How do you solve this titration calculation?

35.0 g of hydrated iron(II) ammonium sulfate $\textsf{F e {\left(N {H}_{4}\right)}_{2} {\left(S {O}_{4}\right)}_{2} .6 {H}_{2} O}$ is dissolved in 1 litre of water. 25.00 ml is added to a flask and acidified. When titrated against $\textsf{K M n {O}_{4}}$ solution an average titre of 21.00 ml was recorded. What is the concentration of the Mn(VII) solution in g/l ?

Jun 10, 2017

You can do it like this:

Explanation:

$\textsf{M n {O}_{4 \left(a q\right)}^{-} + 8 {H}_{\left(a q\right)}^{+} + 5 e \rightarrow M {n}_{\left(a q\right)}^{2 +} + 4 {H}_{2} {O}_{\left(l\right)} \text{ } \textcolor{red}{\left(1\right)}}$

$\textsf{F {e}_{\left(a q\right)}^{2 +} \rightarrow F {e}_{\left(a q\right)}^{3 +} + e \text{ } \textcolor{red}{\left(2\right)}}$

To get the electrons to balance we multiply $\textsf{\textcolor{red}{\left(2\right)}}$ by 5 and add to $\textsf{\textcolor{red}{\left(1\right)} \Rightarrow}$

$\textsf{M n {O}_{4 \left(a q\right)}^{-} + 8 {H}_{\left(a q\right)}^{+} + \cancel{5 e} + 5 F {e}_{\left(a q\right)}^{+} \rightarrow M {n}_{\left(a q\right)}^{2 +} + 4 {H}_{2} {O}_{\left(l\right)} + 5 F {e}_{\left(a q\right)}^{2 +} + \cancel{5 e}}$

This tells us that 1 mole Mn(VII) reacts with 5 moles Fe(II).

The $\textsf{{M}_{r}}$ of the iron(II) salt is 392.13

The no. moles of iron(II) in 1000 ml is therefore $\textsf{\frac{35.0}{392.13} = 0.0892}$

$\therefore$$\textsf{\left[F {e}^{2 +}\right] = 0.0892 \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{n = c \times v}$

$\therefore$$\textsf{{n}_{F {e}^{2 +}} = 0.0892 \times \frac{25.00}{1000} = 2.231 \times {10}^{- 3}}$

From the mole ratio we can say that:

$\textsf{{n}_{M n {O}_{4}^{-}} = \frac{2.231 \times {10}^{-} 3}{5} = 0.4463 \times {10}^{- 3}}$

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{\left[M n {O}_{4}^{-}\right] = \frac{0.4463 \times {10}^{- 3}}{0.021} = 0.0212 \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{{M}_{r} \left[K M n {O}_{4}\right] = 158.034}$

In grams/litre the concentration is given by:

$\textsf{c = 0.0212 \times 158.034 = 3.35 \textcolor{w h i t e}{x} \text{g/l}}$

This is equal to 0.335 g/100 ml or 0.33% (w/v) to 2 sig fig.