Question

Solubility of calcium phosphate (molecular mass, $M$) in water is $W_g$ per 100 mL at $25^\circ C$. Its solubility product at $25^\circ C$ will be approximately:

• A. $10^7 \left( \frac{W}{M} \right)^3$
• B. $10^7 \left( \frac{W}{M} \right)^5$
• C. $10^3 \left( \frac{W}{M} \right)^5$
• D. $10^5 \left( \frac{W}{M} \right)^5$

Answer 1

Answer: (B)

$10^7 \left( \frac{W}{M} \right)^5$

Answer 2

The chemical formula for calcium phosphate is Ca$_3$(PO$_4$)$_2$.
The dissociation of calcium phosphate in water is represented as:
$\text{Ca}_3(\text{PO}_4)_2(s) \iff 3\text{Ca}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)$
Let the molar solubility of Ca$_3$(PO$_4$)$_2$ be $s$ mol/L.
$[\text{Ca}^{2+}] = 3s, \quad [\text{PO}_4^{3-}] = 2s$
The solubility product $K_\text{sp}$ is given by:
$K_\text{sp} = [\text{Ca}^{2+}]^3 \times [\text{PO}_4^{3-}]^2 = (3s)^3 \times (2s)^2 = 27s^3 \times 4s^2 = 108s^5$
Converting mass solubility to molar solubility:
Given that the solubility in grams is $W$ g per 100 mL, the molar solubility $s$ is:
$s = \frac{W}{M} \times \frac{1}{0.1} = 10 \left( \frac{W}{M} \right) \text{mol/L}$
Substituting this value into the expression for $K_\text{sp}$:
$K_\text{sp} \approx 108 \left( 10 \left( \frac{W}{M} \right) \right)^5 \approx 10^7 \left( \frac{W}{M} \right)^5$
Conclusion: The solubility product at 25$^\circ$C is approximately $10^7 \left( \frac{W}{M} \right)^5$.