Compound: Lead(II) chloride (PbCl\u2082)<\/strong><\/h3>\n\n\n\n
\n\n\n\nChemical Equation for the Dissolution Equilibrium:<\/strong><\/h3>\n\n\n\nPbCl2(s)\u21ccPb2+(aq)+2Cl\u2212(aq)\\text{PbCl}_2 (s) \\rightleftharpoons \\text{Pb}^{2+} (aq) + 2\\text{Cl}^- (aq)<\/p>\n\n\n\n
This represents the dissociation of solid lead(II) chloride in water into its ions.<\/p>\n\n\n\n
\n\n\n\nLet the molar solubility be x (mol\/L).<\/strong><\/h3>\n\n\n\nFrom the dissociation equation:<\/p>\n\n\n\n
\n- [Pb\u00b2\u207a] = x<\/li>\n\n\n\n
- [Cl\u207b] = 2x (because 2 moles of Cl\u207b are produced per mole of PbCl\u2082)<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nKsp Expression (Solubility Product):<\/strong><\/h3>\n\n\n\nKsp=[Pb2+][Cl\u2212]2K_{sp} = [\\text{Pb}^{2+}][\\text{Cl}^-]^2 Ksp=(x)(2x)2=4x3K_{sp} = (x)(2x)^2 = 4x^3<\/p>\n\n\n\n
\n\n\n\nGiven: Ksp=1.7\u00d710\u22125K_{sp} = 1.7 \\times 10^{-5} at 25\u00b0C<\/strong><\/h3>\n\n\n\nNow solve for x (molar solubility):<\/strong> Ksp=4×3\u21d2x3=1.7\u00d710\u221254=4.25\u00d710\u22126K_{sp} = 4x^3 \\Rightarrow x^3 = \\frac{1.7 \\times 10^{-5}}{4} = 4.25 \\times 10^{-6} x=4.25\u00d710\u221263\u22480.0162\u2009mol\/Lx = \\sqrt[3]{4.25 \\times 10^{-6}} \\approx 0.0162 \\, \\text{mol\/L}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/strong><\/h3>\n\n\n\n\n- Molar Solubility (x):<\/strong> \u2248 0.0162 mol\/L<\/li>\n\n\n\n
- Ksp:<\/strong> 1.7\u00d710\u221251.7 \\times 10^{-5}<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nExplanation:<\/strong><\/h3>\n\n\n\nThe solubility product constant (Ksp<\/strong>) is an equilibrium constant that describes how much of an ionic compound dissolves in water to form a saturated solution. For sparingly soluble salts like lead(II) chloride (PbCl\u2082)<\/strong>, only a small amount dissolves, establishing a dynamic equilibrium between the undissolved solid and its dissolved ions.<\/p>\n\n\n\nWhen PbCl\u2082 dissolves, it breaks into one Pb\u00b2\u207a ion and two Cl\u207b ions. We represent this as: PbCl2(s)\u21ccPb2+(aq)+2Cl\u2212(aq)\\text{PbCl}_2 (s) \\rightleftharpoons \\text{Pb}^{2+} (aq) + 2\\text{Cl}^- (aq)<\/p>\n\n\n\n
The Ksp expression<\/strong> reflects the concentrations of the dissolved ions: Ksp=[Pb2+][Cl\u2212]2K_{sp} = [\\text{Pb}^{2+}][\\text{Cl}^-]^2<\/p>\n\n\n\nLet x<\/strong> represent the molar solubility \u2014 the number of moles of PbCl\u2082 that dissolve per liter. Since each mole of PbCl\u2082 gives one mole of Pb\u00b2\u207a and two moles of Cl\u207b, the ion concentrations are:
[Pb\u00b2\u207a] = x and [Cl\u207b] = 2x. Substituting into the Ksp equation gives: Ksp=x(2x)2=4x3K_{sp} = x(2x)^2 = 4x^3<\/p>\n\n\n\nGiven the Ksp<\/strong> value is 1.7\u00d710\u221251.7 \\times 10^{-5}, solving this equation gives: x=4.25\u00d710\u221263\u22480.0162\u2009mol\/Lx = \\sqrt[3]{4.25 \\times 10^{-6}} \\approx 0.0162 \\, \\text{mol\/L}<\/p>\n\n\n\nThis means only about 0.0162 moles of PbCl\u2082 dissolve in one liter of water at 25\u00b0C. The solubility is limited by the common ion effect and the low Ksp value. Knowing molar solubility is crucial in predicting precipitation, determining ion concentrations, and understanding chemical equilibria in aqueous systems.<\/p>\n\n\n\n![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner6-29.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":". Determine the molar solubility and value of Ksp for each compound in water at 25\u00b0C. Include the chemical equation for each equilibrium, the expression for Ksp (in terms of molarities and x), and show all work in calculations. Compound 2 PbCl2 Chemical equation: Molar solubility (x) = Ksp = 2. The Correct Answer and […]<\/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-219456","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219456","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=219456"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219456\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219456"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219456"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219456"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
