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

To determine the solubility product (\( K_{sp} \)) of calcium phosphate, its dissolution must be analyzed, and relevant chemical equations and solubility product expressions must be written.

Calcium phosphate, \(\text{Ca}_3(\text{PO}_4)_2\), dissociates in water according to the equilibrium reaction:

\(Ca_3(PO_4)_2 \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-}\)

If the solubility of calcium phosphate is \( W \, \text{g} \) per 100 mL, then its solubility in 1 L is \( 10W \, \text{g} \). The number of moles of \(\text{Ca}_3(\text{PO}_4)_2\) dissolved is:

\(\frac{10W}{M}\)

Given this solubility, the equilibrium concentrations of the respective ions are:

• Concentration of \( Ca^{2+} = 3 \times \frac{10W}{M} \) moles/L.
• Concentration of \( PO_4^{3-} = 2 \times \frac{10W}{M} \) moles/L.

The expression for the solubility product \( K_{sp} \) is:

\(K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2\)

Substituting the above values yields:

\(K_{sp} = \left( 3 \times \frac{10W}{M} \right)^3 \left( 2 \times \frac{10W}{M} \right)^2\)

Simplifying this expression results in:

\(= 27 \left( \frac{10W}{M} \right)^3 \cdot 4 \left( \frac{10W}{M} \right)^2 = 108 \times \left( \frac{10W}{M} \right)^5\)

This can be expressed as:

\(= 108 \times 10^5 \times \left( \frac{W}{M} \right)^5 \approx 10^7 \left( \frac{W}{M} \right)^5\)

Therefore, the solubility product at \( 25^\circ C \) is approximately \(10^7 \left( \frac{W}{M} \right)^5\).

The correct answer is \( 10^7 \left( \frac{W}{M} \right)^5 \).