Question

A solution of $0.2g$ of a compound containing $C{u}^{2+}$ and $C_2{O}_4^{2–}$ ions on titration with $0.02MKMn{O}_4$ in presence of $H_{2}S{O}_4$ consumes $22.6mL$ of the oxidant. The resultant solution is neutralized with $N{a}_{2}C{O}_3$, acidified with dil. acetic acid and treated with excess $KI$.The liberated iodine requires $11.3mL$ of $0.05MN{a}_2{S}_3{O}_3$ solution for complete reduction. Find out the mole ratio of $C{u}^{2+}$ to $C_2{O}_4^{2–}$ in the compound is $a:b$. The sum of $a+b$ is

Answer 1

In the given problem, a solution containing $C{u}^{2+}$ and $C_2{O}_4^{2-}$ is titrated first with $KMn{O}_4$ and then with $N{a}_2{S}_2{O}_3$ in presence of $KI$. In titration with $KMn{O}_4$, it is the $C_2{O}_4^{2-}$ ions that react with the $Mn{O}_4^{-}$ ions. The concerned balanced equation may be written as below.
$2Mn{O}_4^{-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$
Thus according to the above reaction
2 mol $Mn{O}_4^{-}$ reacts with 5 mol of $C_2{O}_4^{2-}$
However,
No. of mmol of $Mn{O}_4^{-}$ and used in titration
$=0.02\times 22.6$
Now,
milimoles of oxalate $=\frac{0.02\times 22.6\times 5}{2}=1.130$
$KI$ reacts with $C{u}^{2+}$ as
$2C{u}^{2+}+4I^{-}\to C{u}_2{I}_2+I_2$
$I_2+2S_2{O}_3^{-}\to 2I^{-}+S_4{O}_6^{2-}$
According to the above reactions
mmoles of thiosulfate =mmoles of $C{u}^{2+}$
$=0.05\times 11.3$
$\therefore ,$
Mole ratio of $C{u}^{2+}$ to $C_2{O}_4^{2-}=\frac{0.565mmol}{1.130mmol}=1:2$
Hence, $a:b=1:2$
So, $a+b=1+2=3$