PbCl2(s)\u21ccPb2+(aq)+2Cl\u2212(aq)\\text{PbCl}_2 (s) \\rightleftharpoons \\text{Pb}^{2+} (aq) + 2\\text{Cl}^- (aq)<\/p>\n\n\n\n
This represents the dissociation of solid lead(II) chloride in water into its ions.<\/p>\n\n\n\n
\n\n\n\n
Let the molar solubility be x (mol\/L).<\/strong><\/h3>\n\n\n\nFrom the dissociation equation:<\/p>\n\n\n\n
\n- [Pb\u00b2\u207a] = x<\/li>\n\n\n\n
- [Cl\u207b] = 2x (because 2 moles of Cl\u207b are produced per mole of PbCl\u2082)<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nKsp Expression (Solubility Product):<\/strong><\/h3>\n\n\n\nKsp=[Pb2+][Cl\u2212]2K_{sp} = [\\text{Pb}^{2+}][\\text{Cl}^-]^2 Ksp=(x)(2x)2=4x3K_{sp} = (x)(2x)^2 = 4x^3<\/p>\n\n\n\n
\n\n\n\nGiven: Ksp=1.7\u00d710\u22125K_{sp} = 1.7 \\times 10^{-5} at 25\u00b0C<\/strong><\/h3>\n\n\n\nNow solve for x (molar solubility):<\/strong> Ksp=4×3\u21d2x3=1.7\u00d710\u221254=4.25\u00d710\u22126K_{sp} = 4x^3 \\Rightarrow x^3 = \\frac{1.7 \\times 10^{-5}}{4} = 4.25 \\times 10^{-6} x=4.25\u00d710\u221263\u22480.0162\u2009mol\/Lx = \\sqrt[3]{4.25 \\times 10^{-6}} \\approx 0.0162 \\, \\text{mol\/L}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/strong><\/h3>\n\n\n\n\n- Molar Solubility (x):<\/strong> \u2248 0.0162 mol\/L<\/li>\n\n\n\n
- Ksp:<\/strong> 1.7\u00d710\u221251.7 \\times 10^{-5}<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nExplanation:<\/strong><\/h3>\n\n\n\nThe solubility product constant (Ksp<\/strong>) is an equilibrium constant that describes how much of an ionic compound dissolves in water to form a saturated solution. For sparingly soluble salts like lead(II) chloride (PbCl\u2082)<\/strong>, only a small amount dissolves, establishing a dynamic equilibrium between the undissolved solid and its dissolved ions.<\/p>\n\n\n\nWhen PbCl\u2082 dissolves, it breaks into one Pb\u00b2\u207a ion and two Cl\u207b ions. We represent this as: PbCl2(s)\u21ccPb2+(aq)+2Cl\u2212(aq)\\text{PbCl}_2 (s) \\rightleftharpoons \\text{Pb}^{2+} (aq) + 2\\text{Cl}^- (aq)<\/p>\n\n\n\n
The Ksp expression<\/strong> reflects the concentrations of the dissolved ions: Ksp=[Pb2+][Cl\u2212]2K_{sp} = [\\text{Pb}^{2+}][\\text{Cl}^-]^2<\/p>\n\n\n\nLet x<\/strong> represent the molar solubility \u2014 the number of moles of PbCl\u2082 that dissolve per liter. Since each mole of PbCl\u2082 gives one mole of Pb\u00b2\u207a and two moles of Cl\u207b, the ion concentrations are:
[Pb\u00b2\u207a] = x and [Cl\u207b] = 2x. Substituting into the Ksp equation gives: Ksp=x(2x)2=4x3K_{sp} = x(2x)^2 = 4x^3<\/p>\n\n\n\nGiven the Ksp<\/strong> value is 1.7\u00d710\u221251.7 \\times 10^{-5}, solving this equation gives: x=4.25\u00d710\u221263\u22480.0162\u2009mol\/Lx = \\sqrt[3]{4.25 \\times 10^{-6}} \\approx 0.0162 \\, \\text{mol\/L}<\/p>\n\n\n\nThis means only about 0.0162 moles of PbCl\u2082 dissolve in one liter of water at 25\u00b0C. The solubility is limited by the common ion effect and the low Ksp value. Knowing molar solubility is crucial in predicting precipitation, determining ion concentrations, and understanding chemical equilibria in aqueous systems.