The solubility product of PbBr(2) is 8.0...

The solubility product of `PbBr_(2)` is `8.0 xx 10^(-5)`. Calculate its solubility.

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We are give that,
Solubility product of silver bromide `= 3.3 xx 10^(-13)`
The solubility of salt of AB type is given as `S = sqrt(K_(sp))`
We are to calculate the solubility of AgBr.
So, solubility of AgBr will be `S = sqrt(3.3 xx 10^(-13)) = sqrt(33 xx 10^(-14)) = 5.74 xx 10^(-7)` mol `L^(-1)`.

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