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

The Correct Option is B

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