Question

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.

Answer 1

`sf(0.0240color(white)(x)"mol/l")`

Explanation:

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

`sf((COOH)_2+2NaOHrarr(COONa)_2+2H_2O)`

Concentration = moles of solute/volume of solution.

`sf(c=n/v)`

`:.` `sf(n=cxxv)`

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

`sf(nNaOH=0.1050xx32.15/1000=3.3757xx10^(-3))`

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

`sf(n(COOH)_2=(3.3757xx10^(-3))/2=1.688xx10^(-3))`

Since `sf(c=n/v)` we get:

`sf([(COOH)_2]=(1.688xx10^(-3))/(25.00/1000)=0.0675color(white)(x)"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 `sf(color(red)((1)))` by 5 and x `sf(color(red)((2)))` by 2 then add both sides of each 1/2 equation together `sf(rArr)`

`sf(5C_2O_4^(2-)+2MnO_4^(-)+16H^(+)+cancel(10e)rarr2Mn^(2+)+8H_2O+10CO_2+cancel(10e))`

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

`:.` `sf(n(COOH)_2=cxxv=0.06752xx25.00/1000=1.688xx10^(-3)`

`:.` `sf(nMnO_4^(-)=1.688xx10^(-3)xx2/5=0.6752xx10^(-3))`

`sf(c=n/v)`

`:.` `sf([MnO_4^(-)]=(0.6752xx10^(-3))/(28.12/1000)=0.0240color(white)(x)"mol/l")`