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. ${10}^9{\left[\frac{W}{M}\right]}^5$
• B. ${10}^7{\left[\frac{W}{M}\right]}^5$
• C. ${10}^5{\left[\frac{W}{M}\right]}^5$
• D. ${10}^3{\left[\frac{W}{M}\right]}^5$

Answer 1

The correct option is B ${10}^7{\left[\frac{W}{M}\right]}^5$

Solubility of calcium phosphate (molecular mass, M) in water is W g per 100 ml. Let M be the molecular weight of calcium phosphate.

Solubility of calcium phosphate (molecular mass, M) in water in terms of moles per litre will be $\frac{10\times W}{M}$ moles per litre.

The expression for the solubility product of $C{a}_3(P{O}_4{)}_2$ is $K_{sp}=\left[C{a}^{2+}{]}^3\right[P{O}_4^{3-}{]}^2$.

Substitute values in the above expression.
$K_{sp}=[3\times \frac{10\times W}{M}{]}^3[2\times \frac{10\times W}{M}{]}^2=108(\frac{10\times W}{M}{)}^5$.

This is approximately equal to ${10}^7{\left[\frac{W}{M}\right]}^5$.