Solubility of calcium phosphate (molecular mass, M) in water is W g per 100 mL at 25°C. Its solubility product at 25°C will be approximately.

1. Convert Solubility to Molarity ($$S$$)

The given solubility is $$W\text{ g}$ per $100\text{ mL}$$. To find the molarity ($$S$$, in $$\text{mol/L}$$):

• Mass in $$1000\text{ mL } (1\text{ L}) = 10 \times W\text{ g}$$
• Moles per liter ($$S$$) = $$\frac{10W}{M}\text{ mol/L}$$

2. Dissociation Equilibrium

Calcium phosphate, $$\text{Ca}_3(\text{PO}_4)_2$$, dissociates in water as follows:

$$\text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Ca}^{2+}(aq) + 2\text{PO}_4^{3-}(aq)$$

If the molar solubility is $$S$$:

• $$[\text{Ca}^{2+}] = 3S$$
• $$[\text{PO}_4^{3-}] = 2S$$

3. Calculate $$K_{sp}$$ Expression

$$K_{sp} = [\text{Ca}^{2+}]^3 \cdot [\text{PO}_4^{3-}]^2$$

$$K_{sp} = (3S)^3 \cdot (2S)^2 = 27S^3 \cdot 4S^2 = 108S^5$$

4. Substitute $$S$$ into the Equation

Now, substitute $$S = \frac{10W}{M}$$:

$$K_{sp} = 108 \left( \frac{10W}{M} \right)^5 = 108 \times 10^5 \times \frac{W^5}{M^5}$$

$$K_{sp} = 1.08 \times 10^7 \times \frac{W^5}{M^5} \approx 10^7 \frac{W^5}{M^5}$$