What is the pH of 0.26 M H\u2083PO\u2084?
The Ka values for phosphoric acid are:
Ka1\u200b=7.5\u00d710\u22123<\/p>\n\n\n\n
Ka2=6.2\u00d710\u22128<\/p>\n\n\n\n
Ka3=3.6\u00d710\u221213<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To calculate the pH of a 0.26 M solution of phosphoric acid (H\u2083PO\u2084), we need to consider its dissociation in water and the respective acid dissociation constants (Ka values). Phosphoric acid is a triprotic acid, meaning it can donate three protons (H\u207a ions). The dissociation steps are as follows:<\/p>\n\n\n\n Given that (Ka1) is significantly larger than (Ka2) and (Ka3), we can assume that the first dissociation step will contribute the most to the concentration of H\u207a ions in solution.<\/p>\n\n\n\n Step 1: Calculate ([H^+]) from the first dissociation<\/strong>.<\/p>\n\n\n\n Using the expression for (Ka1): Step 2: Calculate pH<\/strong>. Conclusion<\/strong>: The pH of a 0.26 M solution of phosphoric acid (H\u2083PO\u2084) is approximately 1.35<\/strong>. The calculation demonstrates the predominance of the first dissociation step, which has the largest (K_a), while the contributions from subsequent dissociations are negligible due to their much smaller (K_a) values.<\/p>\n","protected":false},"excerpt":{"rendered":" What is the pH of 0.26 M H\u2083PO\u2084?The Ka values for phosphoric acid are:Ka1\u200b=7.5\u00d710\u22123 Ka2=6.2\u00d710\u22128 Ka3=3.6\u00d710\u221213 The Correct Answer and Explanation is: To calculate the pH of a 0.26 M solution of phosphoric acid (H\u2083PO\u2084), we need to consider its dissociation in water and the respective acid dissociation constants (Ka values). Phosphoric acid is a […]<\/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-161003","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/161003","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=161003"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/161003\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=161003"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=161003"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=161003"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
[
\\text{H}3\\text{PO}_4 \\rightleftharpoons \\text{H}^+ + \\text{H}_2\\text{PO}_4^- ] [ K<\/em>{a1} = 7.5 \\times 10^{-3}
]<\/li>\n\n\n\n
[
\\text{H}2\\text{PO}_4^- \\rightleftharpoons \\text{H}^+ + \\text{HPO}_4^{2-} ] [ K<\/em>{a2} = 6.2 \\times 10^{-8}
]<\/li>\n\n\n\n
[
\\text{HPO}4^{2-} \\rightleftharpoons \\text{H}^+ + \\text{PO}_4^{3-} ] [ K<\/em>{a3} = 3.6 \\times 10^{-13}
]<\/li>\n<\/ol>\n\n\n\n
[
K_{a1} = \\frac{[\\text{H}^+][\\text{H}2\\text{PO}_4^-]}{[\\text{H}_3\\text{PO}_4]} ] Assuming (x) is the amount that dissociates: [ K<\/em>{a1} = \\frac{x \\cdot x}{0.26 – x} \\approx \\frac{x^2}{0.26}
]
Setting (K_{a1} = 7.5 \\times 10^{-3}), we get:
[
7.5 \\times 10^{-3} = \\frac{x^2}{0.26}
]
[
x^2 = 7.5 \\times 10^{-3} \\times 0.26 \\approx 1.95 \\times 10^{-3}
]
[
x \\approx \\sqrt{1.95 \\times 10^{-3}} \\approx 0.0443 \\text{ M}
]<\/p>\n\n\n\n
Since (x) represents ([H^+]):
[
\\text{pH} = -\\log([H^+]) = -\\log(0.0443) \\approx 1.35
]<\/p>\n\n\n\n