Question

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

Answer 1

Hint : To solve this problem we need to think about how the normality and volumes of the acid and base involved in a titration are equated.

Complete step by step solution :
The normality and volumes of the acid and base involved in any titration are related through the following formula:

(1)${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$
Where,
${{N}_{1}}$ = normality of substance 1 (acid)
${{V}_{1}}$ = volume of substance 1(acid)
${{N}_{2}}$= normality of substance 2 (base)
${{V}_{2}}$ = volume of substance 2 (base)
In this question, we have been given the following:
${{V}_{1}}$ = 10mL (oxalic acid)
${{N}_{2}}$= 0.1N (sodium hydroxide)
We have to find ${{V}_{2}}$ i.e., the volume of $NaOH$ required to completely neutralize the given solution of oxalic acid.
For this we will need ${{N}_{1}}$= normality of oxalic acid
We can find that easily if we use the rest of the given information
We know that,

(2)$\text{Normality(}{{N}_{1}}\text{)= }\dfrac{\text{no}\text{. of gram equivalents (}Eq\text{)}}{\text{volume in litres}}$
Here, note that the volume of the oxalic acid solution is given to us i.e. 250mL
The formula to calculate the number of gram equivalents is
$Eq\text{ = number of moles(}n\text{)}\times \text{n-factor}$

(3)$Eq=\dfrac{\text{given weight}}{\text{molecular weight (}{{M}_{w}}\text{)}}\times \text{n-factor}$
Here, note that the weight of oxalic acid given in the question is 6.3g.
To find the molecular weight consider the molecular formula of oxalic acid i.e. ${{H}_{2}}{{C}_{2}}{{O}_{4}}\cdot 2{{H}_{2}}O$ $$$$
${{M}_{w}}=(6\times \text{atomics weight of }H)+(2\times \text{atomic weight of }C)+(6\times \text{atomic weight of }O)$
${{M}_{w}}=(6\times 1)+(2\times 12)+(6\times 16)$
${{M}_{w}}=6+24+96$
${{M}_{w}}=126$
To find the n-factor, we need to consider how many acidic protons oxalic acid has, or how many $O{{H}^{-}}$ions can 1 molecule of oxalic acid neutralize.
The answer to this is 2, since oxalic acid has the structure $HOOC-COOH$ where the hydrogens at both ends of the carboxylic group is acidic. Therefore, the n-factor of oxalic acid is 2.

Now, plugging these values in equation (3), we get
$Eq=\dfrac{6.3}{126}\times 2$
$Eq=0.1$
Now, putting the value of $Eq$ in equation (2)
${{N}_{1}}=\dfrac{0.1}{0.250}$
${{N}_{1}}=0.4N$
Now putting this value in equation (1), we get
$0.4N\times 10mL=0.1N\times {{V}_{2}}$
${{V}_{2}}=40mL$

Thus, the correct answer is ‘A. 40mL’

Note : Please remember to convert the millilitres into litres while calculating the normality. Also, do not get confused between molarity and normality, always verify the n-factor before calculating the normality.