1.

At 298 K, the solubility of PbCl_(2) is 2 xx 10^(-2) mol/lit, then K_(sp)=

Answer»

`1 XX 10^(-7)`
`3.2 xx 10^(-7)`
`1 xx 10^(-5)`
`3.2 xx 10^(-5)`

SOLUTION :`{:(PbCl_(2)hArr,Pb^(2+),+,2Cl^(-)),(,(S),,(2S)^(2)):}`
`K_(sp) = 4S^(3) = 4 xx (2 xx 10^(-2))^(3) = 3.2 xx 10^(-5)`.


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