{"id":196839,"date":"2025-03-06T05:57:51","date_gmt":"2025-03-06T05:57:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196839"},"modified":"2025-03-06T05:57:53","modified_gmt":"2025-03-06T05:57:53","slug":"how-many-ml-of-4-50m-sodium-hydroxide-must-be-acetic-acid-solutions-to-make-a-buffer-with-phb","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/06\/how-many-ml-of-4-50m-sodium-hydroxide-must-be-acetic-acid-solutions-to-make-a-buffer-with-phb\/","title":{"rendered":"How many mL of 4.50M sodium hydroxide must be acetic acid solutions to make a buffer with pHb"},"content":{"rendered":"\n<p>How many mL of 4.50M sodium hydroxide must be acetic acid solutions to make a buffer with pHb. How many mL of 4.50 M sodium hydroxide must be added to 250.0 mL of a 0.200 Mc. Calculate the pH of a formic acid solution that contains 1.45% formic acid by mass.Question 4a. The pH of the acid and its conjugate base pairbuffer solution can be calculated by using(Assume a density of 1.01 g\/mL for the solution.)acetic acid solution to make a buffer with pH = 5.000?the Anderson Hassel Balch equation:=5.000<\/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 address your questions systematically:<\/p>\n\n\n\n<p><strong>1. Volume of 4.50 M Sodium Hydroxide Needed to Prepare an Acetic Acid Buffer with pH = 5.000<\/strong><\/p>\n\n\n\n<p>The pH of a buffer solution can be estimated using the Henderson-Hasselbalch equation:\ue206<\/p>\n\n\n\n<p>[ \\text{pH} = \\text{p}K_a + \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right) ]\ue206<\/p>\n\n\n\n<p>For acetic acid (CH\u2083COOH), the acid dissociation constant (( K_a )) is ( 1.8 \\times 10^{-5} ), yielding a ( \\text{p}K_a ) of 4.74.\ue206<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Desired pH = 5.000<\/li>\n\n\n\n<li>( \\text{p}K_a ) = 4.74<\/li>\n\n\n\n<li>Initial concentration of acetic acid, [HA] = 0.200 M<\/li>\n\n\n\n<li>Concentration of sodium hydroxide (NaOH) = 4.50 M\ue206<\/li>\n<\/ul>\n\n\n\n<p>First, calculate the required ratio of the conjugate base ([A\u207b]) to the acid ([HA]):\ue206<\/p>\n\n\n\n<p>[ 5.000 = 4.74 + \\log \\left( \\frac{[\\text{A}^-]}{0.200} \\right) ]\ue206<\/p>\n\n\n\n<p>[ 0.26 = \\log \\left( \\frac{[\\text{A}^-]}{0.200} \\right) ]\ue206<\/p>\n\n\n\n<p>[ 10^{0.26} \\approx 1.82 ]\ue206<\/p>\n\n\n\n<p>[ \\frac{[\\text{A}^-]}{0.200} = 1.82 ]\ue206<\/p>\n\n\n\n<p>[ [\\text{A}^-] = 1.82 \\times 0.200 = 0.364 \\, \\text{M} ]\ue206<\/p>\n\n\n\n<p>The amount of acetate ion ([A\u207b]) produced equals the moles of NaOH added. Let ( V ) be the volume of 4.50 M NaOH required:\ue206<\/p>\n\n\n\n<p>[ \\frac{0.364 \\, \\text{mol\/L} \\times 0.250 \\, \\text{L}}{4.50 \\, \\text{mol\/L}} = V ]\ue206<\/p>\n\n\n\n<p>[ V = \\frac{0.091}{4.50} = 0.0202 \\, \\text{L} ]\ue206<\/p>\n\n\n\n<p>[ V = 20.2 \\, \\text{mL} ]\ue206<\/p>\n\n\n\n<p>Therefore, approximately 20.2 mL of 4.50 M NaOH is required to achieve a buffer solution with a pH of 5.000.