To calculate the pH of a 2.5 M aqueous solution of phosphoric acid (H\u2083PO\u2084), we need to consider its stepwise dissociation. Phosphoric acid is a triprotic acid, meaning it can donate three protons (H\u207a) in a stepwise manner. The dissociation constants are given as:<\/p>\n\n\n\n
- \n
- Ka\u2081 = 7.5 \u00d7 10\u207b\u00b3<\/strong> (for the dissociation of H\u2083PO\u2084 to H\u2082PO\u2084\u207b)<\/li>\n\n\n\n
- Ka\u2082 = 6.2 \u00d7 10\u207b\u2078<\/strong> (for the dissociation of H\u2082PO\u2084\u207b to HPO\u2084\u00b2\u207b)<\/li>\n\n\n\n
- Ka\u2083 = 4.2 \u00d7 10\u207b\u00b9\u00b3<\/strong> (for the dissociation of HPO\u2084\u00b2\u207b to PO\u2084\u00b3\u207b)<\/li>\n<\/ul>\n\n\n\n
At high concentrations, the first dissociation is the most significant. So, we will first focus on the first dissociation:<\/p>\n\n\n\n
H\u2083PO\u2084 \u21cc H\u207a + H\u2082PO\u2084\u207b<\/strong><\/p>\n\n\n\n
Step 1: Solve for the concentration of H\u207a from the first dissociation<\/h3>\n\n\n\n
We use the expression for Ka\u2081<\/strong>:<\/p>\n\n\n\n
[
Ka\u2081 = \\frac{[H\u207a][H\u2082PO\u2084\u207b]}{[H\u2083PO\u2084]}
]<\/p>\n\n\n\nLet x be the concentration of H\u207a produced. Initially, [H\u2083PO\u2084] is 2.5 M, and we assume that most of the dissociation comes from the first step. Therefore:<\/p>\n\n\n\n
[
Ka\u2081 = \\frac{x^2}{2.5 – x}
]<\/p>\n\n\n\nGiven Ka\u2081 = 7.5 \u00d7 10\u207b\u00b3, we substitute:<\/p>\n\n\n\n
[
7.5 \\times 10^{-3} = \\frac{x^2}{2.5 – x}
]<\/p>\n\n\n\nAssuming x is small relative to 2.5 M, we can approximate 2.5 – x \u2248 2.5. This simplifies to:<\/p>\n\n\n\n
[
7.5 \\times 10^{-3} = \\frac{x^2}{2.5}
]<\/p>\n\n\n\nSolving for x:<\/p>\n\n\n\n
[
x^2 = (7.5 \\times 10^{-3}) \\times 2.5
]
[
x^2 = 1.875 \\times 10^{-2}
]
[
x = 1.37 \\times 10^{-1} \\, \\text{M}
]<\/p>\n\n\n\nThus, the concentration of H\u207a from the first dissociation is 0.137 M.<\/p>\n\n\n\n
Step 2: Calculate the pH<\/h3>\n\n\n\n
The pH is defined as:<\/p>\n\n\n\n
[
\\text{pH} = -\\log[H\u207a]
]<\/p>\n\n\n\nSubstituting [H\u207a] = 0.137 M:<\/p>\n\n\n\n
[
\\text{pH} = -\\log(0.137) \\approx 0.86
]<\/p>\n\n\n\nThus, the pH of a 2.5 M aqueous solution of phosphoric acid is approximately 0.86<\/strong>.<\/p>\n\n\n\n
The second and third dissociations are negligible at this concentration, so they do not significantly affect the pH. The major contributor is the first dissociation.<\/p>\n\n\n\n
Now, I will generate an image of this calculation.<\/p>\n\n\n\n
Here is the image showing the detailed mathematical calculation for the pH of a 2.5M aqueous solution of phosphoric acid. It includes the steps for solving the first dissociation and calculating the pH. Let me know if you need further clarification!<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-349.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"The acid-dissociation constants of phosphoric acid (H3PO4) at 25 degrees Celsius are Ka1 = 7.510-3, Ka2 =6.2 10-8, and Ka3 =4.210-13. What is the PH of a 2.5M aqueous solution of phosphoric acid? The Correct Answer and Explanation is : To calculate the pH of a 2.5 M aqueous solution of phosphoric acid (H\u2083PO\u2084), we […]<\/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-190280","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190280","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=190280"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190280\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190280"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190280"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190280"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Ka\u2082 = 6.2 \u00d7 10\u207b\u2078<\/strong> (for the dissociation of H\u2082PO\u2084\u207b to HPO\u2084\u00b2\u207b)<\/li>\n\n\n\n