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Find out the solubility of Ni(OH)2 in 0....

Find out the solubility of `Ni(OH)_2` in `0.1 M` NaOH Given that the ionic product of `Ni(OH)_2` is `2xx10^(-15)`.

A

`2xx10^(-8)` M

B

`1xx10^(-13)` M

C

`1xx10^(8)` M

D

`2xx10^(-13)` M

Text Solution

Verified by Experts

The correct Answer is:
D
Let the solubility of `Ni(OH)_2` is s
`underset(s)(Ni(OH)_2) harr underset(s)(Ni^(2+))+underset(2s)(2OH^(-))`
`underset(0.1)(NaOH)rarr underset(0.1)(Na) + underset(0.1)(OH^(-)`
As `K_(sp)` is small `2s lt lt 0.10`
therefore (0.10 + 2s) =0.10
so Total `[OH]^(-) = 0.10`
Ionic product = `[Ni^(2+)][OH^(-)]^2`
`2 xx 10^(-15) = s(0.10)^2`
`s=2xx10^(-13)` M
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Knowledge Check

  • 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) .

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    B
    `2 times 10^(-8)M`
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    `1.11 xx 10^(-2)`
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