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":". Determine the molar solubility and value of Ksp for each compound in water at 25\u00b0C. Include the chemical equation for each equilibrium, the expression for Ksp (in terms of molarities and x), and show all work in calculations. Compound 2 PbCl2 Chemical equation: Molar solubility (x) = Ksp = 2. The Correct Answer and […]<\/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-219456","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219456","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=219456"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219456\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219456"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219456"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219456"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
Ksp Expression (Solubility Product):<\/strong><\/h3>\n\n\n\nKsp=[Pb2+][Cl\u2212]2K_{sp} = [\\text{Pb}^{2+}][\\text{Cl}^-]^2 Ksp=(x)(2x)2=4x3K_{sp} = (x)(2x)^2 = 4x^3<\/p>\n\n\n\n
\n\n\n\nGiven: Ksp=1.7\u00d710\u22125K_{sp} = 1.7 \\times 10^{-5} at 25\u00b0C<\/strong><\/h3>\n\n\n\nNow solve for x (molar solubility):<\/strong> Ksp=4×3\u21d2x3=1.7\u00d710\u221254=4.25\u00d710\u22126K_{sp} = 4x^3 \\Rightarrow x^3 = \\frac{1.7 \\times 10^{-5}}{4} = 4.25 \\times 10^{-6} x=4.25\u00d710\u221263\u22480.0162\u2009mol\/Lx = \\sqrt[3]{4.25 \\times 10^{-6}} \\approx 0.0162 \\, \\text{mol\/L}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/strong><\/h3>\n\n\n\n\n- Molar Solubility (x):<\/strong> \u2248 0.0162 mol\/L<\/li>\n\n\n\n
- Ksp:<\/strong> 1.7\u00d710\u221251.7 \\times 10^{-5}<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nExplanation:<\/strong><\/h3>\n\n\n\nThe solubility product constant (Ksp<\/strong>) is an equilibrium constant that describes how much of an ionic compound dissolves in water to form a saturated solution. For sparingly soluble salts like lead(II) chloride (PbCl\u2082)<\/strong>, only a small amount dissolves, establishing a dynamic equilibrium between the undissolved solid and its dissolved ions.<\/p>\n\n\n\nWhen PbCl\u2082 dissolves, it breaks into one Pb\u00b2\u207a ion and two Cl\u207b ions. We represent this as: PbCl2(s)\u21ccPb2+(aq)+2Cl\u2212(aq)\\text{PbCl}_2 (s) \\rightleftharpoons \\text{Pb}^{2+} (aq) + 2\\text{Cl}^- (aq)<\/p>\n\n\n\n
The Ksp expression<\/strong> reflects the concentrations of the dissolved ions: Ksp=[Pb2+][Cl\u2212]2K_{sp} = [\\text{Pb}^{2+}][\\text{Cl}^-]^2<\/p>\n\n\n\nLet x<\/strong> represent the molar solubility \u2014 the number of moles of PbCl\u2082 that dissolve per liter. Since each mole of PbCl\u2082 gives one mole of Pb\u00b2\u207a and two moles of Cl\u207b, the ion concentrations are:
[Pb\u00b2\u207a] = x and [Cl\u207b] = 2x. Substituting into the Ksp equation gives: Ksp=x(2x)2=4x3K_{sp} = x(2x)^2 = 4x^3<\/p>\n\n\n\nGiven the Ksp<\/strong> value is 1.7\u00d710\u221251.7 \\times 10^{-5}, solving this equation gives: x=4.25\u00d710\u221263\u22480.0162\u2009mol\/Lx = \\sqrt[3]{4.25 \\times 10^{-6}} \\approx 0.0162 \\, \\text{mol\/L}<\/p>\n\n\n\nThis means only about 0.0162 moles of PbCl\u2082 dissolve in one liter of water at 25\u00b0C. The solubility is limited by the common ion effect and the low Ksp value. Knowing molar solubility is crucial in predicting precipitation, determining ion concentrations, and understanding chemical equilibria in aqueous systems.