{"id":197795,"date":"2025-03-07T16:30:10","date_gmt":"2025-03-07T16:30:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197795"},"modified":"2025-03-07T16:30:14","modified_gmt":"2025-03-07T16:30:14","slug":"how-many-ml-of-4-50m-sodium-hydroxide-must-be-acetic-acid-solutions-to-make-a-buffer-with-phb-5","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/07\/how-many-ml-of-4-50m-sodium-hydroxide-must-be-acetic-acid-solutions-to-make-a-buffer-with-phb-5\/","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>Let&#8217;s break down the questions and solve each part step by step:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>How many mL of 4.50 M sodium hydroxide must be added to 250.0 mL of a 0.200 M acetic acid solution to make a buffer with pH = 5.00?<\/strong><\/h3>\n\n\n\n<p>To calculate how much sodium hydroxide (NaOH) is required, we can 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>pH = 5.00<\/li>\n\n\n\n<li>pKa of acetic acid = 4.76 (common value)<\/li>\n\n\n\n<li>([\\text{A}^-]) is the concentration of the acetate ion (CH\u2083COO\u207b), the conjugate base<\/li>\n\n\n\n<li>([\\text{HA}]) is the concentration of acetic acid (CH\u2083COOH)<\/li>\n<\/ul>\n\n\n\n<p>We want to find the ratio ([\\text{A}^-]\/[\\text{HA}]) that gives us a pH of 5.00.<\/p>\n\n\n\n<p>Rearranging the equation:<\/p>\n\n\n\n<p>[<br>\\text{pH} &#8211; \\text{pKa} = \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right)<br>]<\/p>\n\n\n\n<p>[<br>5.00 &#8211; 4.76 = \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right)<br>]<\/p>\n\n\n\n<p>[<br>0.24 = \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right)<br>]<\/p>\n\n\n\n<p>Now, solving for the ratio:<\/p>\n\n\n\n<p>[<br>\\frac{[\\text{A}^-]}{[\\text{HA}]} = 10^{0.24} \\approx 1.737<br>]<\/p>\n\n\n\n<p>Next, we need to determine how much NaOH (sodium hydroxide) to add to the acetic acid solution to achieve this ratio. NaOH will convert acetic acid (HA) into acetate ion (A\u207b). The moles of NaOH required are related to the moles of acetic acid that will be converted to acetate.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Calculation of the volume of NaOH required:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initially, we have 250.0 mL of a 0.200 M acetic acid solution. The moles of acetic acid are:<\/li>\n<\/ul>\n\n\n\n<p>[<br>\\text{moles of HA} = 0.200 \\, \\text{M} \\times 0.250 \\, \\text{L} = 0.0500 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>To reach the ratio ([\\text{A}^-]\/[\\text{HA}] = 1.737), we need to have a certain amount of acetate ion (A\u207b), so the moles of acetate ion required are:<\/li>\n<\/ul>\n\n\n\n<p>[<br>\\text{moles of A}^- = 1.737 \\times \\text{moles of HA} = 1.737 \\times 0.0500 \\, \\text{mol} = 0.08685 \\, \\text{mol}<br>]<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The amount of NaOH needed is equal to the moles of acetate formed, so we need 0.08685 moles of NaOH. Using the concentration of NaOH (4.50 M):<\/li>\n<\/ul>\n\n\n\n<p>[<br>\\text{volume of NaOH} = \\frac{\\text{moles of NaOH}}{\\text{concentration of NaOH}} = \\frac{0.08685 \\, \\text{mol}}{4.50 \\, \\text{M}} = 0.0193 \\, \\text{L} = 19.3 \\, \\text{mL}<br>]<\/p>\n\n\n\n<p>Thus, <strong>19.3 mL of 4.50 M NaOH<\/strong> is required to make the buffer with a pH of 5.00.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">3. <strong>Calculate the pH of a formic acid solution that contains 1.45% formic acid by mass.