The solubility product constant (Ksp) of Mg3(PO4)2 is 1.04 × 10-24. What is its solubility in grams per 100 mL?
The Correct Answer and Explanation is:
To calculate the solubility of magnesium phosphate, Mg₃(PO₄)₂, in grams per 100 mL, we need to follow these steps:
Step 1: Write the dissociation equation
Mg3(PO4)2(s)⇌3Mg2+(aq)+2PO43−(aq)\text{Mg}_3(\text{PO}_4)_2 (s) \rightleftharpoons 3\text{Mg}^{2+} (aq) + 2\text{PO}_4^{3-} (aq)Mg3(PO4)2(s)⇌3Mg2+(aq)+2PO43−(aq)
Let the molar solubility of Mg₃(PO₄)₂ be x mol/L. Then at equilibrium:
- [Mg2+]=3x[\text{Mg}^{2+}] = 3x[Mg2+]=3x
- [PO43−]=2x[\text{PO}_4^{3-}] = 2x[PO43−]=2x
Step 2: Write the Ksp expression
Ksp=[Mg2+]3[PO43−]2=(3x)3(2x)2=27×3⋅4×2=108x5K_{sp} = [\text{Mg}^{2+}]^3[\text{PO}_4^{3-}]^2 = (3x)^3(2x)^2 = 27x^3 \cdot 4x^2 = 108x^5Ksp=[Mg2+]3[PO43−]2=(3x)3(2x)2=27×3⋅4×2=108×51.04×10−24=108×51.04 \times 10^{-24} = 108x^51.04×10−24=108×5
Step 3: Solve for x
x5=1.04×10−24108=9.63×10−27x^5 = \frac{1.04 \times 10^{-24}}{108} = 9.63 \times 10^{-27}x5=1081.04×10−24=9.63×10−27x=(9.63×10−27)1/5≈4.07×10−6 mol/Lx = (9.63 \times 10^{-27})^{1/5} \approx 4.07 \times 10^{-6} \, \text{mol/L}x=(9.63×10−27)1/5≈4.07×10−6mol/L
Step 4: Convert molar solubility to grams per 100 mL
Molar mass of Mg₃(PO₄)₂:
- Mg = 24.305 g/mol × 3 = 72.915
- P = 30.974 g/mol × 2 = 61.948
- O = 16.00 g/mol × 8 = 128.00
Total = 72.915 + 61.948 + 128.00 = 262.86 g/mol
Solubility (g/L)=4.07×10−6 mol/L×262.86 g/mol≈1.07×10−3 g/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)=4.07×10−6mol/L×262.86g/mol≈1.07×10−3g/LSolubility (g/100 mL)=1.07×10−310=1.07×10−4 g/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)=101.07×10−3=1.07×10−4g/100 mL
Explanation
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, Mg₃(PO₄)₂ 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.
To determine solubility, we define x as the molar solubility of Mg₃(PO₄)₂. 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 = 108x⁵. 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.
Once we have the molar solubility, converting it to grams per liter is a matter of multiplying by the compound’s molar mass. To express the solubility per 100 mL, we divide the result by 10. The final answer, 1.07 × 10⁻⁴ g/100 mL, reflects the very low solubility of magnesium phosphate in water.
Such calculations are essential in predicting precipitation, designing chemical reactions, and understanding biological mineralization. Low solubility compounds like Mg₃(PO₄)₂ often play roles in biological systems such as bone and tooth formation or in industrial water treatment processes.
