{"id":261503,"date":"2025-07-20T06:45:44","date_gmt":"2025-07-20T06:45:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=261503"},"modified":"2025-07-20T06:45:47","modified_gmt":"2025-07-20T06:45:47","slug":"what-is-the-solubility-of-copper-hydroxide-cuoh2-in-ml-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/20\/what-is-the-solubility-of-copper-hydroxide-cuoh2-in-ml-2\/","title":{"rendered":"What is the solubility of copper hydroxide Cu(OH)2 in mL"},"content":{"rendered":"\n

What is the solubility of copper hydroxide Cu(OH)2 in mL? Mwt of Cu(OH)2 = 97.57; Ksp = 5.5 x 10^21, Cu^2+ (aq) + 2 OH^- \u00e2\u2020\u2019 Cu(OH)2 (s) (aq)<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To determine the solubility of copper hydroxide (Cu(OH)2) in mL, we first need to use the given solubility product constant (Ksp) and molar mass of Cu(OH)2 to find its solubility in moles per liter (mol\/L) and then convert it into mL.<\/p>\n\n\n\n

Step 1: Write the dissociation equation for Cu(OH)2<\/h3>\n\n\n\n

The dissociation of Cu(OH)2 in water is given by:<\/p>\n\n\n\n

Cu(OH)2(s)\u21ccCu2+(aq)+2OH\u2212(aq)\\text{Cu(OH)}_2 (s) \\rightleftharpoons \\text{Cu}^{2+} (aq) + 2 \\text{OH}^- (aq)Cu(OH)2\u200b(s)\u21ccCu2+(aq)+2OH\u2212(aq)<\/p>\n\n\n\n

The solubility product constant (Ksp) expression for this dissociation is:<\/p>\n\n\n\n

Ksp=[Cu2+][OH\u2212]2K_{sp} = [\\text{Cu}^{2+}] [\\text{OH}^-]^2Ksp\u200b=[Cu2+][OH\u2212]2<\/p>\n\n\n\n

Step 2: Define solubility in terms of molarity<\/h3>\n\n\n\n

Let the solubility of Cu(OH)2 in mol\/L be SSS. When Cu(OH)2 dissolves, for every 1 mole of Cu(OH)2 that dissolves, 1 mole of Cu\u00b2\u207a and 2 moles of OH\u207b are produced. Therefore:<\/p>\n\n\n\n

    \n
  • [Cu\u00b2\u207a] = SSS (in mol\/L)<\/li>\n\n\n\n
  • [OH\u207b] = 2S (in mol\/L)<\/li>\n<\/ul>\n\n\n\n

    Step 3: Substitute values into the Ksp equation<\/h3>\n\n\n\n

    Substitute the expressions for [Cu\u00b2\u207a] and [OH\u207b] into the Ksp equation:<\/p>\n\n\n\n

    Ksp=(S)(2S)2=4S3K_{sp} = (S)(2S)^2 = 4S^3Ksp\u200b=(S)(2S)2=4S3<\/p>\n\n\n\n

    Given Ksp=5.5\u00d710\u221221K_{sp} = 5.5 \\times 10^{-21}Ksp\u200b=5.5\u00d710\u221221, we can solve for SSS:5.5\u00d710\u221221=4S35.5 \\times 10^{-21} = 4S^35.5\u00d710\u221221=4S3S3=5.5\u00d710\u2212214S^3 = \\frac{5.5 \\times 10^{-21}}{4}S3=45.5\u00d710\u221221\u200bS3=1.375\u00d710\u221221S^3 = 1.375 \\times 10^{-21}S3=1.375\u00d710\u221221S=1.375\u00d710\u2212213\u22481.12\u00d710\u22127\u2009mol\/LS = \\sqrt[3]{1.375 \\times 10^{-21}} \\approx 1.12 \\times 10^{-7} \\, \\text{mol\/L}S=31.375\u00d710\u221221\u200b\u22481.12\u00d710\u22127mol\/L<\/p>\n\n\n\n

    Step 4: Convert solubility to grams per liter<\/h3>\n\n\n\n

    To find the solubility in grams per liter, multiply the solubility in moles per liter by the molar mass of Cu(OH)2:Solubility (g\/L)=S\u00d7Molar Mass=1.12\u00d710\u22127\u2009mol\/L\u00d797.57\u2009g\/mol\\text{Solubility (g\/L)} = S \\times \\text{Molar Mass} = 1.12 \\times 10^{-7} \\, \\text{mol\/L} \\times 97.57 \\, \\text{g\/mol}Solubility (g\/L)=S\u00d7Molar Mass=1.12\u00d710\u22127mol\/L\u00d797.57g\/molSolubility (g\/L)\u22481.09\u00d710\u22125\u2009g\/L\\text{Solubility (g\/L)} \\approx 1.09 \\times 10^{-5} \\, \\text{g\/L}Solubility (g\/L)\u22481.09\u00d710\u22125g\/L<\/p>\n\n\n\n

    Step 5: Convert to mL<\/h3>\n\n\n\n

    To express the solubility in mL, we note that 1 L = 1000 mL. Therefore, the solubility in mL is:Solubility (mL)=1.09\u00d710\u22125\u2009g\/L\u00d71000\u2009mL\/L=1.09\u00d710\u22122\u2009mL\/L\\text{Solubility (mL)} = 1.09 \\times 10^{-5} \\, \\text{g\/L} \\times 1000 \\, \\text{mL\/L} = 1.09 \\times 10^{-2} \\, \\text{mL\/L}Solubility (mL)=1.09\u00d710\u22125g\/L\u00d71000mL\/L=1.09\u00d710\u22122mL\/L<\/p>\n\n\n\n

    Thus, the solubility of Cu(OH)2 is approximately 1.09 \u00d7 10^-2 mL\/L<\/strong>, which reflects a very low solubility.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    What is the solubility of copper hydroxide Cu(OH)2 in mL? Mwt of Cu(OH)2 = 97.57; Ksp = 5.5 x 10^21, Cu^2+ (aq) + 2 OH^- \u00e2\u2020\u2019 Cu(OH)2 (s) (aq) The Correct Answer and Explanation is: To determine the solubility of copper hydroxide (Cu(OH)2) in mL, we first need to use the given solubility product constant […]<\/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-261503","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261503","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=261503"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261503\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=261503"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=261503"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=261503"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}