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":". Determine the molar solubility and value of Ksp for each compound in water at 25\u00b0C. Include the chemical equation for each equilibrium, the expression for Ksp (in terms of molarities and x), and show all work in calculations. Compound 2 PbCl2 Chemical equation: Molar solubility (x) = Ksp = 2. The Correct Answer and […]<\/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-219456","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219456","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=219456"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219456\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219456"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219456"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219456"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Now solve for x (molar solubility):<\/strong> Ksp=4×3\u21d2x3=1.7\u00d710\u221254=4.25\u00d710\u22126K_{sp} = 4x^3 \\Rightarrow x^3 = \\frac{1.7 \\times 10^{-5}}{4} = 4.25 \\times 10^{-6} x=4.25\u00d710\u221263\u22480.0162\u2009mol\/Lx = \\sqrt[3]{4.25 \\times 10^{-6}} \\approx 0.0162 \\, \\text{mol\/L}<\/p>\n\n\n\n The solubility product constant (Ksp<\/strong>) is an equilibrium constant that describes how much of an ionic compound dissolves in water to form a saturated solution. For sparingly soluble salts like lead(II) chloride (PbCl\u2082)<\/strong>, only a small amount dissolves, establishing a dynamic equilibrium between the undissolved solid and its dissolved ions.<\/p>\n\n\n\n When PbCl\u2082 dissolves, it breaks into one Pb\u00b2\u207a ion and two Cl\u207b ions. We represent this as: PbCl2(s)\u21ccPb2+(aq)+2Cl\u2212(aq)\\text{PbCl}_2 (s) \\rightleftharpoons \\text{Pb}^{2+} (aq) + 2\\text{Cl}^- (aq)<\/p>\n\n\n\n The Ksp expression<\/strong> reflects the concentrations of the dissolved ions: Ksp=[Pb2+][Cl\u2212]2K_{sp} = [\\text{Pb}^{2+}][\\text{Cl}^-]^2<\/p>\n\n\n\n Let x<\/strong> represent the molar solubility \u2014 the number of moles of PbCl\u2082 that dissolve per liter. Since each mole of PbCl\u2082 gives one mole of Pb\u00b2\u207a and two moles of Cl\u207b, the ion concentrations are: Given the Ksp<\/strong> value is 1.7\u00d710\u221251.7 \\times 10^{-5}, solving this equation gives: x=4.25\u00d710\u221263\u22480.0162\u2009mol\/Lx = \\sqrt[3]{4.25 \\times 10^{-6}} \\approx 0.0162 \\, \\text{mol\/L}<\/p>\n\n\n\n This means only about 0.0162 moles of PbCl\u2082 dissolve in one liter of water at 25\u00b0C. The solubility is limited by the common ion effect and the low Ksp value. Knowing molar solubility is crucial in predicting precipitation, determining ion concentrations, and understanding chemical equilibria in aqueous systems.<\/p>\n\n\n\n . Determine the molar solubility and value of Ksp for each compound in water at 25\u00b0C. Include the chemical equation for each equilibrium, the expression for Ksp (in terms of molarities and x), and show all work in calculations. Compound 2 PbCl2 Chemical equation: Molar solubility (x) = Ksp = 2. The Correct Answer and […]<\/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-219456","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219456","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=219456"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219456\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219456"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219456"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219456"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n\u2705 Final Answers:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nExplanation:<\/strong><\/h3>\n\n\n\n
[Pb\u00b2\u207a] = x and [Cl\u207b] = 2x. Substituting into the Ksp equation gives: Ksp=x(2x)2=4x3K_{sp} = x(2x)^2 = 4x^3<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"