{"id":25467,"date":"2025-06-19T03:00:08","date_gmt":"2025-06-19T03:00:08","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=25467"},"modified":"2025-06-19T03:00:11","modified_gmt":"2025-06-19T03:00:11","slug":"the-solubility-of-barium-carbonate-baco3-is-0-0100-g-l","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-solubility-of-barium-carbonate-baco3-is-0-0100-g-l\/","title":{"rendered":"The solubility of barium carbonate, BaCO3, is 0.0100 g\/L."},"content":{"rendered":"\n

The solubility of barium carbonate, BaCO3, is 0.0100 g\/L. Its molar mass is 197.3 g\/mol. What is the Ksp of barium carbonate? 2. The solubility product constant of calcium sulfate, CaSO4, is 7.10\u00c3\u201410^-5. Its molar mass is 136.1 g\/mol. How many grams of calcium sulfate can dissolve in 61.0 L of pure water?<\/p>\n\n\n\n

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

Let’s solve both parts of the problem step by step.<\/p>\n\n\n\n

\n\n\n\n

1. Solubility of BaCO\u2083 \u2192 Ksp Calculation<\/strong><\/h3>\n\n\n\n

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

    \n
  • Solubility of BaCO\u2083 = 0.0100 g\/L<\/li>\n\n\n\n
  • Molar mass of BaCO\u2083 = 197.3 g\/mol<\/li>\n<\/ul>\n\n\n\n

    Step 1: Convert solubility to mol\/L (molarity)<\/strong>Molar solubility=0.0100 g\/L197.3 g\/mol=5.07\u00d710\u22125 mol\/L\\text{Molar solubility} = \\frac{0.0100 \\text{ g\/L}}{197.3 \\text{ g\/mol}} = 5.07 \\times 10^{-5} \\text{ mol\/L}Molar solubility=197.3 g\/mol0.0100 g\/L\u200b=5.07\u00d710\u22125 mol\/L<\/p>\n\n\n\n

    Step 2: Write the dissociation equation<\/strong>BaCO\u2083 (s)\u21ccBa2+(aq)+CO\u20832\u2212(aq)\\text{BaCO\u2083 (s)} \\rightleftharpoons \\text{Ba}^{2+} (aq) + \\text{CO\u2083}^{2-} (aq)BaCO\u2083 (s)\u21ccBa2+(aq)+CO\u20832\u2212(aq)<\/p>\n\n\n\n

    Let the molar solubility be x<\/strong> mol\/L, then:<\/p>\n\n\n\n

      \n
    • [Ba\u00b2\u207a] = x<\/li>\n\n\n\n
    • [CO\u2083\u00b2\u207b] = x<\/li>\n<\/ul>\n\n\n\n

      Step 3: Write the Ksp expression<\/strong>Ksp=[Ba2+][CO\u20832\u2212]=x\u22c5x=x2K_{sp} = [\\text{Ba}^{2+}][\\text{CO\u2083}^{2-}] = x \\cdot x = x^2Ksp\u200b=[Ba2+][CO\u20832\u2212]=x\u22c5x=x2Ksp=(5.07\u00d710\u22125)2=2.57\u00d710\u22129K_{sp} = (5.07 \\times 10^{-5})^2 = 2.57 \\times 10^{-9}Ksp\u200b=(5.07\u00d710\u22125)2=2.57\u00d710\u22129<\/p>\n\n\n\n

      \u2705 Answer for part 1:<\/strong>
      Ksp of BaCO\u2083 = 2.57 \u00d7 10\u207b\u2079<\/strong><\/p>\n\n\n\n

      \n\n\n\n

      2. Ksp of CaSO\u2084 \u2192 Solubility in grams in 61.0 L<\/strong><\/h3>\n\n\n\n

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

        \n
      • Ksp of CaSO\u2084 = 7.10 \u00d7 10\u207b\u2075<\/li>\n\n\n\n
      • Molar mass = 136.1 g\/mol<\/li>\n\n\n\n
      • Volume = 61.0 L<\/li>\n<\/ul>\n\n\n\n

        Step 1: Write dissociation equation<\/strong>CaSO\u2084 (s)\u21ccCa2+(aq)+SO\u20842\u2212(aq)\\text{CaSO\u2084 (s)} \\rightleftharpoons \\text{Ca}^{2+} (aq) + \\text{SO\u2084}^{2-} (aq)CaSO\u2084 (s)\u21ccCa2+(aq)+SO\u20842\u2212(aq)<\/p>\n\n\n\n

        Let the solubility = s<\/strong> mol\/L, then:<\/p>\n\n\n\n

          \n
        • [Ca\u00b2\u207a] = s<\/li>\n\n\n\n
        • [SO\u2084\u00b2\u207b] = s<\/li>\n<\/ul>\n\n\n\n

          Ksp=s2\u21d2s=7.10\u00d710\u22125=8.43\u00d710\u22123 mol\/LK_{sp} = s^2 \\Rightarrow s = \\sqrt{7.10 \\times 10^{-5}} = 8.43 \\times 10^{-3} \\text{ mol\/L}Ksp\u200b=s2\u21d2s=7.10\u00d710\u22125\u200b=8.43\u00d710\u22123 mol\/L<\/p>\n\n\n\n

          Step 2: Find moles in 61.0 L<\/strong>Moles=s\u22c5V=(8.43\u00d710\u22123)\u22c561.0=0.5142 mol\\text{Moles} = s \\cdot V = (8.43 \\times 10^{-3}) \\cdot 61.0 = 0.5142 \\text{ mol}Moles=s\u22c5V=(8.43\u00d710\u22123)\u22c561.0=0.5142 mol<\/p>\n\n\n\n

          Step 3: Convert to grams<\/strong>Mass=moles\u22c5molar mass=0.5142\u22c5136.1=69.98 g\\text{Mass} = \\text{moles} \\cdot \\text{molar mass} = 0.5142 \\cdot 136.1 = 69.98 \\text{ g}Mass=moles\u22c5molar mass=0.5142\u22c5136.1=69.98 g<\/p>\n\n\n\n

          \u2705 Answer for part 2:<\/strong>
          69.98 grams of CaSO\u2084 can dissolve in 61.0 L of water<\/strong><\/p>\n\n\n\n

          \n\n\n\n

          Summary<\/strong><\/h3>\n\n\n\n
            \n
          1. Ksp of BaCO\u2083<\/strong> is 2.57 \u00d7 10\u207b\u2079<\/strong><\/li>\n\n\n\n
          2. Mass of CaSO\u2084 that can dissolve in 61.0 L<\/strong> is 69.98 g<\/strong><\/li>\n<\/ol>\n\n\n\n

            These calculations demonstrate how solubility and the solubility product constant are connected. Using stoichiometry, molar mass, and basic equilibrium principles, we can easily determine both Ksp and the mass of a salt that can dissolve in a specific volume of water.<\/p>\n\n\n\n

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

            The solubility of barium carbonate, BaCO3, is 0.0100 g\/L. Its molar mass is 197.3 g\/mol. What is the Ksp of barium carbonate? 2. The solubility product constant of calcium sulfate, CaSO4, is 7.10\u00c3\u201410^-5. Its molar mass is 136.1 g\/mol. How many grams of calcium sulfate can dissolve in 61.0 L of pure water? The Correct […]<\/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-25467","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25467","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=25467"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25467\/revisions"}],"predecessor-version":[{"id":25469,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25467\/revisions\/25469"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=25467"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=25467"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=25467"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}