Silver chromate is sparingly soluble in aqueous solutions. The Ksp of Ag2CrO4 is 1.12\u00d710-12. What is the solubility (in mol\/L) of silver chromate in 1.10M potassium chromate aqueous solution? What is the solubility (in mol\/L) of silver chromate in 1.10M silver nitrate aqueous solution? What is the solubility (in mol\/L) of silver chromate in pure water?<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To find the solubility of silver chromate (Ag\u2082CrO\u2084) under various conditions, we use its solubility product constant (Ksp)<\/strong>:<\/p>\n\n\n\n $$ Let solubility = s<\/strong> mol\/L Substitute into the Ksp expression:<\/p>\n\n\n\n $$ $$ \u2705 Solubility in pure water = 6.54 \u00d7 10\u207b\u2075 mol\/L<\/strong><\/p>\n\n\n\n K\u2082CrO\u2084 fully dissociates, providing $[CrO_4^{2-}] = 1.10$ M. Let the solubility of Ag\u2082CrO\u2084 be s<\/strong> mol\/L:<\/p>\n\n\n\n Substitute into Ksp:<\/p>\n\n\n\n $$ \u2705 Solubility in 1.10 M K\u2082CrO\u2084 = 1.59 \u00d7 10\u207b\u2077 mol\/L<\/strong><\/p>\n\n\n\n AgNO\u2083 fully dissociates, providing $[Ag^+] = 1.10$ M. Let the solubility of Ag\u2082CrO\u2084 be s<\/strong> mol\/L:<\/p>\n\n\n\n Substitute into Ksp:<\/p>\n\n\n\n $$ \u2705 Solubility in 1.10 M AgNO\u2083 = 9.26 \u00d7 10\u207b\u00b9\u00b3 mol\/L<\/strong><\/p>\n\n\n\n Silver chromate (Ag\u2082CrO\u2084) is a sparingly soluble salt. Its solubility is governed by its Ksp<\/strong>, which is an equilibrium constant describing the extent to which it dissolves into its ions. When no additional ions are present, we use the stoichiometry of dissociation to calculate solubility directly, finding it to be 6.54 \u00d7 10\u207b\u2075 mol\/L<\/strong> in pure water.<\/p>\n\n\n\n However, the solubility significantly decreases in solutions containing common ions<\/strong> due to the common ion effect<\/strong>. In 1.10 M K\u2082CrO\u2084, the high concentration of chromate ions suppresses dissolution of Ag\u2082CrO\u2084, reducing its solubility to 1.59 \u00d7 10\u207b\u2077 mol\/L<\/strong>. In 1.10 M AgNO\u2083, the abundant silver ions create an even stronger suppression, dropping solubility further to 9.26 \u00d7 10\u207b\u00b9\u00b3 mol\/L<\/strong>. This illustrates how ionic equilibria shift in response to the surrounding environment\u2014a fundamental principle in solution chemistry.<\/p>\n","protected":false},"excerpt":{"rendered":" Silver chromate is sparingly soluble in aqueous solutions. The Ksp of Ag2CrO4 is 1.12\u00d710-12. What is the solubility (in mol\/L) of silver chromate in 1.10M potassium chromate aqueous solution? What is the solubility (in mol\/L) of silver chromate in 1.10M silver nitrate aqueous solution? What is the solubility (in mol\/L) of silver chromate in pure […]<\/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-216128","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216128","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=216128"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216128\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=216128"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=216128"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=216128"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
K_{sp} = [Ag^+]^2[CrO_4^{2-}] = 1.12 \\times 10^{-12}
$$<\/p>\n\n\n\n1. Solubility in pure water:<\/strong><\/h3>\n\n\n\n
Then:<\/p>\n\n\n\n\n
K_{sp} = (2s)^2(s) = 4s^3
$$<\/p>\n\n\n\n
1.12 \\times 10^{-12} = 4s^3
\\Rightarrow s^3 = \\frac{1.12 \\times 10^{-12}}{4} = 2.8 \\times 10^{-13}
\\Rightarrow s = \\sqrt[3]{2.8 \\times 10^{-13}} \\approx 6.54 \\times 10^{-5} \\text{ mol\/L}
$$<\/p>\n\n\n\n
\n\n\n\n2. Solubility in 1.10 M K\u2082CrO\u2084:<\/strong><\/h3>\n\n\n\n
\n
K_{sp} = (2s)^2(1.10) = 4s^2(1.10)
\\Rightarrow 1.12 \\times 10^{-12} = 4.4s^2
\\Rightarrow s^2 = \\frac{1.12 \\times 10^{-12}}{4.4} = 2.545 \\times 10^{-13}
\\Rightarrow s = \\sqrt{2.545 \\times 10^{-13}} \\approx 1.59 \\times 10^{-7} \\text{ mol\/L}
$$<\/p>\n\n\n\n
\n\n\n\n3. Solubility in 1.10 M AgNO\u2083:<\/strong><\/h3>\n\n\n\n
\n
K_{sp} = (1.10)^2(s) = 1.21s
\\Rightarrow s = \\frac{1.12 \\times 10^{-12}}{1.21} \\approx 9.26 \\times 10^{-13} \\text{ mol\/L}
$$<\/p>\n\n\n\n
\n\n\n\nExplanation (Summary)<\/strong><\/h3>\n\n\n\n