{"id":32417,"date":"2025-06-22T13:35:16","date_gmt":"2025-06-22T13:35:16","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32417"},"modified":"2025-06-22T13:35:19","modified_gmt":"2025-06-22T13:35:19","slug":"the-solubility-product-of-calcium-fluoride-caf-is-3-x-10","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-solubility-product-of-calcium-fluoride-caf-is-3-x-10\/","title":{"rendered":"The solubility product of calcium fluoride (CaF) is 3 x 10"},"content":{"rendered":"\n

The solubility product of calcium fluoride (CaF
) is 3 x 10
. Will a fluoride concentration of 1.0 mg\/L be soluble in a water containing 200 mg\/L of calcium?<\/p>\n\n\n\n

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

To determine whether a fluoride concentration of 1.0 mg\/L will remain soluble in water containing 200 mg\/L of calcium, we must compare the ionic product with the solubility product constant (Ksp) of calcium fluoride (CaF\u2082).<\/p>\n\n\n\n

Step 1: Write the dissociation equation<\/h3>\n\n\n\n

CaF2(s)\u2194Ca2+(aq)+2F\u2212(aq)\\text{CaF}_2 (s) \\leftrightarrow \\text{Ca}^{2+} (aq) + 2\\text{F}^- (aq)CaF2\u200b(s)\u2194Ca2+(aq)+2F\u2212(aq)<\/p>\n\n\n\n

Step 2: Use the Ksp value<\/h3>\n\n\n\n

The solubility product constant (Ksp) of CaF\u2082 is:
Ksp=[Ca2+][F\u2212]2=3\u00d710\u221211K_{sp} = [\\text{Ca}^{2+}][\\text{F}^-]^2 = 3 \\times 10^{-11}Ksp\u200b=[Ca2+][F\u2212]2=3\u00d710\u221211<\/p>\n\n\n\n

Step 3: Convert concentrations to mol\/L<\/h3>\n\n\n\n
    \n
  • Calcium (Ca\u00b2\u207a):<\/strong>
    200\u2009mg\/L=20040.08=4.99\u2009mmol\/L=4.99\u00d710\u22123\u2009mol\/L200 \\, \\text{mg\/L} = \\frac{200}{40.08} = 4.99 \\, \\text{mmol\/L} = 4.99 \\times 10^{-3} \\, \\text{mol\/L}200mg\/L=40.08200\u200b=4.99mmol\/L=4.99\u00d710\u22123mol\/L<\/li>\n\n\n\n
  • Fluoride (F\u207b):<\/strong>
    1.0\u2009mg\/L=1.018.998=5.26\u00d710\u22125\u2009mol\/L1.0 \\, \\text{mg\/L} = \\frac{1.0}{18.998} = 5.26 \\times 10^{-5} \\, \\text{mol\/L}1.0mg\/L=18.9981.0\u200b=5.26\u00d710\u22125mol\/L<\/li>\n<\/ul>\n\n\n\n

    Step 4: Calculate the ionic product<\/h3>\n\n\n\n

    Ionic Product=[Ca2+][F\u2212]2=(4.99\u00d710\u22123)(5.26\u00d710\u22125)2\\text{Ionic Product} = [\\text{Ca}^{2+}][\\text{F}^-]^2 = (4.99 \\times 10^{-3})(5.26 \\times 10^{-5})^2Ionic Product=[Ca2+][F\u2212]2=(4.99\u00d710\u22123)(5.26\u00d710\u22125)2
    =(4.99\u00d710\u22123)(2.77\u00d710\u22129)= (4.99 \\times 10^{-3})(2.77 \\times 10^{-9})=(4.99\u00d710\u22123)(2.77\u00d710\u22129)
    =1.38\u00d710\u221211= 1.38 \\times 10^{-11}=1.38\u00d710\u221211<\/p>\n\n\n\n

    Step 5: Compare the ionic product with the Ksp<\/h3>\n\n\n\n
      \n
    • Ionic Product = 1.38\u00d710\u2212111.38 \\times 10^{-11}1.38\u00d710\u221211<\/li>\n\n\n\n
    • Ksp = 3.0\u00d710\u2212113.0 \\times 10^{-11}3.0\u00d710\u221211<\/li>\n<\/ul>\n\n\n\n

      Since the ionic product is less than<\/strong> the Ksp, the solution is unsaturated<\/strong> with respect to calcium fluoride. This means no precipitation will occur, and all of the fluoride will remain dissolved.<\/p>\n\n\n\n

      Conclusion:<\/h3>\n\n\n\n

      Yes, a fluoride concentration of 1.0 mg\/L will be soluble in water that contains 200 mg\/L of calcium. The current ion concentrations do not exceed the solubility limit of CaF\u2082, and therefore, the solution remains stable without forming a precipitate. This calculation ensures water quality compliance where both fluoride and calcium levels coexist safely.<\/p>\n\n\n\n

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

      The solubility product of calcium fluoride (CaF) is 3 x 10. Will a fluoride concentration of 1.0 mg\/L be soluble in a water containing 200 mg\/L of calcium? The Correct Answer and Explanation is: To determine whether a fluoride concentration of 1.0 mg\/L will remain soluble in water containing 200 mg\/L of calcium, we must […]<\/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-32417","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32417","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=32417"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32417\/revisions"}],"predecessor-version":[{"id":32419,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32417\/revisions\/32419"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32417"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32417"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32417"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}