Question

Calculate the mass of anhydrous oxalic acid, which can be oxidized to $C{O_2}$ (g) by 100ml of an $MnO_4^ -$ solution, 10ml of which is capable of oxidizing 50ml of 1N ${I^ - }$ to ${I_2}$:

(A) 45 gm
(B) 22.5 gm
(C) 30 gm
(D) 12.25 gm

Answer 1

First calculate the normality of the $MnO_4^ -$ solution by comparing it with the ${I^ - }$ solution values. Then compare it with the solution of anhydrous oxalic acid and find out the gram equivalent of oxalic acid involved. Then using the gram equivalent of oxalic acid find out the weight of oxalic acid which will be oxidised.

Complete step by step solution:

-First of all we will calculate the normality of the $MnO_4^ -$ solution which is used to neutralise the anhydrous oxalic acid solution.

To calculate normality we will use the following equation
${N_1}{V_1} = {N_2}{V_2}$ (1)

Here we will be comparing the N and V of $MnO_4^ -$ to the N and V of ${I^ - }$ it oxidises.
So, ${N_1}$ = normality of $MnO_4^ -$ (which we will be calculating)
${V_1}$ = Volume of $MnO_4^ -$ used to oxidize ${I^ - }$ = 10 ml
${N_2}$ = normality of ${I^ - }$ = 1N
${V_2}$ = Volume of ${I^ - }$ oxidized = 50 ml

Now putting these values in equation (1):
${N_1} \times 10 = 1 \times 50$
${N_1}$ = 5N
Hence the normality of the given $MnO_4^ -$ solution will be 5N.

-We now know that 100 ml (0.1 L) of 5N $MnO_4^ -$ will be oxidising some given amount of anhydrous oxalic acid.
Using equation (1) where ${N_1}$ and ${V_1}$ are for $MnO_4^ -$ and ${N_2}$ and ${V_2}$ are for the anhydrous oxalic acid.
${N_1}{V_1} = {N_2}{V_2}$
5 × 0.1 = ${N_2}{V_2}$ (2)
We also know that: $N = \dfrac{{g.equivalent}}{{V(l)}}$; and so :
N × V(L) = g.equivalent

Hence, ${N_2}{V_2}$ = gram equivalent of the given anhydrous oxalic acid.
From equation (2): 5 × 0.1 = ${N_2}{V_2}$
${N_2}{V_2}$ = 0.5 gram equivalent of oxalic acid (3)

- The molecular weight of oxalic acid ($HOOC - COOH$) is 90 g/mol and its n-factor is 2.
So, we will first calculate the equivalent weight of oxalic acid using the following formula:
$Eq.wt. = \dfrac{{M.wt.}}{{n - factor}}$
= $\dfrac{{90}}{2}$ = 45

-We also know that the gram equivalent of any substance is equal to the ratio of given weight of the substance to its equivalent weight. Mathematically it can be written a
$g.equivalent = \dfrac{W}{{Eq.wt.}}$ (4)

-Now we will put the values of gram equivalent of oxalic acid from (3) and its equivalent weight in equation (4):
$g.equivalent = \dfrac{W}{{Eq.wt.}}$
0.5 = $\dfrac{W}{{45}}$
W = 0.5 × 45
= 22.5 g
Hence, 22.5 g of anhydrous oxalic acid will be oxidised to $C{O_2}$ by the given amount of $MnO_4^ -$ solution.

So, the correct option will be: (B) 22.5 gm

Additional information: Normality is sometimes used in place of molarity because often 1 mole of acid does not completely neutralize 1 mole of base. Hence, in order to have a one-to-one relationship between acids and bases, many chemists prefer to express the concentration of acids and bases in normality.

Note: The normal concentration of a solution or normality is always equal to or greater than the molar concentration or molarity of a solution. The normal concentration can be calculated by multiplying the molar concentration by the number of equivalents per mole of solute.