{"id":233709,"date":"2025-06-13T10:36:23","date_gmt":"2025-06-13T10:36:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233709"},"modified":"2025-06-13T10:36:26","modified_gmt":"2025-06-13T10:36:26","slug":"consider-two-solutions-the-first-being-50-0-ml-of-1-00-m-cuso4-and-the-second-50-0-ml-of-2-00-m-koh","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/consider-two-solutions-the-first-being-50-0-ml-of-1-00-m-cuso4-and-the-second-50-0-ml-of-2-00-m-koh\/","title":{"rendered":"Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH."},"content":{"rendered":"\n<p>Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 \u2218C to 27.7 \u2218C. C: Write complete equation for the reaction that occurs when the two solutions are mixed. D: Write net ionic equation for the reaction that occurs when the two solutions are mixed. E: From the calorimetric data, calculate \u0394H for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, and that the specific heat and density of the solution after mixing are the same as that of pure water.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>C. Complete Equation:<\/strong><\/p>\n\n\n\n<p>When copper(II) sulfate (CuSO\u2084) and potassium hydroxide (KOH) are mixed, a double displacement (precipitation) reaction occurs:<\/p>\n\n\n\n<p><strong>CuSO\u2084(aq) + 2KOH(aq) \u2192 Cu(OH)\u2082(s) + K\u2082SO\u2084(aq)<\/strong><\/p>\n\n\n\n<p><strong>D. Net Ionic Equation:<\/strong><\/p>\n\n\n\n<p>Remove the spectator ions (K\u207a and SO\u2084\u00b2\u207b) to focus on the actual chemical change:<\/p>\n\n\n\n<p><strong>Cu\u00b2\u207a(aq) + 2OH\u207b(aq) \u2192 Cu(OH)\u2082(s)<\/strong><\/p>\n\n\n\n<p>This is the net ionic equation for the formation of solid copper(II) hydroxide from its ions in solution.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>E. Calculation of \u0394H (Enthalpy Change):<\/strong><\/p>\n\n\n\n<p>To calculate \u0394H, we follow these steps:<\/p>\n\n\n\n<p><strong>Step 1: Determine the heat released (q):<\/strong><\/p>\n\n\n\n<p>We use the formula:<\/p>\n\n\n\n<p><strong>q = m \u00d7 c \u00d7 \u0394T<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><em>m<\/em> = mass of the solution = 100.0 mL \u00d7 1.00 g\/mL = <strong>100.0 g<\/strong><\/li>\n\n\n\n<li><em>c<\/em> = specific heat of water = <strong>4.18 J\/g\u00b7\u00b0C<\/strong><\/li>\n\n\n\n<li><em>\u0394T<\/em> = change in temperature = 27.7\u00b0C \u2013 21.5\u00b0C = <strong>6.2\u00b0C<\/strong><\/li>\n<\/ul>\n\n\n\n<p>So,<\/p>\n\n\n\n<p><strong>q = 100.0 g \u00d7 4.18 J\/g\u00b7\u00b0C \u00d7 6.2\u00b0C = 2,591.6 J = 2.5916 kJ<\/strong><\/p>\n\n\n\n<p>This is the heat released by the reaction. Since the reaction releases heat, \u0394H is <strong>negative<\/strong>:<\/p>\n\n\n\n<p><strong>q = \u20132.5916 kJ<\/strong><\/p>\n\n\n\n<p><strong>Step 2: Determine moles of limiting reactant:<\/strong><\/p>\n\n\n\n<p>From the balanced net ionic equation, 1 mol Cu\u00b2\u207a reacts with 2 mol OH\u207b.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>CuSO\u2084: 50.0 mL \u00d7 1.00 mol\/L = <strong>0.0500 mol Cu\u00b2\u207a<\/strong><\/li>\n\n\n\n<li>KOH: 50.0 mL \u00d7 2.00 mol\/L = <strong>0.100 mol OH\u207b<\/strong><\/li>\n<\/ul>\n\n\n\n<p>The molar ratio is exactly 1:2, so both are completely consumed. The limiting reagent is <strong>Cu\u00b2\u207a<\/strong>, with <strong>0.0500 mol<\/strong> reacting.<\/p>\n\n\n\n<p><strong>Step 3: Calculate \u0394H per mole of Cu\u00b2\u207a:<\/strong><\/p>\n\n\n\n<p>\u0394H = q \/ mol of reaction =<br>\u0394H = (\u20132.5916 kJ) \/ (0.0500 mol) = <strong>\u201351.83 kJ\/mol<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Final Answer:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Complete equation<\/strong>: CuSO\u2084(aq) + 2KOH(aq) \u2192 Cu(OH)\u2082(s) + K\u2082SO\u2084(aq)<\/li>\n\n\n\n<li><strong>Net ionic equation<\/strong>: Cu\u00b2\u207a(aq) + 2OH\u207b(aq) \u2192 Cu(OH)\u2082(s)<\/li>\n\n\n\n<li><strong>\u0394H = \u201351.8 kJ\/mol<\/strong> (to 3 significant figures)<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Explanation<\/strong><\/p>\n\n\n\n<p>When two aqueous solutions\u2014copper(II) sulfate and potassium hydroxide\u2014are combined, a chemical reaction takes place in which a solid precipitate forms and heat is released. This is a typical <strong>double displacement reaction<\/strong>, where Cu\u00b2\u207a cations from CuSO\u2084 react with OH\u207b anions from KOH to form <strong>insoluble copper(II) hydroxide (Cu(OH)\u2082)<\/strong>. The remaining K\u207a and SO\u2084\u00b2\u207b ions remain dissolved in solution and do not participate in the reaction. The complete and net ionic equations reflect this process.<\/p>\n\n\n\n<p>The rise in temperature indicates the reaction is <strong>exothermic<\/strong>, meaning it releases energy into the surroundings\u2014in this case, the solution. The amount of heat released is determined using the formula <strong>q = m \u00d7 c \u00d7 \u0394T<\/strong>, where <strong>m<\/strong> is the mass of the solution (assumed equal to its volume in grams), <strong>c<\/strong> is the specific heat of water, and <strong>\u0394T<\/strong> is the change in temperature. The calculated heat released is about <strong>2.59 kJ<\/strong>.<\/p>\n\n\n\n<p>To find the enthalpy change per mole of reaction (\u0394H), the heat released is divided by the moles of limiting reagent, in this case, Cu\u00b2\u207a, since it completely reacts with the available hydroxide. This gives an enthalpy change of <strong>\u201351.8 kJ\/mol<\/strong>, which represents the energy released when 1 mole of copper(II) ions reacts with hydroxide ions to form the precipitate. This value helps chemists understand the energy profile of the reaction and classify it as energetically favorable due to the exothermic nature.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-309.jpeg\" alt=\"\" class=\"wp-image-233710\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 \u2218C to 27.7 \u2218C. C: Write complete equation for the reaction that [&hellip;]<\/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-233709","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233709","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=233709"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233709\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233709"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233709"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233709"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}