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The solubility of PbCl2 at 25^oC is 6.3 ...

The solubility of `PbCl_2` at `25^oC` is `6.3 xx 10^-3 mol/litre`. The solubility product of `PbCl_2` at `25^oC` is:

A

`(6.3xx10^(-3)) xx (6.3xx10^(-3))`

B

`(6.3xx10^(-3)) xx (12.6xx10^(-3))`

C

`(6.3xx10^(-3)) xx (12.6xx10^(-3))^2`

D

`(12.6xx10^(-3)) xx (12.6xx10^(-3))`

Text Solution

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The correct Answer is:
C
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