Question

Question #71344

Answer 1

The solubility `=9.22xx10^(-10)"mol/l"`

`PbC_2O_(4(s))rightleftharpoonsPb_((aq))^(2+)+C_2O_(4(aq))^(2-)`

`K_(sp)=8.5xx10^(-19)mol^(2).l^(-2)`

`K_(sp)=[Pb_((aq))^(2+)][C_2O_(4(aq))^(2-)]`

Since the concentration of lead(II) and ethandioate are equal we can write:

`K_(sp)=[Pb_((aq))^(2+)]^(2)=8.5xx10^(-19)mol^(2).l^(-2)`

So `[Pb_((aq))^(2+)]=sqrt(8.5xx10^(-19))=9.22xx10^(-10)"mol/l"`

This is the concentration of lead(II) ions in the saturated solution.

Since `PbC_2O_4` is 1 molar with respect to Pb(II) then this is also the solubility of `PbC_2O_4` in `"mol/l"`. I assume this is what you mean by "dissolution value".

You must always quote the temperature for which Ksp refers. You should be given this with the question.

Answer 2

Dissolution is the process by which a solute forms a solution in a solvent. The dissolution value may be the ion concentration after dissolution. This would be `9.2*10^(-10)M`

The Ksp value in this process is the relationship

`K_(sp)=[Pb^(2+)][C_2O_4^(2-)]=8.5*10^(-19)`

For each formula unit of lead II oxalate that dissolves you get one lead II ion and one oxalate ion. Let x = these concentrations.

`x=[Pb^(2+)]=[C_2O_4^(2-)]`
`x^2=8.5*10^(-19)`

so, after dissolution the ion concentrations are

`x=[Pb^(2+)]=[C_2O_4^(2-)]=9.2*10^(-10)M`