Question

Solubility of calcium phosphate (molecular mass, $M$ ) in water is $W$ g per $100mL$ 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. $10^5{\left(\frac{W}{M}\right)}^5$
• D. $10^3{\left(\frac{W}{M}\right)}^5$

Answer 1

Answer: (B)

$\because S=\frac{10W}{M}mol/L$
For $C{a}_3{\left(P{O}_4\right)}_2\rightleftharpoons \underset{3S}{3C{a}^{2+}}+\underset{2S}{2P{O}_4^{3-}}$
$\Rightarrow K_{sp}=(3S{)}^3\times (2S{)}^2$
$\therefore K$ of $C{a}_2{\left(P{O}_4\right)}_2=108S^5$
$\Rightarrow 108{\left(\frac{10W}{M}\right)}^5$
$\Rightarrow 10^7{\left(\frac{W}{M}\right)}^5($ approximately $)$