First slide

Solubility product(KSP)

Question

Molar solubility of $M {(OH)}_{2}$ in 0.2 M $NaOH$ will be $(K_{sp} of M {(OH)}_{2} = 2 \times 10^{- 24} M^{3})$

Moderate

A

$5 \times 10^{- 21} M$
B

$5 \times 10^{- 22} M$
C

$5 \times 10^{- 23} M$
D

$5 \times 10^{- 20} M$

Solution

$M {(OH)}_{2} ⇋ \underset{S}{M^{+ 2}} + \underset{2 S}{2 {OH}^{-}} NaOH \to \underset{0.2 M}{{Na}^{+}} + \underset{0.2 M}{{OH}^{-}} K_{sp} = [M^{+ 2}] {[{OH}^{-}]}^{2} K_{sp} = S {(2 S + 0.2)}^{2};' S' is too small compared to 0.2 M, it can be neglected; S = \frac{K_{sp}}{{(2 \times 10^{- 1})}^{2}} = \frac{2 \times 10^{- 24}}{4 \times 10^{- 2}} = 5 \times 10^{- 23} M$

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