The solubility of `PbI_(2)` is `0.0013`M. Then the solubility product of `PbI_(2)` is

Options

A. `2.2xx10^(-9)`
B. `8.8xx10^(-9)`
C. `6.8xx10^(-6)`
D. `0.8xx10^(-6)`

Answer

B

Explanation

Equilibrium equation is
`PbI_(2)(s)rarrPb^(2+)(aq.)+2I^(-)(aq.)`
The solubility of `PbI_(2)` is `0.0013mol L^(-1)` or `1.3xx10^(-3)mol L^(-1)`
`:. C_(Pb^(2+))=1.3xx10^(-3)M`
`C_(I^(-))=2xx1.3xx10^(-3)mol L^(-1)`
`=2.6xx10^(-3)mol L^(-1)`
`K_(sp)=C_(Pb^(2+)C_(I^(-))`
`=(1.3xx10^(-3))(2.6xx10^(-3))^(2)`
`=8.8xx10^(-9)`