Question

$20$ mL of $H_2{O}_2$ reacts completely with acidified $K_{2}C{r}_2{O}_7$ solution. $40$ mL of $K_{2}C{r}_2{O}_7$ solution was required to oxidise the $H_2{O}_2$ completely. Also, $2.0$ mL of the same $K_{2}C{r}_2{O}_7$ solution required $5.0$ mL of $1.0M$ $H_2{C}_2{O}_4$ solution to reach the equivalence point. Which of the following statements is/are correct?

• A. The $H_2{O}_2$ solution is $5M$.
• B. The volume strength of $H_2{O}_2$ is $56V$.
• C. The volume strength of $H_2{O}_2$ is $112V$.
• D. If $40$ mL of $\frac{M}{8}$ $H_2{O}_2$ is further added to the $10$ mL of the above $H_2{O}_2$ solution, the volume strength of the resulting solution is changed to $16.8V$.

Answer 1

Answer: (A), (B), (D)

The $H_2{O}_2$ solution is $5M$.
If $40$ mL of $\frac{M}{8}$ $H_2{O}_2$ is further added to the $10$ mL of the above $H_2{O}_2$ solution, the volume strength of the resulting solution is changed to $16.8V$.
The volume strength of $H_2{O}_2$ is $56V$.

I. mEq of $\underset{(n=2)}{H_2{O}_2}\equiv$ mEq of $\underset{(n=6)}{C{r}_2{O}_7^{2-}}$
$20mL\times N_1=40mL\times N_2$ .....(i)
a. mEq of $C{r}_2{O}_7^{2-}\equiv$ mEq of $\underset{(n=2)}{C_2{O}_4^{2-}}$
$2.0\times N_2\equiv 5.0\times 1.0\times 2$
$N_2\left(C{r}_2{O}_7^{2-}\right)=5$ ....(ii)
Therefore, substituting the $N_2$ in equation (i)
$20mL\times N_2=40mL\times 5$
$N_1\left(H_2{O}_2\right)=10$
$M_1\left(H_2{O}_2\right)=\frac{10}{2}=5M$
b. $1N$ $H_2{O}_2=5.6V$
$\therefore 10N{H}_2{O}_2=56V$
d. $N_1{V}_1+N_2{V}_2=N_3{V}_3(V_3=10+40=50mL)$
$10\times 10+40\times \frac{5}{8}\times 2=N_3\times 50$
$N_3$(final) $H_2{O}_2=3N$
The volume strength of $H_2{O}_2=5.6\times 3=16.8V$