Question

The concentration of oxalic acid is 'X' $\text{mol li}{{\text{t}}^{\text{-1}}}$. 40mL of this solution reacts with 16mL of 0.05 M acidified $\text{KMn}{{\text{O}}_{\text{4}}}$. What is the pH of 'X' M oxalic acid solution?
[Assume that oxalic acid dissociates completely]
(A) 1.3
(B) 1.699
(C) 1
(D) 2

Answer 1

Hint: The above reaction is not an acid base neutralization reaction. It is an oxidation-reduction process. We will calculate milliequivalents of $\text{KMn}{{\text{O}}_{\text{4}}}$ and then equate it to the milli equivalences of oxalic acid to obtain the concentration of oxalic acid used. Remember, the n-factor for oxalic acid is 2.

Complete step by step answer:
A neutralization reaction is when an acid and a base react to form salt and water as the products. It involves the combination of ${{\text{H}}^{+}}$ions and $\text{O}{{\text{H}}^{-}}$ ions to generate water.
The neutralization reaction of a strong acid and a strong base result in complete neutralization and the pH is 7.

In the same manner redox reactions are carried out. In the above question there is a redox reaction between $\text{KMn}{{\text{O}}_{\text{4}}}$and oxalic acid. The complete reaction is given below:
$\text{2Mn}{{\text{O}}_{\text{4}}}^{-}\text{ + 5}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}^{2-}\text{ + 16}{{\text{H}}^{\text{+}}}\text{ }\to \text{ 10C}{{\text{O}}_{\text{2}}}\text{ + M}{{\text{n}}^{\text{2+}}}\text{ + 8}{{\text{H}}_{\text{2}}}\text{O}$

An equivalent is the amount of substance that is equivalent to or reacts with an arbitrary amount of another substance in a given chemical reaction.
Alternatively equivalent is defined as the number of moles of an ion in a solution, multiplied by the valence of that ion. Milli equivalence is the equivalence of a substance when the volume is taken in milliliters.

Equivalence is also equal to the product of molarity of the solution and n-factor of the molecule.
The formula for milli equivalence is given below:
$\text{mEq = N x V}$
Where,
mEq is the milliequivalents of acid/base,
N is the normality of the solution,
V is the volume of acid/base (in mL).

We will now substitute the values given in the question to find the concentration of oxalic acid used.
Volume of oxalic acid taken = 40 mL
Volume of $\text{KMn}{{\text{O}}_{\text{4}}}$ taken = 16 mL
Concentration of oxalic acid = X
Concentration of $\text{KMn}{{\text{O}}_{\text{4}}}$ = 0.05 M
n-factor of oxalic acid = 2
n-factor of $\text{KMn}{{\text{O}}_{\text{4}}}$ = 5

The milli equivalence of $\text{KMn}{{\text{O}}_{\text{4}}}$ is:
$\text{mE}{{\text{q}}_{\text{KMn}{{\text{O}}_{\text{4}}}}}\text{ = 0}\text{.05 x 5 x 16}$ = 4
We will now equate this to the milli equivalence of oxalic acid:
4 = $\text{40 x 2 x X}$
X = 0.05 M

This means that 0.05 moles of oxalic acid are present in 1L of solution.
$\text{ }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ = 2 x 0}\text{.05 = 0}\text{.1}$
pH = -log($\text{ }\!\![\!\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\!\!]\!\!\text{ }$) = 1
So, the correct answer is “Option C”.

Note: n-factor stands for the number of electrons gained or lost by one mole of reactant. For acids, n -factor is the number of ${{H}^{+}}$ ions that can be replaced in a neutralisation reaction. At time n - factor is equal to the valence of the atom, however it does not hold true when an atom shows variable valency.