# What is the molarity of the potassium permanganate solution?

## In order to standardize an oxalic acid solution, its exact concentration is determined by an acid-base titration. Then, the oxalic acid solution is used to determine the concentration of a potassium permanganate solutionby a redox titration. The titration of 25.00 ml samples of the oxalic acid solution requires 32.15 ml of 0.1050M sodium hydroxide and 28.12 ml of the potassium permanganate solution.

Aug 1, 2016

$\textsf{0.0240 \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

Oxalic (ethandioic) acid is diprotic and can be standardised by titrating against sodium hydroxide solution of known concentration:

$\textsf{{\left(C O O H\right)}_{2} + 2 N a O H \rightarrow {\left(C O O N a\right)}_{2} + 2 {H}_{2} O}$

Concentration = moles of solute/volume of solution.

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

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

So we can get the no. of moles of sodium hydroxide:

$\textsf{n N a O H = 0.1050 \times \frac{32.15}{1000} = 3.3757 \times {10}^{- 3}}$

From the equation we can see from the mole ratio that the number of moles of oxalic acid is given by:

$\textsf{n {\left(C O O H\right)}_{2} = \frac{3.3757 \times {10}^{- 3}}{2} = 1.688 \times {10}^{- 3}}$

Since $\textsf{c = \frac{n}{v}}$ we get:

$\textsf{\left[{\left(C O O H\right)}_{2}\right] = \frac{1.688 \times {10}^{- 3}}{\frac{25.00}{1000}} = 0.0675 \textcolor{w h i t e}{x} \text{mol/l}}$

Note I have converted ml into litre by dividing by 1000.

Now a redox titration is carried out to determine the concentration of the manganate(VII) solution.

The 1/2 equations are:

sf(C_2O_4^(2-)rarr2CO_2+2e" "color(red)((1))

sf(MnO_4^(-)+8H^(+)+5erarrMn^(2+)+4H_2O" "color(red)((2))

To get the electrons to balance we need to x $\textsf{\textcolor{red}{\left(1\right)}}$ by 5 and x $\textsf{\textcolor{red}{\left(2\right)}}$ by 2 then add both sides of each 1/2 equation together $\textsf{\Rightarrow}$

$\textsf{5 {C}_{2} {O}_{4}^{2 -} + 2 M n {O}_{4}^{-} + 16 {H}^{+} + \cancel{10 e} \rightarrow 2 M {n}^{2 +} + 8 {H}_{2} O + 10 C {O}_{2} + \cancel{10 e}}$

This tells us that 2 moles of Mn(VII) react with 5 moles of oxalic acid.

$\therefore$sf(n(COOH)_2=cxxv=0.06752xx25.00/1000=1.688xx10^(-3)

$\therefore$$\textsf{n M n {O}_{4}^{-} = 1.688 \times {10}^{- 3} \times \frac{2}{5} = 0.6752 \times {10}^{- 3}}$

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

$\therefore$$\textsf{\left[M n {O}_{4}^{-}\right] = \frac{0.6752 \times {10}^{- 3}}{\frac{28.12}{1000}} = 0.0240 \textcolor{w h i t e}{x} \text{mol/l}}$