\ue206<\/p>\n\n\n\n<p><strong>2. Volume of 4.50 M Sodium Hydroxide Needed to Add to 250.0 mL of 0.200 M Acetic Acid<\/strong><\/p>\n\n\n\n<p>Assuming the goal is to create a buffer solution with a specific pH, the required volume of NaOH can be calculated as shown above. If the desired pH is 5.000, then 20.2 mL of 4.50 M NaOH should be added to 250.0 mL of 0.200 M acetic acid.\ue206<\/p>\n\n\n\n<p><strong>3. Calculating the pH of a Formic Acid Solution Containing 1.45% Formic Acid by Mass<\/strong><\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass percentage of formic acid (HCOOH) = 1.45%<\/li>\n\n\n\n<li>Density of the solution = 1.01 g\/mL<\/li>\n\n\n\n<li>( K_a ) of formic acid = ( 1.8 \\times 10^{-4} )\ue206<\/li>\n<\/ul>\n\n\n\n<p>Assuming 100 g of the solution:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of formic acid = 1.45 g<\/li>\n\n\n\n<li>Mass of water = 98.55 g\ue206<\/li>\n<\/ul>\n\n\n\n<p>Volume of the solution:\ue206<\/p>\n\n\n\n<p>[ \\text{Volume} = \\frac{100 \\, \\text{g}}{1.01 \\, \\text{g\/mL}} \\approx 99.01 \\, \\text{mL} ]\ue206<\/p>\n\n\n\n<p>Moles of formic acid:\ue206<\/p>\n\n\n\n<p>[ \\text{Moles} = \\frac{1.45 \\, \\text{g}}{46.03 \\, \\text{g\/mol}} \\approx 0.0315 \\, \\text{mol} ]\ue206<\/p>\n\n\n\n<p>Molarity of formic acid:\ue206<\/p>\n\n\n\n<p>[ \\text{Molarity} = \\frac{0.0315 \\, \\text{mol}}{0.09901 \\, \\text{L}} \\approx 0.318 \\, \\text{M} ]\ue206<\/p>\n\n\n\n<p>For a weak acid dissociation:\ue206<\/p>\n\n\n\n<p>[ \\text{HCOOH} \\rightleftharpoons \\text{H}^+ + \\text{HCOO}^- ]\ue206<\/p>\n\n\n\n<p>[ K_a = \\frac{[\\text{H}^+][\\text{HCOO}^-]}{[\\text{HCOOH}]} ]\ue206<\/p>\n\n\n\n<p>Assuming ( [\\text{H}^+] = [\\text{HCOO}^-] = x ) and ( [\\text{HCOOH}] \\approx 0.318 &#8211; x ):\ue206<\/p>\n\n\n\n<p>[ 1.8 \\times 10^{-4} = \\frac{x^2}{0.318 &#8211; x} ]\ue206<\/p>\n\n\n\n<p>Assuming ( x ) is small compared to 0.318:\ue206<\/p>\n\n\n\n<p>[ x^2 \\approx 1.8 \\times 10^{-4} \\times 0.318 ]\ue206<\/p>\n\n\n\n<p>[ x^2 \\approx 5.724 \\times 10^{-5} ]\ue206<\/p>\n\n\n\n<p>[ x \\approx \\sqrt{5.724 \\times 10^{-5}} ]\ue206<\/p>\n\n\n\n<p>[ x \\approx 0.00756 \\, \\text{M} ]\ue206<\/p>\n\n\n\n<p>[ \\text{pH} = -\\log(0.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How many mL of 4.50M sodium hydroxide must be acetic acid solutions to make a buffer with pHb. How many mL of 4.50 M sodium hydroxide must be added to 250.0 mL of a 0.200 Mc. Calculate the pH of a formic acid solution that contains 1.45% formic acid by mass.Question 4a. The pH of [&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-196839","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196839","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=196839"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196839\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196839"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196839"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196839"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}