Answer
It is necessary $1.7 \times 10^{-8}M $ of oxalate.
Work Step by Step
1. Calculate the molar mass:
mm($Ca$) = 40.08g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 5.33}{ 40.08}$
$n(moles) = 0.133$
3. Find the concentration in mol/L:
5.33 g = 0.133 mol
5.33 g/L = 0.133M
4. Write the $K_{sp}$ expression:
$ CaC_2O_4(s) \lt -- \gt 1Ca^{2+}(aq) + 1C_2O_4^{2-}(aq)$
$2.3 \times 10^{-9} = [Ca^{2+}]^ 1[C_2O_4^{2-}]^ 1$
5. Find the $C_2O_4^{2-}$ concentration.
$2.3 \times 10^{-9}= 0.133 \times ( [C_2O_4^{2-}])$
$ \frac{2.3 \times 10^{-9}}{0.133} = ( [C_2O_4^{2-}])$
$1.73 \times 10^{-8} = ( [C_2O_4^{2-}])$
