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
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n We are tasked with determining how many milliliters (mL) of 4.50 M sodium hydroxide (NaOH) must be added to acetic acid (CH\u2083COOH) to make a buffer with a given pH of 5.00.<\/p>\n\n\n\n The Henderson-Hasselbalch equation<\/strong> relates the pH of a buffer solution to the ratio of the concentrations of its conjugate base (acetate, CH\u2083COO\u207b) and its acid (acetic acid, CH\u2083COOH): Rearranging the equation to find the ratio of the conjugate base to acid: This indicates that the ratio of acetate (conjugate base) to acetic acid is approximately 1.737<\/strong>.<\/p>\n\n\n\n To make this buffer, you need to add NaOH to the acetic acid solution, which will convert some of the acetic acid (HA) into acetate (A\u207b). Since the ratio of acetate to acetic acid needs to be 1.737, we can express this relationship in terms of the volumes and concentrations of the solutions.<\/p>\n\n\n\n Given:<\/p>\n\n\n\n We first calculate the amount of acetic acid present: Let ( V_{\\text{NaOH}} ) be the volume of NaOH to be added. The moles of NaOH added will equal the moles of acetate formed, so: Since NaOH dissociates completely, the moles of NaOH added will be equal to the moles of acetate formed. Now calculate the volume of NaOH required: Thus, 19.3 mL<\/strong> of 4.50 M NaOH must be added to 250.0 mL of 0.200 M acetic acid to prepare a buffer with pH = 5.00.<\/p>\n\n\n\n Now, let’s calculate the pH of a formic acid solution containing 1.45% by mass<\/strong> of formic acid.<\/p>\n\n\n\n [ The molar mass of formic acid is approximately 46.03 g\/mol, so: [ Assuming no conjugate base is present, we can calculate the pH using the following equation for a weak acid: So, the pH of the formic acid solution is approximately 2.13<\/strong>.<\/p>\n\n\n\n To make a buffer with pH 5.00 from acetic acid, 19.3 mL of 4.50 M sodium hydroxide<\/strong> must be added to 250.0 mL of 0.200 M acetic acid. The pH of a formic acid solution containing 1.45% formic acid by mass is approximately 2.13<\/strong>.<\/p>\n\n\n\n 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 […]<\/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-197430","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197430","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=197430"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197430\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197430"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197430"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197430"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Question 1: Volume of 4.50M Sodium Hydroxide Needed to Prepare a Buffer Solution<\/h3>\n\n\n\n
Relevant Equation: Henderson-Hasselbalch Equation<\/h4>\n\n\n\n
[
\\text{pH} = \\text{pKa} + \\log \\left(\\frac{[\\text{A}^-]}{[\\text{HA}]}\\right)
]
where:<\/p>\n\n\n\n\n
[
\\text{pH} – \\text{pKa} = \\log \\left(\\frac{[\\text{A}^-]}{[\\text{HA}]}\\right)
]
[
5.00 – 4.76 = \\log \\left(\\frac{[\\text{A}^-]}{[\\text{HA}]}\\right)
]
[
0.24 = \\log \\left(\\frac{[\\text{A}^-]}{[\\text{HA}]}\\right)
]
[
\\frac{[\\text{A}^-]}{[\\text{HA}]} = 10^{0.24} \\approx 1.737
]<\/p>\n\n\n\nPreparing the Buffer:<\/h4>\n\n\n\n
\n
[
\\text{moles of acetic acid} = 0.200 \\, \\text{M} \\times 0.250 \\, \\text{L} = 0.050 \\, \\text{mol}
]<\/p>\n\n\n\n
[
\\text{moles of acetate} = 1.737 \\times \\text{moles of acetic acid}
]
[
\\text{moles of acetate} = 1.737 \\times 0.050 = 0.08685 \\, \\text{mol}
]<\/p>\n\n\n\n
[
V_{\\text{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}
]<\/p>\n\n\n\nQuestion 2: pH of Formic Acid Solution (1.45% by mass)<\/h3>\n\n\n\n
Relevant Information:<\/h4>\n\n\n\n
\n
Step 1: Calculate the mass of formic acid in 1.00 L of solution<\/h5>\n\n\n\n
\\text{mass of formic acid} = \\frac{1.45}{100} \\times 1000 = 14.5 \\, \\text{g}
]<\/p>\n\n\n\nStep 2: Convert mass to moles<\/h5>\n\n\n\n
[
\\text{moles of formic acid} = \\frac{14.5 \\, \\text{g}}{46.03 \\, \\text{g\/mol}} \\approx 0.315 \\, \\text{mol}
]<\/p>\n\n\n\nStep 3: Calculate the concentration of formic acid in the solution<\/h5>\n\n\n\n
[\\text{HCOOH}] = \\frac{0.315 \\, \\text{mol}}{1.00 \\, \\text{L}} = 0.315 \\, \\text{M}
]<\/p>\n\n\n\nStep 4: Use the Henderson-Hasselbalch equation<\/h5>\n\n\n\n
[
\\text{pH} = \\frac{1}{2} (\\text{pKa} – \\log[\\text{HCOOH}])
]
[
\\text{pH} = \\frac{1}{2} (3.75 – \\log[0.315])
]
[
\\text{pH} = \\frac{1}{2} (3.75 – (-0.502)) = \\frac{1}{2} (4.252) = 2.13
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n
\n","protected":false},"excerpt":{"rendered":"