{"id":195557,"date":"2025-02-28T12:44:35","date_gmt":"2025-02-28T12:44:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=195557"},"modified":"2025-02-28T12:44:37","modified_gmt":"2025-02-28T12:44:37","slug":"the-active-ingredient-in-aspirin-is-acetylsalicylic-acid-hc9h7o4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/28\/the-active-ingredient-in-aspirin-is-acetylsalicylic-acid-hc9h7o4\/","title":{"rendered":"The active ingredient in aspirin is acetylsalicylic acid HC9H7O4"},"content":{"rendered":"\n

The active ingredient in aspirin is acetylsalicylic acid HC9H7O4, a monoprotic acid with a Ka=3.3×10^-4 at 25 degrees C . What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 573 mg of acetylsalicylic acid each, in 258 mL of water?<\/p>\n\n\n\n

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

To determine the pH of a solution made by dissolving two extra-strength aspirin tablets, each containing 573 mg of acetylsalicylic acid (ASA), in 258 mL of water, we can follow these steps:<\/p>\n\n\n\n

    \n
  1. Calculate the moles of acetylsalicylic acid:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Molar mass of ASA (HC\u2089H\u2087O\u2084):<\/strong> 180.16 g\/mol<\/li>\n\n\n\n
    • Total mass of ASA:<\/strong> 573 mg\/tablet \u00d7 2 tablets = 1146 mg = 1.146 g<\/li>\n\n\n\n
    • Moles of ASA:<\/strong> 1.146 g \u00f7 180.16 g\/mol \u2248 0.00636 mol<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Determine the concentration of ASA in the solution:<\/strong><\/li>\n<\/ol>\n\n\n\n
          \n
        • Volume of solution:<\/strong> 258 mL = 0.258 L<\/li>\n\n\n\n
        • Concentration (C):<\/strong> 0.00636 mol \u00f7 0.258 L \u2248 0.0247 M<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Set up the ionization equilibrium expression:<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • ASA (HC\u2089H\u2087O\u2084) is a monoprotic weak acid that partially ionizes in water:
              [ \\text{HC}_9\\text{H}_7\\text{O}_4 \\rightleftharpoons \\text{H}^+ + \\text{C}_9\\text{H}_7\\text{O}_4^- ]<\/li>\n\n\n\n
            • Acid dissociation constant (Ka):<\/strong> 3.3 \u00d7 10\u207b\u2074<\/li>\n\n\n\n
            • Expression:<\/strong> [ K_a = \\frac{[\\text{H}^+][\\text{C}_9\\text{H}_7\\text{O}_4^-]}{[\\text{HC}_9\\text{H}_7\\text{O}_4]} ]<\/li>\n<\/ul>\n\n\n\n
                \n
              1. Assume the degree of ionization is small:<\/strong><\/li>\n<\/ol>\n\n\n\n
                  \n
                • Let ( x ) represent the concentration of ionized ASA:\n
                    \n
                  • [ [\\text{H}^+] = x ]<\/li>\n\n\n\n
                  • [ [\\text{C}_9\\text{H}_7\\text{O}_4^-] = x ]<\/li>\n\n\n\n
                  • [ [\\text{HC}_9\\text{H}_7\\text{O}_4] \\approx 0.0247 – x \\approx 0.0247 ]<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                  • Since ( x ) is expected to be much smaller than the initial concentration, the approximation holds.<\/li>\n<\/ul>\n\n\n\n
                      \n
                    1. Solve for ( x ):<\/strong><\/li>\n<\/ol>\n\n\n\n
                        \n
                      • [ K_a = \\frac{x^2}{0.0247} ]<\/li>\n\n\n\n
                      • [ x^2 = K_a \\times 0.0247 ]<\/li>\n\n\n\n
                      • [ x^2 = 3.3 \\times 10^{-4} \\times 0.0247 ]<\/li>\n\n\n\n
                      • [ x^2 = 8.151 \\times 10^{-6} ]<\/li>\n\n\n\n
                      • [ x = \\sqrt{8.151 \\times 10^{-6}} ]<\/li>\n\n\n\n
                      • [ x \\approx 0.00285 \\, \\text{M} ]<\/li>\n<\/ul>\n\n\n\n
                          \n
                        1. Calculate the pH:<\/strong><\/li>\n<\/ol>\n\n\n\n
                            \n
                          • [ \\text{pH} = -\\log[\\text{H}^+] ]<\/li>\n\n\n\n
                          • [ \\text{pH} = -\\log[0.00285] ]<\/li>\n\n\n\n
                          • [ \\text{pH} \\approx 2.55 ]<\/li>\n<\/ul>\n\n\n\n

                            Therefore, the pH of the solution is approximately 2.55.<\/p>\n\n\n\n

                            This calculation assumes that the ionization of acetylsalicylic acid is minimal, allowing the approximation ( [\\text{HC}_9\\text{H}_7\\text{O}_4] \\approx 0.0247 \\, \\text{M} ). Given the relatively small Ka value, this assumption is valid, leading to a slightly acidic solution, as expected for a weak acid like acetylsalicylic acid.<\/p>\n","protected":false},"excerpt":{"rendered":"

                            The active ingredient in aspirin is acetylsalicylic acid HC9H7O4, a monoprotic acid with a Ka=3.3×10^-4 at 25 degrees C . What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 573 mg of acetylsalicylic acid each, in 258 mL of water? The Correct Answer and Explanation is : To determine […]<\/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-195557","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195557","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=195557"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/195557\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=195557"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=195557"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=195557"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}