Questions & Answers

Question

Answers

(A) 40 mL

(B) 20 mL

(C) 10 mL

(D) 4 mL

Answer
Verified

-Normality: $N = \dfrac{{weight}}{{Eq.wt.}} \times \dfrac{{1000}}{{V(mL)}}$ (1)

Where eq.wt. is the equivalent weight.

Complete step by step answer:

-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.