Question

Assertion :If $x$ mL of $0.2M{H}_{2}S{O}_4$ solution requires $10$ mL of $0.24M$ $KOH$ solution, then $x$ mL of $0.1M$ $H_{2}S{O}_3$ would require $20$ mL of $0.01M$ acidified $K_{2}C{r}_2{O}_7$ solution. Reason: $H_{2}S{O}_3$ is a dibasic acid.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Milliequivalents of $\underset{(n=2)}{Fe{C}_2{O}_4}=$ Milliequivalents of $\underset{(n=1)}{C{r}_2{O}_7^{2-}}$

Milliequivalents of $H_{2}S{O}_4=$ Milliequivalents of $KOH$
$x\times 0.2\times 2=10\times 0.24\times 1$
$x=6$ mL

$\underset{(n=2)}{H_{2}S{O}_3}$ $\underset{(n=6)}{K_{2}C{r}_2{O}_7/H^{⨁}}$
$(S{O}_3^{2-}\to S{O}_4^{2-}+2e^{-})(6e^{-}+C{r}_2{O}_7^{2-}\to 2C{r}^{3+})$

Milliequivalents of $H_{2}S{O}_3\equiv$ Milliequivalents of $C{r}_2{O}_7^{2-}$
$x\times 0.1\times 2\equiv 20\times 0.01\times 6$
$x=6$ mL