{"id":236306,"date":"2025-06-16T07:24:09","date_gmt":"2025-06-16T07:24:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=236306"},"modified":"2025-06-16T07:24:11","modified_gmt":"2025-06-16T07:24:11","slug":"vinegar-contains-acetic-acid-ch3cooh","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/16\/vinegar-contains-acetic-acid-ch3cooh\/","title":{"rendered":"Vinegar contains acetic acid, CH3COOH"},"content":{"rendered":"\n

Vinegar contains acetic acid, CH3COOH. Titration of 5.00g of vinegar with 0.100M NaOH requires 33.0 cm^3 to reach the equivalence point. (a) What is the weight percentage of CH3COOH in vinegar? (b) If the vinegar has a density of 1.005 g\/cm^3, what is the molarity of CH3COOH in vinegar?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

(a) Weight percentage of CH\u2083COOH in vinegar<\/strong><\/h3>\n\n\n\n

Step 1: Moles of NaOH used<\/strong>Volume of NaOH=33.0\u2009cm3=0.0330\u2009L\\text{Volume of NaOH} = 33.0 \\, \\text{cm}^3 = 0.0330 \\, \\text{L}Volume of NaOH=33.0cm3=0.0330LMolarity of NaOH=0.100\u2009mol\/L\\text{Molarity of NaOH} = 0.100 \\, \\text{mol\/L}Molarity of NaOH=0.100mol\/LMoles of NaOH=0.100\u00d70.0330=0.00330\u2009mol\\text{Moles of NaOH} = 0.100 \\times 0.0330 = 0.00330 \\, \\text{mol}Moles of NaOH=0.100\u00d70.0330=0.00330mol<\/p>\n\n\n\n

Step 2: Moles of CH\u2083COOH neutralized<\/strong><\/p>\n\n\n\n

Acetic acid reacts with NaOH in a 1:1 molar ratio:CH\u2083COOH+NaOH\u2192CH\u2083COONa+H\u2082O\\text{CH\u2083COOH} + \\text{NaOH} \\rightarrow \\text{CH\u2083COONa} + \\text{H\u2082O}CH\u2083COOH+NaOH\u2192CH\u2083COONa+H\u2082OMoles of CH\u2083COOH=0.00330\u2009mol\\text{Moles of CH\u2083COOH} = 0.00330 \\, \\text{mol}Moles of CH\u2083COOH=0.00330mol<\/p>\n\n\n\n

Step 3: Mass of CH\u2083COOH in sample<\/strong><\/p>\n\n\n\n

Molar mass of CH\u2083COOH:=12.01\u00d72+1.008\u00d74+16.00\u00d72=60.05\u2009g\/mol= 12.01 \\times 2 + 1.008 \\times 4 + 16.00 \\times 2 = 60.05 \\, \\text{g\/mol}=12.01\u00d72+1.008\u00d74+16.00\u00d72=60.05g\/molMass=0.00330\u00d760.05=0.19817\u2009g\\text{Mass} = 0.00330 \\times 60.05 = 0.19817 \\, \\text{g}Mass=0.00330\u00d760.05=0.19817g<\/p>\n\n\n\n

Step 4: Weight percentage<\/strong>Mass of vinegar=5.00\u2009g\\text{Mass of vinegar} = 5.00 \\, \\text{g}Mass of vinegar=5.00gWeight %=(0.198175.00)\u00d7100=3.96%\\text{Weight \\%} = \\left( \\frac{0.19817}{5.00} \\right) \\times 100 = 3.96\\%Weight %=(5.000.19817\u200b)\u00d7100=3.96%<\/p>\n\n\n\n

\n\n\n\n

(b) Molarity of CH\u2083COOH in vinegar<\/strong><\/h3>\n\n\n\n

Step 1: Volume of vinegar sample<\/strong><\/p>\n\n\n\n

Given mass = 5.00 g, density = 1.005 g\/cm\u00b3:Volume=5.001.005=4.9751\u2009cm3=0.0049751\u2009L\\text{Volume} = \\frac{5.00}{1.005} = 4.9751 \\, \\text{cm}^3 = 0.0049751 \\, \\text{L}Volume=1.0055.00\u200b=4.9751cm3=0.0049751L<\/p>\n\n\n\n

Step 2: Molarity<\/strong>Moles of CH\u2083COOH=0.00330\u2009mol\\text{Moles of CH\u2083COOH} = 0.00330 \\, \\text{mol}Moles of CH\u2083COOH=0.00330molMolarity=0.003300.0049751=0.6636\u2009mol\/L\\text{Molarity} = \\frac{0.00330}{0.0049751} = 0.6636 \\, \\text{mol\/L}Molarity=0.00497510.00330\u200b=0.6636mol\/L<\/p>\n\n\n\n

\n\n\n\n

Final Answers:<\/h3>\n\n\n\n
    \n
  • (a) Weight percentage of CH\u2083COOH<\/strong> = 3.96%<\/strong><\/li>\n\n\n\n
  • (b) Molarity of CH\u2083COOH<\/strong> = 0.664 mol\/L<\/strong><\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Explanation<\/h3>\n\n\n\n

    Acetic acid, known by its chemical formula CH\u2083COOH, serves as the active acidic component in vinegar. To determine its concentration, titration is performed using sodium hydroxide (NaOH), a strong base. In the process, a known volume of NaOH is used to neutralize a known mass of vinegar, allowing calculation of the acid content.<\/p>\n\n\n\n

    The neutralization reaction between acetic acid and sodium hydroxide follows a 1:1 molar ratio. Therefore, the number of moles of NaOH used directly reflects the moles of CH\u2083COOH present. By multiplying the volume of NaOH in liters by its molarity, the number of moles involved in the reaction is established.<\/p>\n\n\n\n

    Once the moles of acetic acid are known, converting this to grams is achieved through multiplication by the molar mass of acetic acid. Comparing this mass to the initial mass of the vinegar sample enables calculation of the weight percentage of acetic acid. The result reflects how much of the vinegar\u2019s mass is due to acetic acid content.<\/p>\n\n\n\n

    To determine molarity, the volume of vinegar is required. This is calculated using the density of the solution and its known mass. Converting grams to volume using the density relationship, and then expressing volume in liters, allows the use of the standard molarity formula: moles of solute per liter of solution.<\/p>\n\n\n\n

    These calculations collectively determine both the weight percentage and molarity of acetic acid in vinegar, offering insight into its acidic strength and concentration in the sample.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Vinegar contains acetic acid, CH3COOH. Titration of 5.00g of vinegar with 0.100M NaOH requires 33.0 cm^3 to reach the equivalence point. (a) What is the weight percentage of CH3COOH in vinegar? (b) If the vinegar has a density of 1.005 g\/cm^3, what is the molarity of CH3COOH in vinegar? The Correct Answer and Explanation is: […]<\/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-236306","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236306","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=236306"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236306\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=236306"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=236306"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=236306"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}