Question

The ksp of pbbr2 is 6.60×10−6. What is the molar solubility of pbbr2 in pure water?

Answer 1

Answer : The molar solubility of
`PbBr_2` in pure water is, 0.0118 M

Explanation : Given,

`K_(sp)=6.60* 10^(-6)`

The solubility equilibrium reaction will be:

`PbBr_2\rightleftharpoons Pb^(2+)+2Br^(-)`

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

`K_(sp)=[Pb^(2+)][Br^(-)]^2`

`K_(sp)=(s)* (2s)^2`

Now put all the given values in the above expression, we get:

`6.60* 10^(-6)=4s^3`

`s=0.0118M`

Therefore, the molar solubility of
`PbBr_2` in pure water is, 0.0118 M