Net Ionic Equation and Equilibrium Expression<\/h3>\n\n\n\n
Lead(II) chromate (PbCrO4\\text{PbCrO}_4PbCrO4\u200b) is a slightly soluble salt, and its dissolution in water can be represented by the following net ionic equation<\/strong>: PbCrO4(s)\u21ccPb2+(aq)+CrO42\u2212(aq)\\text{PbCrO}_4(s) \\rightleftharpoons \\text{Pb}^{2+}(aq) + \\text{CrO}_4^{2-}(aq)PbCrO4\u200b(s)\u21ccPb2+(aq)+CrO42\u2212\u200b(aq)<\/p>\n\n\n\n The corresponding equilibrium constant expression<\/strong>, known as the solubility product (Ksp_\\text{sp}sp\u200b)<\/strong>, is: Ksp=[Pb2+][CrO42\u2212]K_{sp} = [\\text{Pb}^{2+}][\\text{CrO}_4^{2-}]Ksp\u200b=[Pb2+][CrO42\u2212\u200b]<\/p>\n\n\n\n This expression quantifies how much PbCrO4\\text{PbCrO}_4PbCrO4\u200b dissolves in water at equilibrium.<\/p>\n\n\n\n To increase the solubility<\/strong> of PbCrO4\\text{PbCrO}_4PbCrO4\u200b, one can add a substance that removes either Pb2+\\text{Pb}^{2+}Pb2+ or CrO42\u2212\\text{CrO}_4^{2-}CrO42\u2212\u200b from solution<\/strong>, thereby shifting the equilibrium to the right (according to Le Ch\u00e2telier\u2019s Principle).<\/p>\n\n\n\n Adding HCl introduces H+\\text{H}^+H+ ions, which react with CrO42\u2212\\text{CrO}_4^{2-}CrO42\u2212\u200b to form HCrO4\u2212\\text{HCrO}_4^-HCrO4\u2212\u200b: CrO42\u2212(aq)+H+(aq)\u21ccHCrO4\u2212(aq)\\text{CrO}_4^{2-}(aq) + \\text{H}^+(aq) \\rightleftharpoons \\text{HCrO}_4^-(aq)CrO42\u2212\u200b(aq)+H+(aq)\u21ccHCrO4\u2212\u200b(aq)<\/p>\n\n\n\n This reaction decreases the concentration of CrO42\u2212\\text{CrO}_4^{2-}CrO42\u2212\u200b, driving the dissolution equilibrium to the right and increasing the solubility of PbCrO4\\text{PbCrO}_4PbCrO4\u200b.<\/p>\n\n\n\n To decrease the concentration of Pb2+\\text{Pb}^{2+}Pb2+<\/strong> in solution, a reagent that forms an insoluble compound with lead can be added.<\/p>\n\n\n\n Sulfate ions react with lead(II) ions to form insoluble lead(II) sulfate: Pb2+(aq)+SO42\u2212(aq)\u2192PbSO4(s)\\text{Pb}^{2+}(aq) + \\text{SO}_4^{2-}(aq) \\rightarrow \\text{PbSO}_4(s)Pb2+(aq)+SO42\u2212\u200b(aq)\u2192PbSO4\u200b(s)<\/p>\n\n\n\n This reaction removes Pb2+\\text{Pb}^{2+}Pb2+ from solution by forming a precipitate, lowering its concentration.<\/p>\n\n\n\n These strategies exploit chemical equilibria to control solubility and ion concentrations in solution<\/p>\n\n\n\n Lead (II) chromate, PbCrO4, is a slightly soluble salt. Write a net ionic equation for the dissolution of solid PbCrO4 in water. Write an equilibrium constant expression for this reaction. K = What might one add to increase the solubility of the PbCrO4? Write equations to describe how this added reagent works to increase the […]<\/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-229530","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229530","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=229530"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229530\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229530"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229530"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229530"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nIncreasing Solubility of PbCrO\u2084<\/h3>\n\n\n\n
Example: Adding Hydrochloric Acid (HCl)<\/h4>\n\n\n\n
\n\n\n\nDecreasing Pb\u00b2\u207a Concentration<\/h3>\n\n\n\n
Example: Adding Sodium Sulfate (Na\u2082SO\u2084)<\/h4>\n\n\n\n
\n\n\n\nSummary<\/h3>\n\n\n\n
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