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:
- \( 10^7 \left( \frac{W}{M} \right)^3 \)
- \( 10^7 \left( \frac{W}{M} \right)^5 \)
- \( 10^3 \left( \frac{W}{M} \right)^5 \)
- \( 10^5 \left( \frac{W}{M} \right)^5 \)
The Correct Option is B
Approach Solution - 1
To determine the solubility product (\( K_{sp} \)) of calcium phosphate, we need to analyze its dissolution and write relevant chemical equations and expressions for solubility product.
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 in 1 L, it 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 concentrations of the respective ions at equilibrium 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 given by:
\(K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2\)
Substituting the values from above, we get:
\(K_{sp} = \left( 3 \times \frac{10W}{M} \right)^3 \left( 2 \times \frac{10W}{M} \right)^2\)
Simplifying this expression gives:
\(= 27 \left( \frac{10W}{M} \right)^3 \cdot 4 \left( \frac{10W}{M} \right)^2 = 108 \times \left( \frac{10W}{M} \right)^5\)
This can be written as:
\(= 108 \times 10^5 \times \left( \frac{W}{M} \right)^5 \approx 10^7 \left( \frac{W}{M} \right)^5\)
Thus, the solubility product at \( 25^\circ C \) is approximately \(10^7 \left( \frac{W}{M} \right)^5\).
The correct answer is \( 10^7 \left( \frac{W}{M} \right)^5 \).
Approach Solution -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 $10^7 \left( \frac{W}{M} \right)^5$.
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