To calculate the solubility of magnesium phosphate, Mg\u2083(PO\u2084)\u2082, in grams per 100 mL, we need to follow these steps:<\/p>\n\n\n\n
\n\n\n\n
Step 1: Write the dissociation equation<\/strong><\/h3>\n\n\n\nMg3(PO4)2(s)\u21cc3Mg2+(aq)+2PO43\u2212(aq)\\text{Mg}_3(\\text{PO}_4)_2 (s) \\rightleftharpoons 3\\text{Mg}^{2+} (aq) + 2\\text{PO}_4^{3-} (aq)Mg3\u200b(PO4\u200b)2\u200b(s)\u21cc3Mg2+(aq)+2PO43\u2212\u200b(aq)<\/p>\n\n\n\n
Let the molar solubility of Mg\u2083(PO\u2084)\u2082 be x<\/strong> mol\/L. Then at equilibrium:<\/p>\n\n\n\n\n- [Mg2+]=3x[\\text{Mg}^{2+}] = 3x[Mg2+]=3x<\/li>\n\n\n\n
- [PO43\u2212]=2x[\\text{PO}_4^{3-}] = 2x[PO43\u2212\u200b]=2x<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 2: Write the Ksp expression<\/strong><\/h3>\n\n\n\nKsp=[Mg2+]3[PO43\u2212]2=(3x)3(2x)2=27×3\u22c54×2=108x5K_{sp} = [\\text{Mg}^{2+}]^3[\\text{PO}_4^{3-}]^2 = (3x)^3(2x)^2 = 27x^3 \\cdot 4x^2 = 108x^5Ksp\u200b=[Mg2+]3[PO43\u2212\u200b]2=(3x)3(2x)2=27×3\u22c54×2=108×51.04\u00d710\u221224=108×51.04 \\times 10^{-24} = 108x^51.04\u00d710\u221224=108×5<\/p>\n\n\n\n
\n\n\n\nStep 3: Solve for x<\/strong><\/h3>\n\n\n\nx5=1.04\u00d710\u221224108=9.63\u00d710\u221227x^5 = \\frac{1.04 \\times 10^{-24}}{108} = 9.63 \\times 10^{-27}x5=1081.04\u00d710\u221224\u200b=9.63\u00d710\u221227x=(9.63\u00d710\u221227)1\/5\u22484.07\u00d710\u22126\u2009mol\/Lx = (9.63 \\times 10^{-27})^{1\/5} \\approx 4.07 \\times 10^{-6} \\, \\text{mol\/L}x=(9.63\u00d710\u221227)1\/5\u22484.07\u00d710\u22126mol\/L<\/p>\n\n\n\n
\n\n\n\nStep 4: Convert molar solubility to grams per 100 mL<\/strong><\/h3>\n\n\n\nMolar mass of Mg\u2083(PO\u2084)\u2082:<\/p>\n\n\n\n
\n- Mg = 24.305 g\/mol \u00d7 3 = 72.915<\/li>\n\n\n\n
- P = 30.974 g\/mol \u00d7 2 = 61.948<\/li>\n\n\n\n
- O = 16.00 g\/mol \u00d7 8 = 128.00
Total = 72.915 + 61.948 + 128.00 = 262.86 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\nSolubility (g\/L)=4.07\u00d710\u22126\u2009mol\/L\u00d7262.86\u2009g\/mol\u22481.07\u00d710\u22123\u2009g\/L\\text{Solubility (g\/L)} = 4.07 \\times 10^{-6} \\, \\text{mol\/L} \\times 262.86 \\, \\text{g\/mol} \\approx 1.07 \\times 10^{-3} \\, \\text{g\/L}Solubility (g\/L)=4.07\u00d710\u22126mol\/L\u00d7262.86g\/mol\u22481.07\u00d710\u22123g\/LSolubility (g\/100 mL)=1.07\u00d710\u2212310=1.07\u00d710\u22124\u2009g\/100 mL\\text{Solubility (g\/100 mL)} = \\frac{1.07 \\times 10^{-3}}{10} = \\boxed{1.07 \\times 10^{-4}} \\, \\text{g\/100 mL}Solubility (g\/100 mL)=101.07\u00d710\u22123\u200b=1.07\u00d710\u22124\u200bg\/100 mL<\/p>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\nThe solubility product constant (Ksp) provides insight into how much of a slightly soluble salt dissolves in water to form a saturated solution. In this case, Mg\u2083(PO\u2084)\u2082 dissociates into three magnesium ions and two phosphate ions for each formula unit that dissolves. This stoichiometric relationship is crucial in constructing the equilibrium expression that connects the concentrations of ions in solution to the Ksp value.<\/p>\n\n\n\n
