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The solubility product of AgCl in water ...

The solubility product of `AgCl` in water is `1.5xx10^(-10)`. Calculate its solubility in `0.01M NaCI`.

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Let solubility be `S` `mol//litre`
`AgCl hArr underset(S)(Ag^(+))+underset((S+0.01))(Cl^(-))`
`:. K_(sp)=Sxx(S+0.1)`
`=Sxx(0.01)` , `( :. Sltlt 0.01)`
`:. S=(1.5xx10^(-10))/(0.1)=1.5xx10^(-9)mol//litre`
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