First slide

Solubility product(KSP)

Question

At $25^{0} C,$ the solubility product of $Mg {(OH)}_{2}$ is $1.0 \times 10^{- 11} .$ At which pH, will ${Mg}^{2 +}$ ions start precipitating in the form of $Mg {(OH)}_{2}$ from a solution of 0.001 M ${Mg}^{2 +}$ ions ?

Moderate

A

9
B

10
C

11
D

8

Solution

$Mg {(OH)}_{2} ⇌ {Mg}^{2 +} + 2 {OH}^{-}$
$K_{SP} = [{Mg}^{2 +}] {[{OH}^{-}]}^{2} [{OH}^{-}] = \sqrt{\frac{K_{SP}}{[{Mg}^{2 +}]}} = \sqrt{\frac{1 \times 10^{- 11}}{0.001}} pOH = - \log [{OH}^{-}] = - \log [10^{- 4}] pOH = 4 and pH = 10 pH = 14 - pOH = 14 - 4 = 10$

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