## Chemistry: The Molecular Science (5th Edition)

It is necessary $1.7 \times 10^{-8}M$ of oxalate.
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. $2.3 \times 10^{-9}= 0.133 \times ( [C_2O_4^{2-}])$ $\frac{2.3 \times 10^{-9}}{0.133} = ( [C_2O_4^{2-}])$ $1.73 \times 10^{-8} = ( [C_2O_4^{2-}])$