Find out the solubility of in 0.1 M NaOH. Given that the ionic product of is . [2020]

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Solution

Q.
Find out the solubility of $Ni {(OH)}_{2}$ in 0.1 M NaOH. Given that the ionic product of $Ni {(OH)}_{2}$ is $2 \times 10^{- 15}$ . [2020]

1 $2 \times 10^{- 13}$ M

2 $2 \times 10^{- 8}$ M

3 $1 \times 10^{- 13}$ M

4 $1 \times 10^{8}$ M

Ans.
(1)

$\underset{s}{{Ni(OH)}_{2}} ⇌ \underset{s}{{Ni}^{2 +}} + \underset{2 s}{2 {OH}^{-}}$

where $s$ is the solubility of $Ni {(OH)}_{2}$ .

$\underset{0.1 M}{NaOH} ⇌ \underset{0.1 M}{{Na}^{+}} + \underset{0.1 M}{{OH}^{-}}$

$[{OH}^{-}] = 2 s + 0.1 \approx 0.1 (∵ 2 s < < < 0.1)$

Ionic product of $Ni {(OH)}_{2}$ $= [{Ni}^{2 +}] {[{OH}^{-}]}^{2} = 2 \times 10^{- 15} = s {(0.1)}^{2}$

$s = \frac{2 \times 10^{- 15}}{0.1 \times 0.1} = 2 \times 10^{- 13}$ $M$

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