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When 1 L of a saturated solution of PbCI...

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`?

A

`1.70 xx 10^(-5) `

B

`2.70 xx 10^(-6)`

C

`3.20 xx 10^(-5)`

D

`1.80 xx 10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
A
Let us suppose the solubility of `PbCI_2 ` be S mol /L
` PbCI_2 hArr pB^(2+) +2CI^-`
` therefore k_(sp ) = [Pb^(2+) + 2CI^(-)`
` therefore K_(sp ) = [Pb^(2+)][CI^(-)]^2 =S xx (2S)^2 =4S^2`
Solubility `=4.5 g //lit =(4.5 ) /( 278 )= 1.62 xx 10^(-2)` Mol / lit
( molar mass of `PbCI_2 =278` )
` therefore K_(sp ) = 4 xx ( 1.62 xx 10^(-2))^3`
` or K_(sp ) = 1.70 xx 10^(-5)`
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