1.

The solubility of PbCl_(2) at 25^(@)C is 6.3 xx 10^(-3) mole/litre. Its solubility product at that temperature is

Answer»

`(6.3 xx 10^(-3)) xx (6.3 xx 10^(-3))`
`(6.3 xx 10^(-3)) xx (12.6 xx 10^(-3))`
`(6.3 xx 10^(-3)) xx (12.6 xx 10^(-3))^(2)`
`(12.6 xx 10^(-3)) xx (12.6 xx 10^(-3))`

Solution :`PbCl_(2) rarr Punderset(S)(b^(++)) + UNDERSET(2S)(2Cl^(-))`
`K_(SP) = S xx (2S)^(2) = [6.3 xx 10^(-3)] xx [12.6 xx 10^(-3)]^(2)`.


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