Question

In the standardization of $N{a}_2{S}_2{O}_3$ using $K_{2}C{r}_2{O}_7$ by iodometry, the equivalent weight of $K_{2}C{r}_2{O}_7$ is:

• A. $\frac{\text{Molecular weight}}{2}$
• B. $\frac{\text{Molecular weight}}{6}$
• C. $\frac{\text{Molecular weight}}{3}$
• D. $\text{Same as its molecular weight}$

Answer 1

The correct option is B $\frac{\text{Molecular weight}}{6}$

The reaction between $N{a}_2{S}_2{O}_3$ and $K_{2}C{r}_2{O}_7$ occurs as follow:
$S_2{O}_3^{2-}+C{r}_2{O}_7^{2-}\to S{O}_4^{2-}+C{r}^{3+}$
The oxidation state of chromium changes from $+6$ to $+3$. The net change in the oxidation number per formula unit is 6 or n-factor is 6.
$\therefore \text{Equivalent weight}=\frac{\text{Molecular weight}}{6}$