$[⊝ O H]$ of a solution is $2.8 \times 10^{- x} M$ after $100$ mL of $0.1 M$ $M g C l_{2}$ is added to $100$ mL of $0.2 M$ NaOH. The value of $x$ is
$K_{s p} M g (O H)_{2} = 1.2 \times 10^{- 11}$

Open in App

Solution

$M g C l_{2} + 2 N a O H ⟶ M g (O H)_{2} + 2 N a C l$
mmoles before reaction $⟹$ 10 20 0 0
mmoles after reaction $⟹$ 0 0 10 20
Thus, $10$ mmol of $M g (O H)_{2}$ are formed. The product of $[M g^{2 +}] [O H^{⊝}]^{2}$ is, therefore, $[\frac{10}{200}] \times {[\frac{20}{200}]}^{2}$
$= 5 \times 10^{- 4}$ which is more than $K_{s p}$ of $M g (O H)_{2}$ .
Now solubility (S) of $M g (O H)_{2}$ can be derived by $K_{s p} = 4 S^{3}$
$∴ S = \sqrt{\frac{K_{s p}}{4}} = 3 \sqrt{\frac{1. \times 10^{- 11}}{4}} = 4.1 \times 10^{- 4}$
$∴ [O H^{⊝}] = 2 S = 2.8 \times 10^{- 4}$
Hence, the value of $x$ is $4$ .

Suggest Corrections

Similar questions

20

x M

H C l

10

0.1 M

N a H C O_{3}

5

0.2 M

N a_{2} C O_{3}

x

100

0.1 M

H_{2} S O_{4}

100

0.1 M

N a O H

100 m L

0.1 M M g C l_{2}

100 m L

0.2 M N a O H

[K_{s p}

M g (O H)_{2} = 1.2 \times 10^{- 11}]

Join BYJU'S Learning Program