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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^(-13)` M

B

` 2 xx 10^(-8)` M

C

`1 xx 10^(-13)` M

D

`1 xx 10^(8)` M

Text Solution

Verified by Experts

The correct Answer is:
A
`underset (S) (Ni(OH)_(2)) " "underset (S) (Ni^(2+)) + overset (Theta) underset (2S) (2OH)`
`underset (0.1)(NaOH) to underset (0.1)(Na^(+)) + overset (Theta) underset(0.1)(OH)`
Total `overset (Theta)([OH])= 2S +0.1 approx 0.1 `
Ionic product = `[Ni^(2+)] [overset(-)(OH)]^(2)`
`2 xx 10^(-15) = S(0.1)^(2)`
`S = 2 xx 10^(-13)`
Solubility of `Ni(OH)_(2) = 2 xx 10^(-13) M `
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