{"id":207339,"date":"2025-04-21T13:55:48","date_gmt":"2025-04-21T13:55:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=207339"},"modified":"2025-04-21T13:55:50","modified_gmt":"2025-04-21T13:55:50","slug":"enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-11","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/21\/enough-of-a-monoprotic-acid-is-dissolved-in-water-to-produce-a-0-0192-m-solution-11\/","title":{"rendered":"Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution"},"content":{"rendered":"\n

Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To calculate the acid dissociation constant (Ka)<\/strong> for the monoprotic acid<\/strong>, we follow these steps:<\/p>\n\n\n\n

\n\n\n\n

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

Ka = 1.83 \u00d7 10\u207b\u2075<\/strong><\/p>\n\n\n\n

\n\n\n\n

\ud83e\uddea Step-by-Step Explanation:<\/h3>\n\n\n\n

1. Given Data:<\/strong><\/h4>\n\n\n\n
    \n
  • Initial concentration of acid, [HA] = 0.0192 M<\/strong><\/li>\n\n\n\n
  • pH of solution = 2.53<\/strong><\/li>\n\n\n\n
  • Monoprotic acid (donates 1 proton<\/strong> per molecule)<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    2. Find ([H\u207a]) from pH:<\/strong><\/h4>\n\n\n\n

    We use the formula:<\/p>\n\n\n\n

    [
    [H^+] = 10^{-\\text{pH}} = 10^{-2.53}
    ]<\/p>\n\n\n\n

    [
    [H^+] = 2.95 \\times 10^{-3} \\, \\text{M}
    ]<\/p>\n\n\n\n

    This is the amount of acid that dissociated<\/strong> into H\u207a and A\u207b.<\/p>\n\n\n\n

    \n\n\n\n

    3. ICE Table Setup:<\/strong><\/h4>\n\n\n\n
    Species<\/th>Initial (M)<\/th>Change (M)<\/th>Equilibrium (M)<\/th><\/tr><\/thead>
    HA<\/td>0.0192<\/td>\u2013x<\/td>0.0192 \u2013 x<\/td><\/tr>
    H\u207a<\/td>0<\/td>+x<\/td>x = 2.95 \u00d7 10\u207b\u00b3<\/td><\/tr>
    A\u207b<\/td>0<\/td>+x<\/td>x = 2.95 \u00d7 10\u207b\u00b3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Since ([H\u207a] = x), we can substitute into the Ka<\/strong> expression:<\/p>\n\n\n\n

    [
    K_a = \\frac{[H^+][A^-]}{[HA]} = \\frac{x^2}{0.0192 – x}
    ]<\/p>\n\n\n\n

    \n\n\n\n

    4. Substitute values:<\/strong><\/h4>\n\n\n\n

    [
    K_a = \\frac{(2.95 \\times 10^{-3})^2}{0.0192 – 2.95 \\times 10^{-3}}
    ]<\/p>\n\n\n\n

    [
    = \\frac{8.70 \\times 10^{-6}}{0.01625}
    ]<\/p>\n\n\n\n

    [
    K_a = 1.83 \\times 10^{-4}
    ]<\/p>\n\n\n\n

    \n\n\n\n

    \u2757 Correction:<\/h3>\n\n\n\n

    Oops! Wait \u2014 there’s a typo. Let’s recalculate that denominator properly:<\/p>\n\n\n\n

    [
    K_a = \\frac{8.70 \\times 10^{-6}}{0.01625} = 5.35 \\times 10^{-4}
    ]<\/p>\n\n\n\n

    BUT this contradicts our earlier value. Let’s check once again using precise values:<\/p>\n\n\n\n

    [
    (2.95 \\times 10^{-3})^2 = 8.70 \\times 10^{-6}
    ]<\/p>\n\n\n\n

    [
    0.0192 – 0.00295 = 0.01625
    ]<\/p>\n\n\n\n

    [
    Ka = \\frac{8.70 \\times 10^{-6}}{0.01625} \u2248 5.35 \\times 10^{-4}
    ]<\/p>\n\n\n\n

    \u2714\ufe0f Final Corrected Ka = 5.35 \u00d7 10\u207b\u2074<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    \ud83e\udde0 Explanation:<\/h3>\n\n\n\n

    Ka measures how much the acid dissociates in water. A stronger acid<\/strong> has a higher Ka, meaning more ionization. By knowing the initial concentration<\/strong> and the pH<\/strong>, we calculate how much H\u207a is in solution, and use that to find Ka using the equilibrium expression.<\/p>\n\n\n\n

    This problem is classic for weak acids\u2014since it\u2019s monoprotic, the math stays cleaner. The most important step is accurately converting pH to [H\u207a]<\/strong>, and then applying it to the Ka expression<\/strong>.<\/p>\n\n\n\n

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

    Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid. The correct answer and explanation is : To calculate the acid dissociation constant (Ka) for the monoprotic acid, we follow these steps: \u2705 Final Answer: Ka […]<\/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-207339","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207339","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=207339"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207339\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207339"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207339"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207339"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}