First slide

Solubility product(KSP)

Question

If the solubility product of MOH is $1 \times 10^{- 10} {mol}^{2} {dm}^{- 6}$ , then pH of its aqueous solution will be

Moderate

A

12
B

9
C

6
D

3

Solution

$s = \sqrt{K_{sp}} = \sqrt{1 \times 10^{- 10}} = 1 \times 10^{- 5} M [{OH}^{-}] = [MOH] = 1 \times 10^{- 5} [H_{3} O^{+}] = \frac{1 \times 10^{- 14}}{1 \times 10^{- 5}} = 1 \times 10^{- 9} pH = - \log [H_{3} O^{+}] = - \log 10^{- 9} = 9$

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