Question

If 50 mL of 0.5 M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is _____g.

Answer 1

Since the neutralization reaction occurs, the equivalents of oxalic acid will be equal to the equivalents of NaOH.

- Volume of oxalic acid = 50 mL
- Molarity of oxalic acid = 0.5 M
- Volume of NaOH solution = 25 mL

Step 1. Calculate the equivalents of oxalic acid:
$\text{Equivalents of oxalic acid} = \text{Volume} \times \text{Molarity} \times \text{Basicity of oxalic acid}$
$= 50 \times 0.5 \times 2 = 25 \, \text{meq}$

Step 2. Since the equivalents of NaOH are equal to the equivalents of oxalic acid, we have:
$\text{Equivalents of NaOH} = 25 \, \text{meq}$

Step 3. Molarity of NaOH:
$M_{\text{NaOH}} = 2 \, \text{M}$

Step 4. Calculate the weight of NaOH in 50 mL:
$W_{\text{NaOH}} = \text{Normality} \times \text{Volume} \times \text{Molar mass of NaOH}$
$= 2 \times 50 \times 40 \times 10^{-3} = 4 \, \text{g}$

The Correct Answer is: 4g