The compound lead(II) chromate (PbCrO4, molar mass = 323.194 g\/mol) is only slightly soluble in water. Its Ksp at 25\u00b0C is 1.8\u00d710-14. Estimate the solubility of lead(II) chromate in water in units of grams per 100 mL of water. g per 100 mL<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To estimate the solubility of lead(II) chromate (PbCrO\u2084) in water in grams per 100 mL<\/strong>, we first determine its solubility in mol\/L (molar solubility)<\/strong> using its Ksp<\/strong> value, then convert that to grams per 100 mL.<\/p>\n\n\n\n Lead(II) chromate dissociates in water as: 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 Let the solubility be s mol\/L<\/strong>, then:<\/p>\n\n\n\n Ksp=[Pb2+][CrO42\u2212]=s\u22c5s=s2K_{sp} = [\\text{Pb}^{2+}][\\text{CrO}_4^{2-}] = s \\cdot s = s^2Ksp\u200b=[Pb2+][CrO42\u2212\u200b]=s\u22c5s=s2 1.8\u00d710\u221214=s21.8 \\times 10^{-14} = s^21.8\u00d710\u221214=s2 s=1.8\u00d710\u221214=1.34\u00d710\u22127 mol\/Ls = \\sqrt{1.8 \\times 10^{-14}} = 1.34 \\times 10^{-7} \\text{ mol\/L}s=1.8\u00d710\u221214\u200b=1.34\u00d710\u22127 mol\/L<\/p>\n\n\n\n Molar mass of PbCrO4=323.194 g\/mol\\text{Molar mass of PbCrO}_4 = 323.194 \\text{ g\/mol}Molar mass of PbCrO4\u200b=323.194 g\/mol Solubility (g\/L)=1.34\u00d710\u22127 mol\/L\u00d7323.194 g\/mol\u22484.33\u00d710\u22125 g\/L\\text{Solubility (g\/L)} = 1.34 \\times 10^{-7} \\text{ mol\/L} \\times 323.194 \\text{ g\/mol} \\approx 4.33 \\times 10^{-5} \\text{ g\/L}Solubility (g\/L)=1.34\u00d710\u22127 mol\/L\u00d7323.194 g\/mol\u22484.33\u00d710\u22125 g\/L<\/p>\n\n\n\n Solubility (g\/100 mL)=4.33\u00d710\u22125 g1000 mL\u00d7100 mL=4.33\u00d710\u22126 g\/100 mL\\text{Solubility (g\/100 mL)} = \\frac{4.33 \\times 10^{-5} \\text{ g}}{1000 \\text{ mL}} \\times 100 \\text{ mL} = 4.33 \\times 10^{-6} \\text{ g\/100 mL}Solubility (g\/100 mL)=1000 mL4.33\u00d710\u22125 g\u200b\u00d7100 mL=4.33\u00d710\u22126 g\/100 mL<\/p>\n\n\n\n 4.33\u00d710\u22126 g per 100 mL\\boxed{4.33 \\times 10^{-6} \\text{ g per 100 mL}}4.33\u00d710\u22126 g per 100 mL\u200b<\/p>\n\n\n\n Solubility is a measure of how much of a substance can dissolve in a given amount of solvent. For sparingly soluble salts like lead(II) chromate (PbCrO\u2084), solubility is often calculated using the solubility product constant (Ksp). The Ksp is an equilibrium constant that reflects the extent to which a salt dissociates into its ions in solution.<\/p>\n\n\n\n When PbCrO\u2084 dissolves, it breaks into Pb\u00b2\u207a and CrO\u2084\u00b2\u207b ions in a 1:1 molar ratio. If we let \u201cs\u201d represent the molar solubility (in mol\/L), then the equilibrium concentrations of both Pb\u00b2\u207a and CrO\u2084\u00b2\u207b will be \u201cs.\u201d The Ksp expression for this equilibrium is then Ksp=[Pb2+][CrO42\u2212]=s2K_{sp} = [Pb^{2+}][CrO_4^{2-}] = s^2Ksp\u200b=[Pb2+][CrO42\u2212\u200b]=s2. By substituting the known Ksp value of 1.8\u00d710\u2212141.8 \\times 10^{-14}1.8\u00d710\u221214, we can solve for \u201cs,\u201d the molar solubility.<\/p>\n\n\n\n Once the molar solubility is determined (approximately 1.34\u00d710\u221271.34 \\times 10^{-7}1.34\u00d710\u22127 mol\/L), we convert it to mass by multiplying it by the molar mass of PbCrO\u2084 (323.194 g\/mol), resulting in about 4.33\u00d710\u221254.33 \\times 10^{-5}4.33\u00d710\u22125 g\/L. Since the question asks for grams per 100 mL, we scale this down by a factor of 10 (because 100 mL is one-tenth of a liter), yielding 4.33\u00d710\u221264.33 \\times 10^{-6}4.33\u00d710\u22126 g per 100 mL.<\/p>\n\n\n\n This very low solubility reflects the fact that PbCrO\u2084 is highly insoluble in water\u2014a property common among heavy metal chromates. Such data is important in environmental chemistry, where understanding the mobility and bioavailability of heavy metal compounds is critical for pollution control and water safety.<\/p>\n","protected":false},"excerpt":{"rendered":" The compound lead(II) chromate (PbCrO4, molar mass = 323.194 g\/mol) is only slightly soluble in water. Its Ksp at 25\u00b0C is 1.8\u00d710-14. Estimate the solubility of lead(II) chromate in water in units of grams per 100 mL of water. g per 100 mL The Correct Answer and Explanation is: To estimate the solubility of lead(II) […]<\/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-212739","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212739","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=212739"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212739\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=212739"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=212739"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=212739"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Dissociation Equation<\/h3>\n\n\n\n
\n
\n\n\n\nStep 2: Ksp Expression<\/h3>\n\n\n\n
\n\n\n\nStep 3: Convert molar solubility to grams per liter<\/h3>\n\n\n\n
\n\n\n\nStep 4: Convert g\/L to g per 100 mL<\/h3>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/h3>\n\n\n\n
\n\n\n\n\ud83e\uddea Explanation (300+ words):<\/h3>\n\n\n\n