Question

In the standardization of $N{a_2}S{O_3}$ using ${K_2}C{r_2}{O_7}$ by iodometry, the equivalent weight of ${K_2}C{r_2}{O_7}$ is:
A: $\dfrac{{Mw}}{2}$
B: $\dfrac{{Mw}}{6}$
C: $\dfrac{{Mw}}{3}$
D: Same as the molecular weight

Answer 1

Hint: Equivalent mass of a substance can be calculated by dividing molecular mass of the substance with the valence or oxidation state or change in oxidation state. Valency is the charge that an atom or ion will possess if the compound was ionic.

Formula used: Equivalent mass$ = \dfrac{{{\text{Molecular mass}}}}{{{\text{Change in valency}}}} = \dfrac{{Mw}}{{{\text{n factor}}}}$

Complete step by step answer:
 In this question we have to find the equivalent weight of ${K_2}C{r_2}{O_7}$ when ${K_2}C{r_2}{O_7}$ reacts with $N{a_2}S{O_3}$. For this we have to write the reaction which will take place.
$26{H^ + } + 3{S_2}O_3^{2 - } + 4C{r_2}O_7^{2 - }\xrightarrow[{}]{}6SO_4^{2 - } + 8C{r^{3 + }} + 13{H_2}O$
We have to find an equivalent weight of ${K_2}C{r_2}{O_7}$. For this we have to find the change in oxidation number of chromium using the above reaction. Oxidation state of two atoms of chromium on reactant side is $ + 12$ (as oxidation state of oxygen is $ - 2$ and there are seven atoms of oxygen so total charge is $ - 14$ and on whole compound charge is $ - 2$ calculating this we get charge on chromium as $ + 12$).
Charge on two atoms of chromium$ = + 12$
There are two atoms of chromium in a given compound. So, we will calculate change in oxidation state with respect to two atoms of chromium
So, charge on one atom of chromium on reactant side is $ + 12$
Charge on one atom of chromium on product side is $ + 3$
So, charge on two atoms of chromium on product side is $2 \times 3 = 6$
This means charge on chromium atoms on product side is $ + 6$
Therefore change in oxidation state$ = $ oxidation state on reactant side $ - $ oxidation state on reactant side
Change in oxidation state$ = 12 - 6 = 6$
Using the formula for equivalent weight written above that is:
 Equivalent mass$ = \dfrac{{{\text{Molecular mass}}}}{{{\text{Change in valency}}}}$
Substituting the values we get:
Equivalent mass$ = \dfrac{{Mw}}{6}$
So, option (B) is correct.

Note:
Normality of a solution is calculated using equivalent mass. Normality is defined as the number of gram equivalents present per liter of the solution. Gram equivalents are calculated by multiplying the number of moles with change in oxidation state.