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The solubilituy of PbCI(2) in water is 0...

The solubilituy of `PbCI_(2)` in water is `0.01M 25^(@)C`. Its maximum concentration in `0.1M NaCI` will be:

A

`2xx10^(-3)M`

B

`1xx10^(-4)M`

C

`1.6xx10^(-2)M`

D

`4xx10^(-4)M`

Text Solution

Verified by Experts

The correct Answer is:
D
`K_(SP)of PbCI_(2)=4S^(3)=4xx(0.01)^(3)=4xx10^(-6)`
In NaCI solution for `PbCI_(2)`,
`K_(SP)= [Pb^(2+)][CI^(-)]^(2)`
or `4xx10^(-6)=[Pb^(2+)][0.1]^(2)`
`:. [Pb^(2+)]=4xx10^(-4)M`
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