Question

The 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

Answer: (B)

$S \, =\frac{10 W}{M} $ mole per litre
$K_{s p}$ of $C a_{3}\left(P O_{4}\right)_{2}= \, 108 \, S^{5}$ $= \, 108\left(\frac{10 W}{M}\right)^{5}$
$=10^{7}\left(\frac{W}{M}\right)^{5}$ (approximately)