Question

. Calculate Ksp for copper oxalate (CuC2O4). Use your calculated copper ion concentration (0.0000027904 M) from the copper oxalate half-cell. Hint: This problem is also similar to practice problem 3.

Answer 1

Answer : The value of
`K_(sp)` is
`7.7863* 10^(-12)`

Explanation :

The solubility equilibrium reaction will be:

`CuC_2O_4\rightleftharpoons Cu^(+)+C_2O_4^(-)`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Cu^(+)][C_2O_4^(-)]`

Concentration of copper ion = Concentration of oxalate ion = 0.0000027904 M

`K_(sp)=(0.0000027904M)* (0.0000027904M)`

`K_(sp)=7.7863* 10^(-12)`

Therefore, the value of
`K_(sp)` is
`7.7863* 10^(-12)`