1.

The solubility product of a sparingly soluble salt AB at room temperature is 1.21xx10^(-6)M^(2). Its molar solubility is :

Answer»

`1.21xx10^(-6)M`
`1.1xx10^(-4)M`
`1.1xx10^(-3)M`
`1.1xx10^(-5)M`

Solution :`AB hArr A^(+)+B^(-)`
Let its molar solubility be s
`[A^(+)]=s [B^(+)]=s`
`K_(sp)=[A^(+)][B^(-)]=s xx s`
`1.21xx 10^(-6)=s^(2)`
or `s=1.1xx10^(-3)M`


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