{"id":257351,"date":"2025-07-17T02:57:48","date_gmt":"2025-07-17T02:57:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257351"},"modified":"2025-07-17T02:57:50","modified_gmt":"2025-07-17T02:57:50","slug":"what-is-the-solubility-of-copper-hydroxide-cuoh2-in-ml","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/what-is-the-solubility-of-copper-hydroxide-cuoh2-in-ml\/","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 calculate the solubility of copper hydroxide (Cu(OH)\u2082), we need to use its solubility product constant (Ksp) and apply stoichiometry. Here’s the step-by-step approach:<\/p>\n\n\n\n
Given Data:<\/h3>\n\n\n\n
\n
Molar mass of Cu(OH)\u2082: 97.57 g\/mol<\/li>\n\n\n\n
The dissolution reaction for Cu(OH)\u2082 is: Cu\u00b2\u207a(aq) + 2OH\u207b(aq) \u21cc Cu(OH)\u2082(s)<\/li>\n<\/ul>\n\n\n\n
Let the solubility of Cu(OH)\u2082 be “S” mol\/L. This represents the amount of Cu(OH)\u2082 that dissolves in water. When Cu(OH)\u2082 dissolves, it dissociates into Cu\u00b2\u207a and OH\u207b ions according to the reaction:<\/p>\n\n\n\n
\n
Cu(OH)\u2082 (s) \u21cc Cu\u00b2\u207a (aq) + 2 OH\u207b (aq)<\/li>\n<\/ul>\n\n\n\n
From this, we can see that for every 1 mole of Cu(OH)\u2082 that dissolves, 1 mole of Cu\u00b2\u207a and 2 moles of OH\u207b are produced. So, if the solubility of Cu(OH)\u2082 is “S” mol\/L:<\/p>\n\n\n\n
\n
[Cu\u00b2\u207a] = S mol\/L<\/li>\n\n\n\n
[OH\u207b] = 2S mol\/L<\/li>\n<\/ul>\n\n\n\n
Now, use the Ksp expression for the dissociation equilibrium: Ksp=[Cu2+][OH\u2212]2Ksp = [Cu\u00b2\u207a][OH\u207b]^2Ksp=[Cu2+][OH\u2212]2<\/p>\n\n\n\n
Substituting the concentrations into the Ksp expression: Ksp=(S)(2S)2=4S3Ksp = (S)(2S)\u00b2 = 4S\u00b3Ksp=(S)(2S)2=4S3<\/p>\n\n\n\n
Given that Ksp = 5.5 \u00d7 10\u207b\u00b2\u00b9, we can solve for S: 5.5\u00d710\u221221=4S35.5 \u00d7 10\u207b\u00b2\u00b9 = 4S\u00b35.5\u00d710\u221221=4S3 S3=5.5\u00d710\u2212214S\u00b3 = \\frac{5.5 \u00d7 10\u207b\u00b2\u00b9}{4}S3=45.5\u00d710\u221221\u200b S3=1.375\u00d710\u221221S\u00b3 = 1.375 \u00d7 10\u207b\u00b2\u00b9S3=1.375\u00d710\u221221 S=1.375\u00d710\u2212213\u22485.13\u00d710\u22128mol\/LS = \\sqrt[3]{1.375 \u00d7 10\u207b\u00b2\u00b9} \\approx 5.13 \u00d7 10\u207b\u2078 mol\/LS=31.375\u00d710\u221221\u200b\u22485.13\u00d710\u22128mol\/L<\/p>\n\n\n\n
So the solubility of Cu(OH)\u2082 in moles per liter is approximately 5.13 \u00d7 10\u207b\u2078 mol\/L<\/strong>.<\/p>\n\n\n\n
To convert the solubility to grams per liter:<\/h3>\n\n\n\n
Now, to find the solubility in grams per liter, we multiply the molar solubility by the molar mass of Cu(OH)\u2082: Solubility (g\/L)=S\u00d7Molar mass of Cu(OH)\u2082\\text{Solubility (g\/L)} = S \\times \\text{Molar mass of Cu(OH)\u2082}Solubility (g\/L)=S\u00d7Molar mass of Cu(OH)\u2082 Solubility (g\/L)=(5.13\u00d710\u22128mol\/L)\u00d797.57g\/mol\\text{Solubility (g\/L)} = (5.13 \u00d7 10\u207b\u2078 mol\/L) \u00d7 97.57 g\/molSolubility (g\/L)=(5.13\u00d710\u22128mol\/L)\u00d797.57g\/mol Solubility (g\/L)\u22485.0\u00d710\u22126g\/L\\text{Solubility (g\/L)} \\approx 5.0 \u00d7 10\u207b\u2076 g\/LSolubility (g\/L)\u22485.0\u00d710\u22126g\/L<\/p>\n\n\n\n
This means the solubility of Cu(OH)\u2082 in water is approximately 5.0 \u00d7 10\u207b\u2076 g\/L<\/strong>, or 5.0 \u03bcg\/L.<\/p>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
Copper hydroxide (Cu(OH)\u2082) is extremely insoluble in water, as indicated by its very low solubility value of approximately 5.0 \u03bcg\/L<\/strong>. This is consistent with the very small Ksp value, which reflects the limited dissociation of Cu(OH)\u2082 in aqueous solution.<\/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 calculate the solubility of copper hydroxide (Cu(OH)\u2082), we need to use its solubility product constant (Ksp) and apply stoichiometry. […]<\/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-257351","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257351","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=257351"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257351\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257351"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257351"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257351"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-752.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"