{"id":234342,"date":"2025-06-14T04:46:14","date_gmt":"2025-06-14T04:46:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=234342"},"modified":"2025-06-14T04:46:16","modified_gmt":"2025-06-14T04:46:16","slug":"formic-acid-hcooh-is-a-weak-acid-with-a-pka-of-3-74","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/14\/formic-acid-hcooh-is-a-weak-acid-with-a-pka-of-3-74\/","title":{"rendered":"Formic acid, HCOOH, is a weak acid with a pKa of 3.74."},"content":{"rendered":"\n

Formic acid, HCOOH, is a weak acid with a pKa of 3.74. Draw up a reaction table for the reaction of 0.300 M formic acid with water and calculate the pH of this solution.<\/p>\n\n\n\n

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

To solve this problem, we will analyze the ionization of formic acid (HCOOH), a weak acid, in water using an ICE table<\/strong> (Initial, Change, Equilibrium) and the acid dissociation constant<\/strong> KaK_aKa\u200b, which is derived from the given pKa=3.74pK_a = 3.74pKa\u200b=3.74.<\/p>\n\n\n\n

\n\n\n\n

1. Reaction Equation<\/strong><\/h3>\n\n\n\n

HCOOH (aq)\u21ccH+(aq)+HCOO\u2212(aq)\\text{HCOOH (aq)} \\rightleftharpoons \\text{H}^+ (aq) + \\text{HCOO}^- (aq)HCOOH (aq)\u21ccH+(aq)+HCOO\u2212(aq)<\/p>\n\n\n\n

\n\n\n\n

2. Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Initial concentration of formic acid: [HCOOH]0=0.300\u2009M[HCOOH]_0 = 0.300\\, \\text{M}[HCOOH]0\u200b=0.300M<\/li>\n\n\n\n
  • pKa=3.74pK_a = 3.74pKa\u200b=3.74
    Convert to KaK_aKa\u200b: Ka=10\u2212pKa=10\u22123.74\u22481.82\u00d710\u22124K_a = 10^{-pK_a} = 10^{-3.74} \\approx 1.82 \\times 10^{-4}Ka\u200b=10\u2212pKa\u200b=10\u22123.74\u22481.82\u00d710\u22124<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    3. ICE Table<\/strong><\/h3>\n\n\n\n
    Species<\/th>Initial (M)<\/th>Change (M)<\/th>Equilibrium (M)<\/th><\/tr><\/thead>
    HCOOH<\/td>0.300<\/td>\u2212x-x\u2212x<\/td>0.300\u2212x0.300 – x0.300\u2212x<\/td><\/tr>
    H\u207a<\/td>0<\/td>+x+x+x<\/td>xxx<\/td><\/tr>
    HCOO\u207b<\/td>0<\/td>+x+x+x<\/td>xxx<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
    \n\n\n\n

    4. Apply Equilibrium Expression<\/strong><\/h3>\n\n\n\n

    Ka=[H+][HCOO\u2212][HCOOH]=x20.300\u2212xK_a = \\frac{[H^+][HCOO^-]}{[HCOOH]} = \\frac{x^2}{0.300 – x}Ka\u200b=[HCOOH][H+][HCOO\u2212]\u200b=0.300\u2212xx2\u200b<\/p>\n\n\n\n

    Assuming xxx is small compared to 0.300, we simplify:Ka\u2248x20.300\u21d2x2=Ka\u00d70.300=(1.82\u00d710\u22124)(0.300)=5.46\u00d710\u22125K_a \\approx \\frac{x^2}{0.300} \\Rightarrow x^2 = K_a \\times 0.300 = (1.82 \\times 10^{-4})(0.300) = 5.46 \\times 10^{-5}Ka\u200b\u22480.300×2\u200b\u21d2x2=Ka\u200b\u00d70.300=(1.82\u00d710\u22124)(0.300)=5.46\u00d710\u22125x=[H+]=5.46\u00d710\u22125\u22487.39\u00d710\u22123x = [H^+] = \\sqrt{5.46 \\times 10^{-5}} \\approx 7.39 \\times 10^{-3}x=[H+]=5.46\u00d710\u22125\u200b\u22487.39\u00d710\u22123<\/p>\n\n\n\n

    \n\n\n\n

    5. Calculate pH<\/strong><\/h3>\n\n\n\n

    pH=\u2212log\u2061[H+]=\u2212log\u2061(7.39\u00d710\u22123)\u22482.13\\text{pH} = -\\log[H^+] = -\\log(7.39 \\times 10^{-3}) \\approx 2.13pH=\u2212log[H+]=\u2212log(7.39\u00d710\u22123)\u22482.13<\/p>\n\n\n\n

    \n\n\n\n

    \u2705 Final Answer:<\/h3>\n\n\n\n

    pH = 2.13<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    \ud83d\udcd8 Explanation<\/h3>\n\n\n\n

    Formic acid (HCOOH) is a weak monoprotic acid<\/strong>, meaning it only partially ionizes in aqueous solution. This partial dissociation is governed by its acid dissociation constant KaK_aKa\u200b, which quantifies the extent of ionization. The given pKapK_apKa\u200b of 3.74 helps us calculate the KaK_aKa\u200b using the inverse log function:Ka=10\u2212pKa=10\u22123.74\u22481.82\u00d710\u22124K_a = 10^{-pK_a} = 10^{-3.74} \\approx 1.82 \\times 10^{-4}Ka\u200b=10\u2212pKa\u200b=10\u22123.74\u22481.82\u00d710\u22124<\/p>\n\n\n\n

    We begin by writing the balanced chemical equation:HCOOH\u21ccH++HCOO\u2212\\text{HCOOH} \\rightleftharpoons \\text{H}^+ + \\text{HCOO}^-HCOOH\u21ccH++HCOO\u2212<\/p>\n\n\n\n

    Since the initial concentration of formic acid is 0.300 M and none of the products are present initially, we set up an ICE table<\/strong> to track changes in concentration. Assuming the change in the concentration of formic acid is small relative to the initial concentration (a safe assumption for weak acids), we approximate the equilibrium concentration of the acid as 0.300\u2212x\u22480.3000.300 – x \\approx 0.3000.300\u2212x\u22480.300.<\/p>\n\n\n\n

    We then apply the expression for the equilibrium constant:Ka=[H+][HCOO\u2212][HCOOH]=x20.300K_a = \\frac{[H^+][HCOO^-]}{[HCOOH]} = \\frac{x^2}{0.300}Ka\u200b=[HCOOH][H+][HCOO\u2212]\u200b=0.300×2\u200b<\/p>\n\n\n\n

    Solving for xxx, we find the concentration of hydrogen ions, then calculate the pH using the definition of pH:pH=\u2212log\u2061[H+]\\text{pH} = -\\log[H^+]pH=\u2212log[H+]<\/p>\n\n\n\n

    The final pH of the 0.300 M formic acid solution is approximately 2.13<\/strong>, reflecting its weak acidic nature and partial ionization in water.<\/p>\n\n\n\n

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

    Formic acid, HCOOH, is a weak acid with a pKa of 3.74. Draw up a reaction table for the reaction of 0.300 M formic acid with water and calculate the pH of this solution. The Correct Answer and Explanation is: To solve this problem, we will analyze the ionization of formic acid (HCOOH), a weak […]<\/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-234342","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234342","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=234342"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234342\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=234342"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=234342"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=234342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}