Question

Strength of oxalic acid in the solution is :

• A. $4.5g{L}^{-1}$
• B. $4.9g{L}^{-1}$
• C. $2.25g{L}^{-1}$
• D. $2.45g{L}^{-1}$

Answer 1

Answer: (A)

$4.5g{L}^{-1}$

In the first titration, both $H_{2}S{O}_4$ and $H_2{C}_2{O}_4$ react with base it is an acid-base titration.

$n$ factor for both is $2$ due to $2H^{⨁}$ per mol in each case.

Let $x=$ mEq. of $H_2{C}_2{O}_4$ and $y=$ mEq of $H_{2}S{O}_4$.

$\therefore$ mEq. of $H_{2}S{O}_4$ $+$ mEq. of $H_{2}S{O}_4$ $=$ mEq. of $KOH$

$\therefore x+y=20\times 0.1\times 1$ ($n$-factor)
$x+y=2$ ..........(i)

In second titration, only $H_2{C}_2{O}_4$(being reducing agent) reacts with $K_{2}C{r}_2{O}_7$.
$\therefore \underset{(n=2)}{mEq.of{H}_{2}C{r}_2{O}_4}=\underset{(n=6)}{mEq.of{K}_{2}C{r}_2{O}_7}$

$(C_2{O}_4^{2-}\to 2C{O}_2+2e^{-})(6e^{-}+C{r}_2{O}_7^{2-}\to 2C{r}^{3+})$

$x=$ mEq. of $H_2{C}_2{O}_4=50\times \frac{1}{300}\times 6$
$\therefore x=1$ mEq ......(ii)

From equations $\left(i\right)$ and $\left(ii\right)$, we get
$x=1$ mEq., $y=1$ mEq.

Weight of $H_2{C}_2{O}_4=1\times {10}^{-3}\times 45$ $(Eq.wt.of{H}_2{C}_2{O}_4=\frac{90}{2}=45)$
$=0.045$ g per $10$ mL $=\frac{0.045\times 1000}{10}g{L}^{-1}$ $=4.5g{L}^{-1}$

Therefore, strength of $H_2{C}_2{O}_4=4.5g{L}^{-1}$
Note:
In the above reaction, $H_{2}S{O}_4$ also reacts with $K_{2}C{r}_2{O}_7$ but in the same reaction with $H_2{C}_2{O}_4$, the mEq. of $H_{2}S{O}_4$ should not be added separately.

$H_2{C}_2{O}_4+K_{2}C{r}_2{O}_7+H_{2}S{O}_4\to 2C{r}^{3+}+C{O}_2$