{"id":191045,"date":"2025-02-14T05:35:51","date_gmt":"2025-02-14T05:35:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=191045"},"modified":"2025-02-14T05:35:53","modified_gmt":"2025-02-14T05:35:53","slug":"copper-ii-sulfate-reacts-with-sodium-hydroxide-in-a-double-displacement-reaction","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/14\/copper-ii-sulfate-reacts-with-sodium-hydroxide-in-a-double-displacement-reaction\/","title":{"rendered":"copper (II) sulfate reacts with sodium hydroxide in a double displacement reaction"},"content":{"rendered":"\n

copper (II) sulfate reacts with sodium hydroxide in a double displacement reaction. 50 mL of 0.30 M CuSO4 (aq)\u00c3\u201a\u00c2 is mixed with and equal volume of 0.699 M of NaOH(aq) solution. the initial temperature of bith solutions was 21.4 degree. after mixing the solutions in a styroform calorimeter, the highest temperature that is reached is 24.6 degree. determine the molar enthalpy for copper (II) sulfate. write the balanced chemical equation and include the energy as a term in the equation. we can the heat capacity of water as the both solutions are aqueous. which is 4.19 J\\g.C<\/p>\n\n\n\n

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

To determine the molar enthalpy ((\\Delta H)) for copper(II) sulfate ((\\ce{CuSO4})), follow these steps:<\/p>\n\n\n\n

Step 1: Write the Balanced Chemical Equation<\/strong><\/h3>\n\n\n\n

The reaction between (\\ce{CuSO4}) and (\\ce{NaOH}) is a double displacement reaction, forming copper(II) hydroxide precipitate and sodium sulfate:<\/p>\n\n\n\n

[
\\ce{CuSO4 (aq) + 2NaOH (aq) -> Cu(OH)2 (s) + Na2SO4 (aq)}
]<\/p>\n\n\n\n

Since this reaction releases energy, it is exothermic.<\/p>\n\n\n\n

Step 2: Calculate the Heat Energy Released<\/strong><\/h3>\n\n\n\n

The heat ((q)) released in the reaction is calculated using the formula:<\/p>\n\n\n\n

[
q = mc\\Delta T
]<\/p>\n\n\n\n

where:<\/p>\n\n\n\n

    \n
  • (m) is the mass of the solution (assuming the density of the solution is similar to water, 1.00 g\/mL),<\/li>\n\n\n\n
  • (c) is the specific heat capacity of water ((4.19 \\, \\text{J\/g\u00b0C})),<\/li>\n\n\n\n
  • (\\Delta T) is the temperature change.<\/li>\n<\/ul>\n\n\n\n

    Given:<\/p>\n\n\n\n

      \n
    • Initial temperature: 21.4\u00b0C<\/strong><\/li>\n\n\n\n
    • Final temperature: 24.6\u00b0C<\/strong><\/li>\n\n\n\n
    • Volume of (\\ce{CuSO4}) solution: 50 mL<\/strong><\/li>\n\n\n\n
    • Volume of (\\ce{NaOH}) solution: 50 mL<\/strong><\/li>\n\n\n\n
    • Total volume of solution: 100 mL<\/strong><\/li>\n\n\n\n
    • (\\Delta T = 24.6 – 21.4 = 3.2\u00b0C)<\/li>\n\n\n\n
    • Since 100 mL of solution is approximately 100 g<\/strong> (assuming water density),<\/li>\n<\/ul>\n\n\n\n

      [
      q = (100 \\, \\text{g}) (4.19 \\, \\text{J\/g\u00b0C}) (3.2 \\, \\text{\u00b0C})
      ]<\/p>\n\n\n\n

      [
      q = 1340.8 \\, \\text{J} \\approx 1.34 \\, \\text{kJ}
      ]<\/p>\n\n\n\n

      Step 3: Determine the Limiting Reactant<\/strong><\/h3>\n\n\n\n

      Moles of (\\ce{CuSO4})<\/strong>:<\/h4>\n\n\n\n

      [
      n_{\\ce{CuSO4}} = 0.30 \\, M \\times 0.050 \\, L = 0.015 \\, \\text{mol}
      ]<\/p>\n\n\n\n

      Moles of (\\ce{NaOH})<\/strong>:<\/h4>\n\n\n\n

      [
      n_{\\ce{NaOH}} = 0.699 \\, M \\times 0.050 \\, L = 0.03495 \\, \\text{mol}
      ]<\/p>\n\n\n\n

