1.

The solubility of PbCl_2 will be given by the equation

Answer»

`(K_(sp))^(1/3)`
`ROOT3(K_(sp)/4)`
`(8K_(sp))^(1/2)`
`sqrt(K_(sp))`

Solution :`PbCl_2 HARR PB^(2+) + 2Cl^(-)`
`K_(SP)=[Pb^(24)]XX[2Cl^-]^2`
`K_(SP)=(S)(2S)^2`
`K_(SP)=4S^3`
`therefore S=root3((K_(SP))/4)`


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