\/strong> is made up of a weak acid (acetic acid) and its conjugate base (acetate ion from sodium acetate). The pH of the buffer is determined using the <strong>Henderson-Hasselbalch equation<\/strong>, which calculates the pH based on the acid\/base ratio.<\/p>\n\n\n\n<p>In this case:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Acetic acid partially ionizes in water.<\/li>\n\n\n\n<li>Sodium acetate completely dissociates into acetate ions (CH\u2083COO\u207b) and Na\u207a.<\/li>\n\n\n\n<li>The higher concentration of acetate ions shifts the equilibrium to the left, resisting pH changes when small amounts of acid or base are added.<\/li>\n<\/ul>\n\n\n\n<p>Since <strong>[A\u207b] &gt; [HA]<\/strong>, the solution is slightly more basic than the pKa of acetic acid, giving a final pH of <strong>5.07<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-367-1024x681.png\" alt=\"\" class=\"wp-image-197048\"\/><\/figure>\n\n\n\n<p>The pH of the buffer solution is <strong>5.07<\/strong>, as determined using the Henderson-Hasselbalch equation. The pie chart visually represents the composition of the buffer system, showing the relative amounts of <strong>acetic acid (HAc) and sodium acetate (NaAc)<\/strong>, which together maintain the buffer&#8217;s resistance to pH changes.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>HAc is the abbreviation for acetic acid (CH3COOH); it is by random chance that the numerical value of the acid dissociation constant for acetic acid is equal to the base hydrolysis constant of ammonia. 1.00L of buffer is prepared in which [HAc] = 0.35M and [NaAc] = 0.75M. What is the pH of this buffer? [&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-197047","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197047","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=197047"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197047\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197047"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197047"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197047"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}