Question

500 ml of a saturated clear solution of $Ca{C}_2{O}_4$(aq) requires 12.6 ml of 0.00102M $KMn{O}_{4\left(aq\right)}$ in acidic medium for complete oxidation of $C_2{O}_4^{2-}$ ions. The $K_{sp}$ of $Ca{C}_2{O}_4$ is:

• A. $3\times {10}^{-9}$
• B. $4\times {10}^{-9}$
• C. $5\times {10}^{-9}$
• D. $6\times {10}^{-9}$

Answer 1

The correct option is B $4\times {10}^{-9}$

$M{n}^{7+}+5e^{-}\to M{n}^{2+}$

$M.eq.of{C}_2{O}_4^{2-}=M.eq.ofKMn{O}_4$ $=0.00102\times 12.6\times 5=0.064$.

Molarity of $C_2{O}_4^{2-}=\frac{M.eq.}{V_{ml}\times Valencyfactor}$

i.e., solubility $=\frac{0.064}{500\times 2}=6.4\times {10}^{-5}M$

$C_2^{3+}\to 2C^{4+}+2e^{-}$

$K_{sp}$ of $Ca{C}_2{O}_4=\left[C{a}^{2+}\right]\left[C_2{O}_4^{2-}\right]=s^2=(6.4\times {10}^{-5}{)}^2=4.09\times {10}^{-9}mo{l}^2{L}^{-2}$.