Question

The solubility of calcium phosphate (molecular mass $=M)$ in water is $Wg$ per $100mL$ 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

Answer: (B)

$S=\frac{10W}{M}$ mole per litre
$K_{sp}$ of $C{a}_3{\left(P{O}_4\right)}_2=108S^5$ $=108{\left(\frac{10W}{M}\right)}^5$
$=10^7{\left(\frac{W}{M}\right)}^5$ (approximately)