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 $7.7863\times 10^{-12}$

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

$K_{sp}=(0.0000027904M)\times (0.0000027904M)$

$K_{sp}=7.7863\times 10^{-12}$

Therefore, the value of $K_{sp}$ is $7.7863\times 10^{-12}$