Antimony(III) sulfide (Sb2S3, FW 339.68 g\/mole) is a very insoluble, ionic compound with a Kap of 1.7 x 10. Calculate the mass, in grams, of Sb,S; dissolved in Lake Erie (V4.58 x 10 L).<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the mass of antimony(III) sulfide (Sb\u2082S\u2083) that dissolves in Lake Erie, we can use the solubility product constant (Ksp), the volume of the lake, and the molar mass of Sb\u2082S\u2083.<\/p>\n\n\n\n Antimony(III) sulfide dissolves in water according to the following dissociation equation:<\/p>\n\n\n\n [ The solubility product (Ksp) is given by the equilibrium expression:<\/p>\n\n\n\n [ Let the molar solubility of Sb\u2082S\u2083 be denoted as ( s ) (in moles per liter). From the dissociation equation, for every mole of Sb\u2082S\u2083 that dissolves, 2 moles of Sb\u00b3\u207a and 3 moles of S\u00b2\u207b are produced. Thus, at equilibrium:<\/p>\n\n\n\n [ Substitute these expressions into the Ksp expression:<\/p>\n\n\n\n [ Simplifying this:<\/p>\n\n\n\n [ Given that the Ksp for Sb\u2082S\u2083 is ( 1.7 \\times 10^{-15} ), we can now solve for ( s ):<\/p>\n\n\n\n [ Solve for ( s ):<\/p>\n\n\n\n [ [ [ The volume of Lake Erie is ( 4.58 \\times 10^9 \\, \\text{L} ). To find the moles of Sb\u2082S\u2083 that can dissolve in this volume:<\/p>\n\n\n\n [ [ [ To find the mass, multiply the moles of Sb\u2082S\u2083 by its molar mass (339.68 g\/mol):<\/p>\n\n\n\n [ [ [ The mass of Sb\u2082S\u2083 that dissolves in Lake Erie is approximately ( 2.68 \\times 10^8 \\, \\text{g} ).<\/p>\n\n\n\n We calculated the mass of Sb\u2082S\u2083 that dissolves by first using the Ksp value to find the molar solubility (( s )) of Sb\u2082S\u2083 in water. Using this value, we determined how many moles of Sb\u2082S\u2083 would dissolve in the volume of Lake Erie, and finally, we multiplied by the molar mass of Sb\u2082S\u2083 to find the total mass.<\/p>\n","protected":false},"excerpt":{"rendered":" Antimony(III) sulfide (Sb2S3, FW 339.68 g\/mole) is a very insoluble, ionic compound with a Kap of 1.7 x 10. Calculate the mass, in grams, of Sb,S; dissolved in Lake Erie (V4.58 x 10 L). The Correct Answer and Explanation is : To calculate the mass of antimony(III) sulfide (Sb\u2082S\u2083) that dissolves in Lake Erie, we […]<\/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-186100","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186100","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=186100"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186100\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186100"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186100"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186100"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Write the dissociation equation for Sb\u2082S\u2083<\/h3>\n\n\n\n
\\text{Sb}_2\\text{S}_3 (s) \\rightleftharpoons 2\\text{Sb}^{3+} (aq) + 3\\text{S}^{2-} (aq)
]<\/p>\n\n\n\nStep 2: Set up the expression for the solubility product constant (Ksp)<\/h3>\n\n\n\n
K_{sp} = [\\text{Sb}^{3+}]^2 [\\text{S}^{2-}]^3
]<\/p>\n\n\n\n
[\\text{Sb}^{3+}] = 2s \\quad \\text{and} \\quad [\\text{S}^{2-}] = 3s
]<\/p>\n\n\n\n
K_{sp} = (2s)^2 (3s)^3
]<\/p>\n\n\n\n
K_{sp} = 4s^2 \\times 27s^3 = 108s^5
]<\/p>\n\n\n\n
1.7 \\times 10^{-15} = 108s^5
]<\/p>\n\n\n\n
s^5 = \\frac{1.7 \\times 10^{-15}}{108}
]<\/p>\n\n\n\n
s^5 = 1.57 \\times 10^{-17}
]<\/p>\n\n\n\n
s = \\sqrt[5]{1.57 \\times 10^{-17}} \\approx 1.72 \\times 10^{-4} \\, \\text{mol\/L}
]<\/p>\n\n\n\nStep 3: Calculate the moles of Sb\u2082S\u2083 dissolved in Lake Erie<\/h3>\n\n\n\n
\\text{moles of Sb}_2\\text{S}_3 = s \\times \\text{volume of Lake Erie}
]<\/p>\n\n\n\n
\\text{moles of Sb}_2\\text{S}_3 = (1.72 \\times 10^{-4} \\, \\text{mol\/L}) \\times (4.58 \\times 10^9 \\, \\text{L})
]<\/p>\n\n\n\n
\\text{moles of Sb}_2\\text{S}_3 \\approx 7.88 \\times 10^5 \\, \\text{mol}
]<\/p>\n\n\n\nStep 4: Calculate the mass of Sb\u2082S\u2083<\/h3>\n\n\n\n
\\text{mass of Sb}_2\\text{S}_3 = \\text{moles of Sb}_2\\text{S}_3 \\times \\text{molar mass of Sb}_2\\text{S}_3
]<\/p>\n\n\n\n
\\text{mass of Sb}_2\\text{S}_3 = (7.88 \\times 10^5 \\, \\text{mol}) \\times (339.68 \\, \\text{g\/mol})
]<\/p>\n\n\n\n
\\text{mass of Sb}_2\\text{S}_3 \\approx 2.68 \\times 10^8 \\, \\text{g}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n