Question

# An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is?                   (A) 40 mL                   (B) 20 mL                   (C) 10 mL                   (D) 4 mL

Hint: First find out the normality concentration of oxalic acid. Then using the formula ${N_1}{V_1} = {N_2}{V_2}$ you can find out the volume of NaOH required for complete neutralisation.

-Normality: $N = \dfrac{{weight}}{{Eq.wt.}} \times \dfrac{{1000}}{{V(mL)}}$ (1)
Where eq.wt. is the equivalent weight.
-Oxalic acid has a molecular formula of: $HOOC - COOH$ or ${C_2}{H_2}{O_4}$. Oxalic acid dihydrate has a formula of ${C_2}{H_2}{O_4}.2{H_2}O$

Molecular weight of ${C_2}{H_2}{O_4}.2{H_2}O$ is = 90 + 18(2) = 126 g/mole
-To find the volume of NaOH required to neutralise 10 mL oxalic acid solution we will be using the following formula:
${N_1}{V_1} = {N_2}{V_2}$ (2)
Where ${N_1}$, ${V_1}$ are normality and volume for oxalic acid and ${N_2}$, ${V_2}$ are normality and volume for NaOH base.
-So, first we need to find out the normality of oxalic acid.
Since oxalic acid has 2 H atoms that it can lose to show acidic character, it means that n-factor for oxalic acid dihydrate is = 2. Its molecular weight is 126 g/mole.
Equivalent weight of oxalic acid = molecular weight / n-factor
= 126 / 2
= 63
The weight given in the question = 6.3 g
Using equation (1) we will find out the normality of oxalic acid dihydrate:
$N = \dfrac{{weight}}{{Eq.wt.}} \times \dfrac{{1000}}{{V(mL)}}$
= $\dfrac{{6.3}}{{63}} \times \dfrac{{1000}}{{250}}$
= 0.4 N
-Now we will use equation (2) to find out the volume of NaOH required for complete neutralisation.
The question gives the value of: ${N_1}$ = 0.4 N (calculated above), ${V_1}$ = 10 mL and ${N_2}$= 0.1 N. We need to find out the value of ${V_2}$.
${N_1}{V_1} = {N_2}{V_2}$
0.4×10 = 0.1×${V_2}$
${V_2} = \dfrac{{0.4 \times 10}}{{0.1}}$
= 40 mL
So, the volume of 0.1 N NaOH required to neutralise 0.4 N and 10 mL solution of oxalic acid dihydrate is 40 mL.
So, the correct answer is “Option A”.

Note: The molecular formula of oxalic acid is ${C_2}{H_2}{O_4}$ and its molecular weight is 90 g/mole while that of oxalic acid dihydrate is ${C_2}{H_2}{O_4}.2{H_2}O$ and molecular weight is 126 g/mole. Although both will have the same value of n-factor to be 2, they will have different equivalent weights due to difference in molecular weight. So while calculating the normality in this question use molecular weight to be 126 g/mole.