1.

At 298 K, the solubility product of PbCl_(2) is 1.0 xx 10^(-6). What will be the solubility of PbCl_(2) in moles/litre

Answer»

`6.3 xx 10^(-3)`
`1.0 xx 10^(-3)`
`3.0 xx 10^(-3)`
`4.6 xx 10^(-14)`

SOLUTION :`K_(SP) = 4S^(3)`
`S = 3sqrt((K_(sp))/(4)) = 3sqrt((1.0 xx 10^(-6))/(4)) = 6.3 xx 10^(-3)`


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