1.

When 1 L of a saturated solution of PbCI_2, is evaporated to dryness, the residue is found to weigh 4.5 g. What will be the K_(sp) for PbCl_2?

Answer»

`1.70 XX 10^(-5) `
`2.70 xx 10^(-6)`
`3.20 xx 10^(-5)`
`1.80 xx 10^(-6)`

Solution :LETUS supposethe solubilityof `PbCI_2 ` beS mol /L
` PbCI_2 hArrpB^(2+) +2CI^-`
` thereforek_(sp )= [Pb^(2+) + 2CI^(-)`
` thereforeK_(sp )= [Pb^(2+)][CI^(-)]^2 =S xx (2S)^2 =4S^2`
Solubility`=4.5g//lit =(4.5 ) /( 278 )= 1.62 xx 10^(-2)` Mol / lit
( molarmassof`PbCI_2 =278` )
` thereforeK_(sp ) = 4 xx ( 1.62 xx 10^(-2))^3`
` or K_(sp )= 1.70 xx 10^(-5)`


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