{"id":190277,"date":"2025-02-12T05:19:54","date_gmt":"2025-02-12T05:19:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190277"},"modified":"2025-02-12T05:19:58","modified_gmt":"2025-02-12T05:19:58","slug":"the-acid-dissociation-constants-of-phosphoric-acid-h3po4-are-ka1-7-5x10-3-ka2-6-2x10-8-and-ka3-4-2x10-13-at-25-0-degrees-celcius","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/12\/the-acid-dissociation-constants-of-phosphoric-acid-h3po4-are-ka1-7-5x10-3-ka2-6-2x10-8-and-ka3-4-2x10-13-at-25-0-degrees-celcius\/","title":{"rendered":"The acid-dissociation constants of phosphoric acid (H3PO4) are Ka1= 7.5×10-3, Ka2 =6.2×10-8, and Ka3 =4.2×10-13 at 25.0 degrees celcius"},"content":{"rendered":"\n

The acid-dissociation constants of phosphoric acid (H3PO4) are Ka1= 7.5×10-3, Ka2 =6.2×10-8, and Ka3 =4.2×10-13 at 25.0 degrees celcius. What is the pH of a 3.0M aqueous solution of phosphoric acid?<\/p>\n\n\n\n

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

To calculate the pH of a 3.0 M aqueous solution of phosphoric acid (H\u2083PO\u2084), we must consider the acid dissociation constants (Ka1, Ka2, and Ka3) and recognize that phosphoric acid undergoes stepwise dissociation in water. The dissociation reactions and the given values for Ka are:<\/p>\n\n\n\n

    \n
  1. First dissociation:<\/strong>
    [
    H\u2083PO\u2084 (aq) \\rightleftharpoons H\u207a (aq) + H\u2082PO\u2084\u207b (aq)
    ]
    Ka1 = ( 7.5 \\times 10^{-3} )<\/li>\n\n\n\n
  2. Second dissociation:<\/strong>
    [
    H\u2082PO\u2084\u207b (aq) \\rightleftharpoons H\u207a (aq) + HPO\u2084\u00b2\u207b (aq)
    ]
    Ka2 = ( 6.2 \\times 10^{-8} )<\/li>\n\n\n\n
  3. Third dissociation:<\/strong>
    [
    HPO\u2084\u00b2\u207b (aq) \\rightleftharpoons H\u207a (aq) + PO\u2084\u00b3\u207b (aq)
    ]
    Ka3 = ( 4.2 \\times 10^{-13} )<\/li>\n<\/ol>\n\n\n\n

    Since the solution is initially 3.0 M in H\u2083PO\u2084, the first dissociation is the most significant, as Ka1 is much larger than Ka2 and Ka3. This means the concentration of H\u207a from the first dissociation will dominate the pH.<\/p>\n\n\n\n

    Step 1: Set up an ICE table for the first dissociation<\/h3>\n\n\n\n
    Species<\/th>Initial (M)<\/th>Change (M)<\/th>Equilibrium (M)<\/th><\/tr><\/thead>
    H\u2083PO\u2084<\/td>3.0<\/td>-x<\/td>3.0 – x<\/td><\/tr>
    H\u207a<\/td>0<\/td>+x<\/td>x<\/td><\/tr>
    H\u2082PO\u2084\u207b<\/td>0<\/td>+x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Using the expression for Ka1:
    [
    Ka1 = \\frac{[H\u207a][H\u2082PO\u2084\u207b]}{[H\u2083PO\u2084]} = 7.5 \\times 10^{-3}
    ]<\/p>\n\n\n\n

    Substituting the equilibrium concentrations:
    [
    7.5 \\times 10^{-3} = \\frac{x^2}{3.0 – x}
    ]<\/p>\n\n\n\n

    Since Ka1 is relatively large compared to Ka2 and Ka3, we can assume that the change in concentration (x) is small relative to 3.0 M, so (3.0 – x \\approx 3.0). This simplifies the equation to:
    [
    7.5 \\times 10^{-3} = \\frac{x^2}{3.0}
    ]
    [
    x^2 = 7.5 \\times 10^{-3} \\times 3.0 = 2.25 \\times 10^{-2}
    ]
    [
    x = \\sqrt{2.25 \\times 10^{-2}} = 0.15 \\, \\text{M}
    ]<\/p>\n\n\n\n

    Step 2: Calculate pH<\/h3>\n\n\n\n

    The concentration of H\u207a ions is approximately 0.15 M. The pH is calculated as:
    [
    pH = -\\log[H\u207a] = -\\log(0.15) \\approx 0.82
    ]<\/p>\n\n\n\n

    Thus, the pH of the 3.0 M solution of phosphoric acid is approximately 0.82<\/strong>.<\/p>\n\n\n\n

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

    Phosphoric acid dissociates in steps, with the first dissociation being the most important for determining the pH. Since Ka1 is significantly larger than Ka2 and Ka3, the contribution of H\u207a ions from the first dissociation far outweighs those from the second and third dissociation steps. This results in a relatively low pH for a concentrated phosphoric acid solution, typical of strong acids.<\/p>\n\n\n\n

    Let me also generate an image to visually represent this concept.<\/p>\n\n\n\n

    Here is the image showing the dissociation of phosphoric acid (H\u2083PO\u2084) in water, with the first dissociation step and the resulting pH calculation. This visual helps illustrate the equilibrium process and the pH level of the 3.0 M solution of phosphoric acid. Let me know if you need any further details!<\/p>\n\n\n\n

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

    The acid-dissociation constants of phosphoric acid (H3PO4) are Ka1= 7.5×10-3, Ka2 =6.2×10-8, and Ka3 =4.2×10-13 at 25.0 degrees celcius. What is the pH of a 3.0M aqueous solution of phosphoric acid? The Correct Answer and Explanation is : To calculate the pH of a 3.0 M aqueous solution of phosphoric acid (H\u2083PO\u2084), we must consider […]<\/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-190277","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190277","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=190277"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190277\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190277"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190277"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190277"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}