Question

A solution containing $C{u}^{2+}$ and $C_2{O}_4^{-}$ ions which is titrated with $20ml$ of $M/4KMn{O}_4$ solution in acidic medium. The resulting solution is treated with excess of $KI$ after neutralization. The evolved $I_2$ is then absorbs in $25ml$ of $N/10$ hypo solution, then the difference of the number millimoles of $C{u}^{2+}$ and $C_2{O}_4^{2-}$ ions in the solution is:

Answer 1

Reaction of $C_2{O}_4^{2-}$

$C_2{O}_4^{2-}+Mn{O}_4^{-}+H^{+}\to M{n}^{2+}+C{O}_2+H_{2}O$

n for $C_2{O}_4^{2-}=2$

n for $Mn{O}_4^{-}=5$

Milliequivalents of $C_2{O}_4^{2-}=$Milliequivalents of $Mn{O}_4^{-}$

meq$=20\times \frac{M}{4}\times 5$

meq of $C_2{O}_4^{2-}=25$

Millimoles of $C_2{O}_4^{2-}=\frac{meq}{nfactor}=\frac{25}{2}=12.5$ millimoles.

$C{u}^{2+}$ reacts with KI

Reaction of $C{u}^{2+}$:

$2C{u}^{2+}+4KI\to C{u}_2{I}_2+I_2$

$I_2+2S_2{O}_3^{2-}\to S_4{O}_6^{2-}+2I^{-}$

meq of $S_2{O}_3^{2-}=25\times \frac{M}{10}\times 1$

$=2.5$ meq

meq of $S_2{O}_3^{2-}=$meq of $C{u}^{2+}$

$\therefore$ m moles of $C{u}^{2+}=\frac{2.5}{1}=2.5$ millimoles

Here n factor of $C{u}^{2+}$ is $1$

$C{u}^{2+}+e^{-}\to C{u}^{+}$

$\therefore$ Difference in the millimoles of $C_2{O}_4^{2-}$ & $C{u}^{2+}$

is $=12.5-2.5$

$=10$ millimoles.