First slide

Solubility product(KSP)

Question

Solubility of calcium phosphate (molecular mas s, M) in water is W g per 100 mL at 25°C, Its solubility product at 25°C will be approximately.

Moderate

A

$10^{7} {(\frac{W}{M})}^{3}$
B

$10^{7} {(\frac{W}{M})}^{5}$
C

$10^{5} {(\frac{W}{M})}^{5}$
D

$10^{3} {(\frac{W}{M})}^{5}$

Solution

$S = \frac{10 W}{M} mol L$
$K_{sp} of {Ca}_{3} {({PO}_{4})}_{2} = 108 S^{5}$
$= 108 {(\frac{10 W}{M})}^{5}$ $= 108 \times 10^{5} (\frac{W}{M})$
$= 10^{7} {(\frac{W}{M})}^{5} (approximately)$

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