1.

Solubility product of PbCl_(2) at 298 K is 1.0xx10^(-6). At this temperature solubility of PbCl_(2) in moles per litre is :

Answer»

`(1.0xx10^(-6))^(1//2)`
`(1.0xx10^(-6))^(1//3)`
`(0.25xx10^(-6))^(1//3)`
`(0.25xx10^(-6))^(1//2)`

Solution :Let SOLUBILITY of `PbCl_(2)=x`
`K_(sp)=[Pb^(2+)][Cl^(-)]^(2)=(x)(2x)^(2)=4x^(3)`
`THEREFORE x = ((1.0xx10^(-8))/(4))^(1//3)`
`=(0.25xx10^(-6))1//3`


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