![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-432.png\")
<\/figure>\n\n\n\nThe Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Based on the Lewis structure for propanoic acid depicted in the provided image, let’s address each of the questions:<\/p>\n\n\n\n i. Propanoic Acid as a Monoprotic, Diprotic, or Triprotic Acid<\/strong><\/p>\n\n\n\n Propanoic acid, which has the chemical formula C2H5COOH, is a monoprotic acid. This means it can donate only one proton (H+) per molecule during the dissociation process. The presence of one carboxylic acid group (-COOH) in its structure is responsible for its ability to release one hydrogen ion in solution.<\/p>\n\n\n\n ii. Acid Dissociation of Propanoic Acid<\/strong><\/p>\n\n\n\n The dissociation of propanoic acid in water can be written as: iii. Determination of the pH of a 0.015 M Propanoic Acid Solution<\/strong><\/p>\n\n\n\n To find the pH, we first use the acid dissociation constant (Ka) and the initial concentration of propanoic acid to set up the expression based on the equilibrium constant: Given that Ka = (1.35 \\times 10^{-5}) and assuming that the change in concentration of H+ at equilibrium is x, the equation becomes: Approximating that (0.015 – x \\approx 0.015) (since x will be small), the equation simplifies to: Thus, the pH of the solution is: iv. Determination of the ([OH^-]) of the Solution<\/strong><\/p>\n\n\n\n The relationship between the concentration of hydrogen ions ([\\text{H}^+]) and hydroxide ions ([\\text{OH}^-]) in a solution is governed by the water ion-product, (K_w = [\\text{H}^+][\\text{OH}^-] = 1.0 \\times 10^{-14}) at 25\u00b0C. Using the concentration of H+ calculated: Therefore, the concentration of hydroxide ions in the 0.015 M propanoic acid solution is approximately (7.04 \\times 10^{-11}) M.<\/p>\n","protected":false},"excerpt":{"rendered":" Shown below is the Lewis Structure for propanoic acid. i. Is propanoic acid a monoprotic, diprotic, or triprotic acid? ii. Write the acid dissociation of propanoic acid. iii. Determine the pH of a 0.015 M propanoic acid solution (Ka = 1.35×10-5). iv. Determine the [OH-1] of the solution from iii. The Correct Answer and Explanation […]<\/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-190641","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190641","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=190641"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190641\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190641"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190641"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190641"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
[ \\text{CH}_3\\text{CH}_2\\text{COOH} \\rightleftharpoons \\text{CH}_3\\text{CH}_2\\text{COO}^- + \\text{H}^+ ]
This reaction represents the release of a hydrogen ion from the carboxylic acid group, leaving behind the propanoate ion.<\/p>\n\n\n\n
[ \\text{Ka} = \\frac{[\\text{CH}_3\\text{CH}_2\\text{COO}^-][\\text{H}^+]}{[\\text{CH}_3\\text{CH}_2\\text{COOH}]} ]<\/p>\n\n\n\n
[ 1.35 \\times 10^{-5} = \\frac{x^2}{0.015 – x} ]<\/p>\n\n\n\n
[ x^2 = 1.35 \\times 10^{-5} \\times 0.015 ]
[ x^2 = 2.025 \\times 10^{-7} ]
[ x = \\sqrt{2.025 \\times 10^{-7}} ]
[ x \\approx 1.42 \\times 10^{-4} ]<\/p>\n\n\n\n
[ \\text{pH} = -\\log[\\text{H}^+] ]
[ \\text{pH} = -\\log[1.42 \\times 10^{-4}] ]
[ \\text{pH} \\approx 3.85 ]<\/p>\n\n\n\n
[ [\\text{OH}^-] = \\frac{1.0 \\times 10^{-14}}{1.42 \\times 10^{-4}} ]
[ [\\text{OH}^-] = 7.04 \\times 10^{-11} ]<\/p>\n\n\n\n