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The solubility product of AgCI in water ...

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

Text Solution

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Let solubility be S mol/litre
`AgCl Leftrifhtarrow Ag^(+) +Cl^(-1)`
`S (S+0.01)`
`K_(SP)=S xx (S+0.01)`
`=S xx (0.01) (therefore S lt lt 0.01)`
`S=(1.5 xx 10^(-10))/(0.01)=1.5 xx 10^(-8)"mol/litre".`
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