      Since (\\ce{CuSO4}) reacts with (\\ce{NaOH}) in a 1:2 ratio<\/strong>, the amount of (\\ce{NaOH}) needed for 0.015 mol (\\ce{CuSO4}) is:<\/p>\n\n\n\n

      [
      0.015 \\times 2 = 0.030 \\text{ mol NaOH}
      ]<\/p>\n\n\n\n

      Since there is 0.03495 mol<\/strong> of (\\ce{NaOH}) available, which is more than required, (\\ce{CuSO4}) is the limiting reactant.<\/p>\n\n\n\n

      Step 4: Calculate the Molar Enthalpy Change<\/strong><\/h3>\n\n\n\n

      [
      \\Delta H = \\frac{q}{n_{\\ce{CuSO4}}}
      ]<\/p>\n\n\n\n

      [
      \\Delta H = \\frac{1.34 \\, \\text{kJ}}{0.015 \\, \\text{mol}}
      ]<\/p>\n\n\n\n

      [
      \\Delta H = -89.33 \\, \\text{kJ\/mol}
      ]<\/p>\n\n\n\n

      (Since heat is released, (\\Delta H) is negative<\/strong>, indicating an exothermic reaction<\/strong>.)<\/p>\n\n\n\n

      Step 5: Write the Balanced Equation with Energy<\/strong><\/h3>\n\n\n\n

      [
      \\ce{CuSO4 (aq) + 2NaOH (aq) -> Cu(OH)2 (s) + Na2SO4 (aq) \\quad \\Delta H = -89.3 \\, kJ\/mol}
      ]<\/p>\n\n\n\n

      \n\n\n\n

      Explanation (300 Words)<\/strong><\/h3>\n\n\n\n

      This reaction is a double displacement reaction<\/strong> where copper(II) sulfate ((\\ce{CuSO4}))<\/strong> reacts with sodium hydroxide ((\\ce{NaOH}))<\/strong> to form copper(II) hydroxide ((\\ce{Cu(OH)2})), a solid precipitate, and sodium sulfate ((\\ce{Na2SO4})), which remains in solution<\/strong>. Since heat is released in this reaction, it is an exothermic reaction<\/strong>.<\/p>\n\n\n\n

      To determine the molar enthalpy change<\/strong>, we first calculate the heat energy released using the formula (q = mc\\Delta T). Assuming that the solution behaves like water, we take the total mass as 100 g (since 1 mL \u2248 1 g), the specific heat capacity as 4.19 J\/g\u00b0C, and the temperature change as (3.2^\\circ C). This gives us a heat release of 1.34 kJ<\/strong>.<\/p>\n\n\n\n

      Next, we find the limiting reactant. (\\ce{CuSO4}) is the limiting reactant<\/strong> since it fully reacts with excess (\\ce{NaOH}). We then use the moles of (\\ce{CuSO4})<\/strong> (0.015 mol) to calculate the enthalpy change per mole:<\/p>\n\n\n\n

      [
      \\Delta H = \\frac{1.34 \\, \\text{kJ}}{0.015 \\, \\text{mol}} = -89.3 \\, \\text{kJ\/mol}
      ]<\/p>\n\n\n\n

      This means that when one mole of (\\ce{CuSO4}) reacts with sodium hydroxide, 89.3 kJ of energy is released<\/strong>. The negative sign indicates that the reaction is exothermic<\/strong>.<\/p>\n\n\n\n

      Thus, the balanced reaction including energy is:<\/p>\n\n\n\n

      [
      \\ce{CuSO4 (aq) + 2NaOH (aq) -> Cu(OH)2 (s) + Na2SO4 (aq) \\quad \\Delta H = -89.3 \\, kJ\/mol}
      ]<\/p>\n\n\n\n

      This reaction is often performed in calorimetry experiments to study enthalpy changes in precipitation reactions.<\/p>\n","protected":false},"excerpt":{"rendered":"

      copper (II) sulfate reacts with sodium hydroxide in a double displacement reaction. 50 mL of 0.30 M CuSO4 (aq)\u00c3\u201a\u00c2 is mixed with and equal volume of 0.699 M of NaOH(aq) solution. the initial temperature of bith solutions was 21.4 degree. after mixing the solutions in a styroform calorimeter, the highest temperature that is reached is […]<\/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-191045","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191045","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=191045"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191045\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191045"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191045"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191045"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}