Silver carbonate, Ag2CO3, has a solubility product constant of 8.46 x 10-12 at 25 \u00b0C.<\/p>\n\n\n\n
(a) A sample of solid silver carbonate is added to pure water. Calculate the molar solubility (mol\/L) and gram solubility (g Ag2CO3\/100 mL H20) of silver carbonate in water and determine [Agt) and (CO3).<\/p>\n\n\n\n
(b) Suppose 100.0 mL of 0.100 M AgNO3(aq) and 60.0 mL of 0.0600 M K2CO3(aq) are mixed. Will a precipitate form? Show your reasoning.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The solubility product constant (Ksp) of silver carbonate, ( \\text{Ag}_2\\text{CO}_3 ), is given as:<\/p>\n\n\n\n [ The dissociation of silver carbonate in water is as follows:<\/p>\n\n\n\n [ Let the molar solubility of ( \\text{Ag}_2\\text{CO}_3 ) be denoted by ( s ). From the dissociation equation, we can express the concentrations of the ions in solution:<\/p>\n\n\n\n Now, using the expression for the solubility product:<\/p>\n\n\n\n [ Substitute the concentrations:<\/p>\n\n\n\n [ Now solve for ( s ):<\/p>\n\n\n\n [ [ [ So, the molar solubility of silver carbonate in water is ( 1.28 \\times 10^{-4} \\, \\text{mol\/L} ).<\/p>\n\n\n\n To convert the molar solubility to gram solubility, we use the molar mass of ( \\text{Ag}_2\\text{CO}_3 ):<\/p>\n\n\n\n [ Now calculate the gram solubility:<\/p>\n\n\n\n [ Convert this to g per 100 mL:<\/p>\n\n\n\n [ From the molar solubility, the concentrations of the ions are:<\/p>\n\n\n\n [ [ We are mixing 100.0 mL of 0.100 M AgNO\u2083 with 60.0 mL of 0.0600 M K\u2082CO\u2083. We first calculate the concentrations of ( \\text{Ag}^+ ) and ( \\text{CO}_3^{2-} ) after mixing.<\/p>\n\n\n\n [ Concentration of ( \\text{Ag}^+ ) after mixing (total volume = 100.0 mL + 60.0 mL = 160.0 mL = 0.160 L):<\/p>\n\n\n\n [ [ Concentration of ( \\text{CO}_3^{2-} ) after mixing:<\/p>\n\n\n\n [ Now, calculate the reaction quotient ( Q_{sp} ) and compare it to ( K_{sp} ) to determine if a precipitate will form:<\/p>\n\n\n\n [ Since ( K_{sp} ) of ( \\text{Ag}2\\text{CO}_3 ) is ( 8.46 \\times 10^{-12} ), and ( Q<\/em>{sp} ) is much greater than ( K_{sp} ), a precipitate of ( \\text{Ag}_2\\text{CO}_3 ) will form.<\/p>\n\n\n\n Silver carbonate, Ag2CO3, has a solubility product constant of 8.46 x 10-12 at 25 \u00b0C. (a) A sample of solid silver carbonate is added to pure water. Calculate the molar solubility (mol\/L) and gram solubility (g Ag2CO3\/100 mL H20) of silver carbonate in water and determine [Agt) and (CO3). (b) Suppose 100.0 mL of 0.100 […]<\/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-186550","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186550","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=186550"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186550\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186550"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186550"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186550"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(a) Molar Solubility of Silver Carbonate in Water<\/h3>\n\n\n\n
K_{sp} = 8.46 \\times 10^{-12}
]<\/p>\n\n\n\n
\\text{Ag}_2\\text{CO}_3 (s) \\rightleftharpoons 2\\text{Ag}^+ (aq) + \\text{CO}_3^{2-} (aq)
]<\/p>\n\n\n\n\n
K_{sp} = [\\text{Ag}^+]^2 [\\text{CO}_3^{2-}]
]<\/p>\n\n\n\n
K_{sp} = (2s)^2 \\cdot s = 4s^3
]<\/p>\n\n\n\n
8.46 \\times 10^{-12} = 4s^3
]<\/p>\n\n\n\n
s^3 = \\frac{8.46 \\times 10^{-12}}{4} = 2.115 \\times 10^{-12}
]<\/p>\n\n\n\n
s = \\sqrt[3]{2.115 \\times 10^{-12}} = 1.28 \\times 10^{-4} \\, \\text{mol\/L}
]<\/p>\n\n\n\nGram Solubility of Silver Carbonate in Water<\/h3>\n\n\n\n
\\text{Molar mass of } \\text{Ag}_2\\text{CO}_3 = 2(107.87) + 12.01 + 3(16.00) = 275.87 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{Gram solubility} = 1.28 \\times 10^{-4} \\, \\text{mol\/L} \\times 275.87 \\, \\text{g\/mol} = 0.0353 \\, \\text{g\/L}
]<\/p>\n\n\n\n
\\text{Gram solubility} = 0.0353 \\, \\text{g\/L} \\times 0.1 = 0.00353 \\, \\text{g\/100 mL}
]<\/p>\n\n\n\nConcentrations of Ions in Solution<\/h3>\n\n\n\n
[\\text{Ag}^+] = 2s = 2 \\times 1.28 \\times 10^{-4} = 2.56 \\times 10^{-4} \\, \\text{mol\/L}
]<\/p>\n\n\n\n
[\\text{CO}_3^{2-}] = s = 1.28 \\times 10^{-4} \\, \\text{mol\/L}
]<\/p>\n\n\n\n(b) Will a Precipitate Form?<\/h3>\n\n\n\n
\n
\\text{moles of Ag}^+ = 0.100 \\, \\text{M} \\times 0.100 \\, \\text{L} = 0.0100 \\, \\text{mol}
]<\/p>\n\n\n\n
[\\text{Ag}^+] = \\frac{0.0100 \\, \\text{mol}}{0.160 \\, \\text{L}} = 0.0625 \\, \\text{M}
]<\/p>\n\n\n\n\n
\\text{moles of CO}_3^{2-} = 0.0600 \\, \\text{M} \\times 0.0600 \\, \\text{L} = 0.00360 \\, \\text{mol}
]<\/p>\n\n\n\n
[\\text{CO}_3^{2-}] = \\frac{0.00360 \\, \\text{mol}}{0.160 \\, \\text{L}} = 0.0225 \\, \\text{M}
]<\/p>\n\n\n\n
Q_{sp} = [\\text{Ag}^+]^2 [\\text{CO}_3^{2-}] = (0.0625)^2 \\times 0.0225 = 2.44 \\times 10^{-4}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n
\n