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Solubility product (K(sp)) of saturated...

Solubility product `(K_(sp))` of saturated `PbCI_(2)` in water is `1.8xx10^(-4) mol^(3) dm ^(-9)` what is the concentration of `Pb^(2+)` in the solution

A

`(0.45 xx10^(-4))^(1//3) "mol dm"^(-3)`

B

`(1.8xx10^(-4))^(1//3) "mol dm"^(-3)`

C

`(0.9 xx10^(-4))^(1//3) "mol dm"^(-3)`

D

`(2.0 xx10^(-4))^(1//3) "mol dm"^(-3)`

Text Solution

Verified by Experts

The correct Answer is:
A
For the reaction of the `AB_(2)`
Given `K_(sp)=1.8xx10^(-4) mol^(3) dm^(-9)`
`therefore` solubility of `Pb^(+2)` ions will be
`therefore S=[(1.8xx10^(-4))/(4)]^(1/3)`
`=[0.45xx10^(-4)]^(1/3` mol `dm^(-3)`
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