Question

A titration is carried out using $0.010MKMn{O}_4\left(aq\right)$, in the presence of excess sulfuric acid, to determine the concentration of iron(II) ion in a solution of iron (II) sulfate, $FeS{O}_4$. The balanced equation for the reaction is
$2KMn{O}_4\left(aq\right)+10FeS{O}_4\left(aq\right)+8H_{2}S{O}_4\left(aq\right)\to 5F{e}_2\left(S{O}_4{)}_3\right(aq)+2MnS{O}_4(aq)+K_{2}S{O}_4(aq)$
$+8H_{2}O\left(l\right)$
If $25mL$ of the $FeS{O}_4$ solution requires $37mL$ of the permanganate titrant, which of the following expressions CORRECTLY gives the concentration of $F{e}^{2+}$, in mol/L, in the iron(II) sulfate solution?
(The molar masses of the reactants and products are: $KMn{O}_4,158g/mol;FeS{O}_4,152g/mol;F{e}_2(S{O}_4{)}_3,400g/mol$; and $MnS{O}_4,151g/mol$.)

• A. $\frac{0.037\times 0.010\times 5}{0.025}$
• B. $\frac{0.025\times 0.010\times 5}{0.037}$
• C. $\frac{0.025\times 5}{0.037\times 0.010}$
• D. $\frac{0.037\times 5\times 400}{158\times 0.025}$

Answer 1

The correct option is A $\frac{0.037\times 0.010\times 5}{0.025}$

$KMn{O}_4\left(aq\right)+10FeS{O}_4\left(aq\right)+8H_{2}S{O}_4\left(aq\right)⟶5F{e}_2\left(S{O}_4{)}_3\right(aq)+2MnS{O}_4(aq)+K_{2}S{O}_4(aq)+8H_{2}O(l)$

Now, $\left[KMn{O}_4\right]\equiv \left[FeS{O}_4\right]\times \frac{2}{10}$

Now, $\left[KMn{O}_4\right]=0.01M$ and $V_{KMn{O}_4}=37ml$

$\therefore \left[FeS{O}_4\right]\times V_{FeS{O}_4}=\left[KMn{O}_4\right]\times V_{KMn{O}_4}$

$\Rightarrow \left[FeS{O}_4\right]=\frac{10}{2}\times \frac{0.01\times 37}{25}=\frac{0.037\times 0.010\times 5}{0.025}$ .