<\/strong><\/h3>\n\n\n\n<p>The formula for formic acid is <strong>HCOOH<\/strong>, and its dissociation constant (Ka) is approximately (1.77 \\times 10^{-4}).<\/p>\n\n\n\n<p>To find the pH of the formic acid solution, we will follow these steps:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate the molarity of formic acid in the solution:<\/strong><\/li>\n<\/ol>\n\n\n\n<p>The density of the solution is 1.01 g\/mL, and we have 1.45% formic acid by mass. This means 1.45 g of formic acid in 100 g of solution. For 1 L of solution (1000 mL), the mass of the solution is:<\/p>\n\n\n\n<p>[<br>\\text{mass of solution} = 1.01 \\, \\text{g\/mL} \\times 1000 \\, \\text{mL} = 1010 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>The mass of formic acid in 1 L of solution is:<\/p>\n\n\n\n<p>[<br>\\text{mass of formic acid} = 1.45\\% \\times 1010 \\, \\text{g} = 14.645 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>Now, convert the mass of formic acid to moles:<\/p>\n\n\n\n<p>[<br>\\text{moles of formic acid} = \\frac{14.645 \\, \\text{g}}{46.03 \\, \\text{g\/mol}} = 0.318 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p>The molarity of formic acid is:<\/p>\n\n\n\n<p>[<br>\\text{Molarity of HCOOH} = \\frac{0.318 \\, \\text{mol}}{1 \\, \\text{L}} = 0.318 \\, \\text{M}<br>]<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Set up the equilibrium expression for formic acid dissociation:<\/strong><\/li>\n<\/ol>\n\n\n\n<p>[<br>\\text{HCOOH} \\rightleftharpoons \\text{H}^+ + \\text{HCOO}^-<br>]<\/p>\n\n\n\n<p>Let (x) be the concentration of H\u207a (and HCOO\u207b) at equilibrium. The Ka expression is:<\/p>\n\n\n\n<p>[<br>K_a = \\frac{[\\text{H}^+][\\text{HCOO}^-]}{[\\text{HCOOH}]}<br>]<\/p>\n\n\n\n<p>Substituting the values:<\/p>\n\n\n\n<p>[<br>K_a = \\frac{x^2}{0.318 &#8211; x}<br>]<\/p>\n\n\n\n<p>[<br>1.77 \\times 10^{-4} = \\frac{x^2}{0.318 &#8211; x}<br>]<\/p>\n\n\n\n<p>Assuming (x) is small compared to 0.318 M, we approximate:<\/p>\n\n\n\n<p>[<br>1.77 \\times 10^{-4} \\approx \\frac{x^2}{0.318}<br>]<\/p>\n\n\n\n<p>Solving for (x):<\/p>\n\n\n\n<p>[<br>x^2 = (1.77 \\times 10^{-4}) \\times 0.318 = 5.63 \\times 10^{-5}<br>]<\/p>\n\n\n\n<p>[<br>x = \\sqrt{5.63 \\times 10^{-5}} = 7.51 \\times 10^{-3} \\, \\text{M}<br>]<\/p>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li><strong>Calculate the pH:<\/strong><\/li>\n<\/ol>\n\n\n\n<p>The pH is given by:<\/p>\n\n\n\n<p>[<br>\\text{pH} = -\\log [\\text{H}^+] = -\\log (7.51 \\times 10^{-3}) \\approx 2.12<br>]<\/p>\n\n\n\n<p>Thus, the <strong>pH of the formic acid solution is approximately 2.12<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>To create a buffer with pH = 5.00 using acetic acid and sodium hydroxide, 19.3 mL of 4.50 M NaOH is required.<\/li>\n\n\n\n<li>The pH of the formic acid solution with 1.45% formic acid by mass is approximately 2.12.<\/li>\n<\/ul>\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-197795","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197795","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=197795"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197795\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197795"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197795"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197795"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}