# If 26.23 mL of potassium permanganate solution is required to titrate 1.041 g of ferrous ammonium sulfate hexahydrate, FeSO_4(NH_4)_2SO_4 * 6H_2O, chow do you calculate the molarity of the KMnO_4 solution.?

Feb 21, 2017

You can do it like this:

#### Explanation:

We have 1.041 g of ferrous ammonium sulfate hexahydrate.

The $\textsf{{M}_{r} = 392.14}$

$\therefore$$\textsf{{n}_{F {e}^{2 +}} = \frac{m}{M} _ \left(r\right) = \frac{1.041}{392.14} = 0.002654 \textcolor{w h i t e}{x} \text{mol}}$

Set up the equation from the two 1/2 equations:

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

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

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

$\textsf{M n {O}_{4}^{-} + 8 {H}^{+} + \cancel{5 e} + 5 F {e}^{2 +} \rightarrow M {n}^{2 +} + 4 {H}_{2} O + 5 F {e}^{3 +} + \cancel{5 e}}$

This tells that 5 moles of $\textsf{F {e}^{2 +}}$ react with 1 mole of $\textsf{M n {O}_{4}^{-}}$

$\therefore$$\textsf{{n}_{M n {O}_{4}^{-}} = \frac{0.002654}{5} = 5.3093 \times {10}^{- 4}}$

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

$\therefore$$\textsf{\left[M n {O}_{4}^{-}\right] = \frac{5.3093 \times {10}^{- 4}}{\frac{26.23}{1000}} = 0.02024 \textcolor{w h i t e}{x} \text{mol/l}}$