Question

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is:
(The formula for Mohr's salt is $FeS{O}_4.(N{H}_4{)}_{2}S{O}_4.6H_{2}O$)

• A. 6
• B. 5
• C. 4
• D. 3

Answer 1

The correct option is A 6

$\text{Mohr's salt}\equiv FeS{O}_4.(N{H}_4{)}_{2}S{O}_4.6H_{2}O$
Dichromate will oxidise $F{e}^{2+}$ from Mohr's salt to $F{e}^{3+}$.

$\stackrel{+6}{C}{r}_2{O}_7^{2-}+\stackrel{+2}{F}{e}^{2+}\stackrel{H^{+}}{⟶}\stackrel{+3}{C}{r}^{3+}+\stackrel{+3}{F}{e}^{3+}$

$n_f=\left(|\text{Change in O.S.}|\right)\times \text{Number of atoms}$
$n_{f_{C{r}_2{O}_7^{2-}}}=|+3-(+6)|\times 2=6$
$n_{f_{F{e}^{2+}}}=|+3-(+2)|\times 1=1$

From the law of equivalence,
$\text{Equivalents of}F{e}^{2+}=\text{Equivalents of}C{r}_2{O}_7^{2-}$
$n_{F{e}^{2+}}\times 1=1\times 6$
$\therefore n_{\text{Mohr's salt}}=n_{F{e}^{2+}}=6mol$