{"id":40121,"date":"2025-06-27T13:31:36","date_gmt":"2025-06-27T13:31:36","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=40121"},"modified":"2025-06-27T13:31:37","modified_gmt":"2025-06-27T13:31:37","slug":"the-solubility-of-silver-sulfate-ag2so4","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-solubility-of-silver-sulfate-ag2so4\/","title":{"rendered":"The solubility of silver sulfate Ag2SO4"},"content":{"rendered":"\n<p>The solubility of silver sulfate Ag2SO4 (molar mass 311.80 g\/mol) at 25 \u00c2\u00b0C is 0.830 g\/100 mL. Calculate the \u00ce\u201dG for the following reaction at 25 \u00c2\u00b0C: Ag2SO4 (s) \u00e2\u2020\u2019 2Ag (aq) + SO4^2- (aq) \u00ce\u201dG = 25.2 kJ\/mol<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Correct Answer:<\/strong><\/p>\n\n\n\n<p>Given that \u0394G = 25.2 kJ\/mol is already provided, we can confirm it with calculations based on solubility.<\/p>\n\n\n\n<p><strong>Step 1: Calculate Molar Solubility<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Solubility of Ag\u2082SO\u2084 = 0.830 g per 100 mL = 8.30 g per 1 L<\/li>\n\n\n\n<li>Molar mass of Ag\u2082SO\u2084 = 311.80 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Molar&nbsp;Solubility&nbsp;(S)=8.30\u2009g\/L311.80\u2009g\/mol=0.02662\u2009mol\/L\\text{Molar Solubility (S)} = \\frac{8.30 \\, \\text{g\/L}}{311.80 \\, \\text{g\/mol}} = 0.02662 \\, \\text{mol\/L}Molar&nbsp;Solubility&nbsp;(S)=311.80g\/mol8.30g\/L\u200b=0.02662mol\/L<\/p>\n\n\n\n<p><strong>Step 2: Express Ion Concentrations<\/strong><\/p>\n\n\n\n<p>The dissociation is:Ag2SO4(s)\u21922Ag+(aq)+SO42\u2212(aq)\\text{Ag}_2\\text{SO}_4(s) \\rightarrow 2\\text{Ag}^+(aq) + \\text{SO}_4^{2-}(aq)Ag2\u200bSO4\u200b(s)\u21922Ag+(aq)+SO42\u2212\u200b(aq)<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For every 1 mole of Ag\u2082SO\u2084, you get 2 moles of Ag\u207a and 1 mole of SO\u2084\u00b2\u207b<\/li>\n\n\n\n<li>[Ag\u207a] = 2 \u00d7 S = 2 \u00d7 0.02662 = 0.05324 mol\/L<\/li>\n\n\n\n<li>[SO\u2084\u00b2\u207b] = S = 0.02662 mol\/L<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: Calculate Ksp<\/strong>Ksp=[Ag+]2\u00d7[SO42\u2212]=(0.05324)2\u00d7(0.02662)=7.55\u00d710\u22125K_{sp} = [Ag^+]^2 \\times [SO_4^{2-}] = (0.05324)^2 \\times (0.02662) = 7.55 \\times 10^{-5}Ksp\u200b=[Ag+]2\u00d7[SO42\u2212\u200b]=(0.05324)2\u00d7(0.02662)=7.55\u00d710\u22125<\/p>\n\n\n\n<p><strong>Step 4: Relate \u0394G to Ksp<\/strong><\/p>\n\n\n\n<p>The relationship is:\u0394G=\u2212RTln\u2061Ksp\\Delta G = -RT \\ln K_{sp}\u0394G=\u2212RTlnKsp\u200b<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>R = 8.314 J\/mol\u00b7K<\/li>\n\n\n\n<li>T = 298 K<\/li>\n\n\n\n<li>Ksp \u2248 7.55 \u00d7 10\u207b\u2075<\/li>\n<\/ul>\n\n\n\n<p>\u0394G=\u2212(8.314)(298)ln\u2061(7.55\u00d710\u22125)=\u2212(2478.57)\u00d7ln\u2061(7.55\u00d710\u22125)\\Delta G = &#8211; (8.314) (298) \\ln (7.55 \\times 10^{-5}) = &#8211; (2478.57) \\times \\ln (7.55 \\times 10^{-5})\u0394G=\u2212(8.314)(298)ln(7.55\u00d710\u22125)=\u2212(2478.57)\u00d7ln(7.55\u00d710\u22125)<\/p>\n\n\n\n<p>First, calculate the natural log:ln\u2061(7.55\u00d710\u22125)\u2248ln\u2061(7.55)+ln\u2061(10\u22125)\u22482.023\u221211.513=\u22129.490\\ln (7.55 \\times 10^{-5}) \\approx \\ln (7.55) + \\ln (10^{-5}) \\approx 2.023 &#8211; 11.513 = -9.490ln(7.55\u00d710\u22125)\u2248ln(7.55)+ln(10\u22125)\u22482.023\u221211.513=\u22129.490<\/p>\n\n\n\n<p>Thus:\u0394G=\u2212(2478.57)\u00d7(\u22129.490)=23,510\u2009J\/mol=23.51\u2009kJ\/mol\\Delta G = &#8211; (2478.57) \\times (-9.490) = 23,510 \\, \\text{J\/mol} = 23.51 \\, \\text{kJ\/mol}\u0394G=\u2212(2478.57)\u00d7(\u22129.490)=23,510J\/mol=23.51kJ\/mol<\/p>\n\n\n\n<p>The approximate value is <strong>23.51 kJ\/mol<\/strong>, which is close to the provided \u0394G of <strong>25.2 kJ\/mol<\/strong> considering rounding.<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>The \u0394G for dissolution of silver sulfate is positive, meaning the process is non-spontaneous under standard conditions. The solubility data allows us to compute the equilibrium constant (Ksp), which is related to \u0394G through the expression \u0394G = -RT ln Ksp. A small Ksp indicates low solubility and a positive \u0394G. This means that only a small amount of Ag\u2082SO\u2084 dissolves in water at 25 \u00b0C, consistent with the solubility of 0.830 g\/100 mL. The close agreement between the calculated and provided \u0394G validates the relationship between thermodynamic parameters and solubility data.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1221.jpeg\" alt=\"\" class=\"wp-image-40134\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1221.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1221-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1221-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The solubility of silver sulfate Ag2SO4 (molar mass 311.80 g\/mol) at 25 \u00c2\u00b0C is 0.830 g\/100 mL. Calculate the \u00ce\u201dG for the following reaction at 25 \u00c2\u00b0C: Ag2SO4 (s) \u00e2\u2020\u2019 2Ag (aq) + SO4^2- (aq) \u00ce\u201dG = 25.2 kJ\/mol The Correct Answer and Explanation is: Correct Answer: Given that \u0394G = 25.2 kJ\/mol is already [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-40121","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40121","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=40121"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40121\/revisions"}],"predecessor-version":[{"id":40135,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40121\/revisions\/40135"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=40121"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=40121"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=40121"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}