To determine solubility, we define x<\/strong> as the molar solubility of Mg\u2083(PO\u2084)\u2082. Since three magnesium ions and two phosphate ions are produced per unit of salt, their concentrations become 3x and 2x, respectively. Substituting into the Ksp expression, we obtain an equation in terms of x: Ksp = 108x\u2075. Solving this equation requires taking the fifth root of the quotient of Ksp and 108, yielding a very small molar solubility due to the extremely low Ksp value.<\/p>\n\n\n\nOnce we have the molar solubility, converting it to grams per liter is a matter of multiplying by the compound\u2019s molar mass. To express the solubility per 100 mL, we divide the result by 10. The final answer, 1.07 \u00d7 10\u207b\u2074 g\/100 mL, reflects the very low solubility of magnesium phosphate in water.<\/p>\n\n\n\n
Such calculations are essential in predicting precipitation, designing chemical reactions, and understanding biological mineralization. Low solubility compounds like Mg\u2083(PO\u2084)\u2082 often play roles in biological systems such as bone and tooth formation or in industrial water treatment processes.<\/p>\n\n\n\n![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-246.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"The solubility product constant (Ksp) of Mg3(PO4)2 is 1.04 \u00d7 10-24. What is its solubility in grams per 100 mL? The Correct Answer and Explanation is: To calculate the solubility of magnesium phosphate, Mg\u2083(PO\u2084)\u2082, in grams per 100 mL, we need to follow these steps: Step 1: Write the dissociation equation Mg3(PO4)2(s)\u21cc3Mg2+(aq)+2PO43\u2212(aq)\\text{Mg}_3(\\text{PO}_4)_2 (s) \\rightleftharpoons 3\\text{Mg}^{2+} […]<\/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-39255","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39255","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=39255"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39255\/revisions"}],"predecessor-version":[{"id":39258,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39255\/revisions\/39258"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39255"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39255"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39255"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- \n
- [Mg2+]=3x[\\text{Mg}^{2+}] = 3x[Mg2+]=3x<\/li>\n\n\n\n
- [PO43\u2212]=2x[\\text{PO}_4^{3-}] = 2x[PO43\u2212\u200b]=2x<\/li>\n<\/ul>\n\n\n\n
\n\n\n\nStep 2: Write the Ksp expression<\/strong><\/h3>\n\n\n\n
Ksp=[Mg2+]3[PO43\u2212]2=(3x)3(2x)2=27×3\u22c54×2=108x5K_{sp} = [\\text{Mg}^{2+}]^3[\\text{PO}_4^{3-}]^2 = (3x)^3(2x)^2 = 27x^3 \\cdot 4x^2 = 108x^5Ksp\u200b=[Mg2+]3[PO43\u2212\u200b]2=(3x)3(2x)2=27×3\u22c54×2=108×51.04\u00d710\u221224=108×51.04 \\times 10^{-24} = 108x^51.04\u00d710\u221224=108×5<\/p>\n\n\n\n
\n\n\n\nStep 3: Solve for x<\/strong><\/h3>\n\n\n\n
x5=1.04\u00d710\u221224108=9.63\u00d710\u221227x^5 = \\frac{1.04 \\times 10^{-24}}{108} = 9.63 \\times 10^{-27}x5=1081.04\u00d710\u221224\u200b=9.63\u00d710\u221227x=(9.63\u00d710\u221227)1\/5\u22484.07\u00d710\u22126\u2009mol\/Lx = (9.63 \\times 10^{-27})^{1\/5} \\approx 4.07 \\times 10^{-6} \\, \\text{mol\/L}x=(9.63\u00d710\u221227)1\/5\u22484.07\u00d710\u22126mol\/L<\/p>\n\n\n\n
\n\n\n\nStep 4: Convert molar solubility to grams per 100 mL<\/strong><\/h3>\n\n\n\n
Molar mass of Mg\u2083(PO\u2084)\u2082:<\/p>\n\n\n\n
- \n
- Mg = 24.305 g\/mol \u00d7 3 = 72.915<\/li>\n\n\n\n
- P = 30.974 g\/mol \u00d7 2 = 61.948<\/li>\n\n\n\n
- O = 16.00 g\/mol \u00d7 8 = 128.00
