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)$$\text{Normality(}{N}_1\text{)= }{\displaystyle \frac{\text{no}\text{. of gram equivalents (}Eq\text{)}}{\text{volume in litres}}}$$
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)$$Eq={\displaystyle \frac{\text{given weight}}{\text{molecular weight (}{M}_w\text{)}}}\times \text{n-factor}$$
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. $H_2{C}_2{O}_4\cdot 2H_{2}O$ $$$$
$$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
$$Eq={\displaystyle \frac{6.3}{126}}\times 2$$
$$Eq=0.1$$
Now, putting the value of $Eq$ in equation (2)
$$N_1={\displaystyle \frac{0.1}{0.250}}$$
$$N_1=0.4N$$
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.