JEE Main 2019 — Redox Reactions Question with Solution

The reaction between sodium hydroxide (NaOH) and oxalic acid (H₂C₂O₄) is as follows :

$${\text{H}}_2{\text{C}}_2{\text{O}}_4+2\text{NaOH}\to {\text{Na}}_2{\text{C}}_2{\text{O}}_4+2{\text{H}}_2\text{O}$$

We can see that 1 mole of oxalic acid (${\text{H}}_2{\text{C}}_2{\text{O}}_4$) reacts with 2 moles of sodium hydroxide (NaOH).

Given that 50 mL of 0.5 M oxalic acid is needed to neutralize the sodium hydroxide, we can find the number of moles of oxalic acid that reacted :

$${\text{Moles of H}}_2{\text{C}}_2{\text{O}}_4=\text{Molarity}\times \text{Volume (L)}=0.5\text{mol/L}\times 0.05\text{L}=0.025\text{mol}$$

Since 1 mole of ${\text{H}}_2{\text{C}}_2{\text{O}}_4$ reacts with 2 moles of $\text{NaOH}$, the number of moles of $\text{NaOH}$ in 25 mL solution would be twice that of ${\text{H}}_2{\text{C}}_2{\text{O}}_4$ :

$$\text{Moles of NaOH}=0.025\text{mol}\times 2=0.05\text{mol}$$

Now, to find the mass of NaOH, we multiply the number of moles by the molar mass of NaOH (40 g/mol) :

$$\text{Mass of NaOH}=\text{Number of moles}\times \text{Molar mass}=0.05\text{mol}\times 40\text{g/mol}=2\text{g}$$

However, the question asks for the amount of $\text{NaOH}$ in 50 mL of the given solution. Since we've found the amount in 25 mL, we just need to double our result to find the amount in 50 mL :

$$2\text{g}\times 2=4\text{g}$$

So, the correct answer is Option B : 4 g.