Total = 72.915 + 61.948 + 128.00 = 262.86 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\nSolubility (g\/L)=4.07\u00d710\u22126\u2009mol\/L\u00d7262.86\u2009g\/mol\u22481.07\u00d710\u22123\u2009g\/L\\text{Solubility (g\/L)} = 4.07 \\times 10^{-6} \\, \\text{mol\/L} \\times 262.86 \\, \\text{g\/mol} \\approx 1.07 \\times 10^{-3} \\, \\text{g\/L}Solubility (g\/L)=4.07\u00d710\u22126mol\/L\u00d7262.86g\/mol\u22481.07\u00d710\u22123g\/LSolubility (g\/100 mL)=1.07\u00d710\u2212310=1.07\u00d710\u22124\u2009g\/100 mL\\text{Solubility (g\/100 mL)} = \\frac{1.07 \\times 10^{-3}}{10} = \\boxed{1.07 \\times 10^{-4}} \\, \\text{g\/100 mL}Solubility (g\/100 mL)=101.07\u00d710\u22123\u200b=1.07\u00d710\u22124\u200bg\/100 mL<\/p>\n\n\n\n
\n\n\n\nExplanation<\/strong><\/h3>\n\n\n\n
The solubility product constant (Ksp) provides insight into how much of a slightly soluble salt dissolves in water to form a saturated solution. In this case, Mg\u2083(PO\u2084)\u2082 dissociates into three magnesium ions and two phosphate ions for each formula unit that dissolves. This stoichiometric relationship is crucial in constructing the equilibrium expression that connects the concentrations of ions in solution to the Ksp value.<\/p>\n\n\n\n
To determine solubility, we define x<\/strong> as the molar solubility of Mg\u2083(PO\u2084)\u2082. Since three magnesium ions and two phosphate ions are produced per unit of salt, their concentrations become 3x and 2x, respectively. Substituting into the Ksp expression, we obtain an equation in terms of x: Ksp = 108x\u2075. Solving this equation requires taking the fifth root of the quotient of Ksp and 108, yielding a very small molar solubility due to the extremely low Ksp value.<\/p>\n\n\n\n
Once we have the molar solubility, converting it to grams per liter is a matter of multiplying by the compound\u2019s molar mass. To express the solubility per 100 mL, we divide the result by 10. The final answer, 1.07 \u00d7 10\u207b\u2074 g\/100 mL, reflects the very low solubility of magnesium phosphate in water.<\/p>\n\n\n\n
Such calculations are essential in predicting precipitation, designing chemical reactions, and understanding biological mineralization. Low solubility compounds like Mg\u2083(PO\u2084)\u2082 often play roles in biological systems such as bone and tooth formation or in industrial water treatment processes.<\/p>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"The solubility product constant (Ksp) of Mg3(PO4)2 is 1.04 \u00d7 10-24. What is its solubility in grams per 100 mL? The Correct Answer and Explanation is: To calculate the solubility of magnesium phosphate, Mg\u2083(PO\u2084)\u2082, in grams per 100 mL, we need to follow these steps: Step 1: Write the dissociation equation Mg3(PO4)2(s)\u21cc3Mg2+(aq)+2PO43\u2212(aq)\\text{Mg}_3(\\text{PO}_4)_2 (s) \\rightleftharpoons 3\\text{Mg}^{2+} […]<\/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-39255","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39255","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=39255"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39255\/revisions"}],"predecessor-version":[{"id":39258,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39255\/revisions\/39258"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39255"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39255"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39255"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}