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Q.

If the solubility product of MOH is $1 \times 10^{- 10} m o l^{2} . d m^{- 2}$ . Then the $P^{H}$ of its aqueous solution will be

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a

12

b

6

c

9

d

3

answer is B.

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Detailed Solution

$K_{s p} = S^{2} = 10^{- 10}$

$(o r)$

$[O H^{-}] = S = \sqrt{10^{- 10}} = 10^{- 5}$

$∴ P^{O H} = 5 (o r) P^{H} = 9$

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If the solubility product of MOH is 1×10−10mol2.dm−2 . Then the PH of its aqueous solution will be