Question

In the titration ${K_2}C{r_2}{O_7}$ and $FeS{O_4}$, the following data is obtained, ${V_1}mL$ of ${M_1}{K_2}C{r_2}{O_7}$ requires \[{V_2}mL\] of \[{M_2}FeS{O_4}\]. Which of the following relations is true for the above titration?
A) \[6{M_1}{V_1} = \;{M_2}{V_2}\]
B) \[{M_1}{V_1} = \;6{M_2}{V_2}\]
C) \[{M_1}{V_1} = \;{M_2}{V_2}\]
D) \[3{M_1}{V_1} = \;4{M_2}{V_2}\]

Answer 1

Hint: Write the balanced equation including sulphuric acid in the reactants and equate the ratio of calculated number of moles (Concentration x volume) of ${K_2}C{r_2}{O_7}$ and $FeS{O_4}$ to the ratio of their coefficients from the balanced chemical equation.

Complete step by step answer:
This is an oxidation-reduction titration as Potassium dichromate (${K_2}C{r_2}{O_7}$) is an oxidizing agent and oxidized Iron(II) sulphate ($FeS{O_4}$) to Iron(III) sulphate \[\left\{ {F{e_2}{{(S{O_4})}_3}} \right\}\] in the presence of sulphuric acid \[({H_2}S{O_4})\] in the aqueous medium in which ions are free to move and react.
The balanced chemical equation for this reaction is:
\[{K_2}C{r_2}{O_7}\; + 7\;{H_2}S{O_4}\; + 6\;FeS{O_4}\; \to {{\text{K}}_2}{\text{S}}{{\text{O}}_4}\; + \;{\text{C}}{{\text{r}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3}\; + 3\;{\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3}\; + 7\;{{\text{H}}_2}{\text{O}}\].
From the above balanced equation, it is clear from the coefficients of substances that for every one mole of Potassium dichromate, six moles of Iron (II) sulphate are used.
Finding the number of moles of Potassium dichromate (${K_2}C{r_2}{O_7}$)
= concentration × volume = \[{M_1}{V_1}\] ---------(i)
 Finding the number of moles of Iron (II) sulphate ($FeS{O_4}$)
= concentration × volume = \[{M_2}{V_2}\] ---------(ii)
From the balanced chemical equation:
\[\dfrac{{The\;number\;of\;moles\;of\;{K_2}C{r_2}{O_7}}}{{The\;number\;of\;moles\;of\;FeS{O_4}}} = \dfrac{1}{6}\]
The number of moles of Iron (II) sulphate $(FeS{O_4})$ = 6 x The number of moles of Potassium dichromate $({K_2}C{r_2}{O_7})$
\[ \Rightarrow {M_2}{V_2} = 6{M_1}{V_1}\;\]

Therefore, option A. is the correct answer.

Note: Do not proceed without knowing the coefficients of \[{K_2}C{r_2}{O_7}\;\] and \[FeS{O_4}\;\] from the balanced chemical equation. The calculated moles from the product of molarity and volume should be in the ratio of their coefficients in the balanced chemical equation.