Volume of 0.1 M NaOH needed for the neutralisation of 20 mL of 0.05 M oxalic acid is

Options

A. 10 mL
B. 15 mL
C. 20 mL
D. 30 mL

Answer

To find the volume of 0.1 M NaOH needed to neutralize 20 mL of 0.05 M oxalic acid, we can follow these steps:

### Step 1: Write the neutralization reaction
Oxalic acid (H₂C₂O₄) is a diprotic acid, meaning it can donate two protons (H⁺). The balanced neutralization reaction with sodium hydroxide (NaOH) is:
\[ \text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \]

### Step 2: Calculate the moles of oxalic acid
Using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]
Convert the volume of oxalic acid from mL to L:
\[ 20 \, \text{mL} = 0.020 \, \text{L} \]
Now calculate the moles of oxalic acid:
\[ \text{Moles of H}_2\text{C}_2\text{O}_4 = 0.05 \, \text{M} \times 0.020 \, \text{L} = 0.001 \, \text{moles} \]

### Step 3: Determine the moles of NaOH required
From the balanced equation, we see that 1 mole of oxalic acid reacts with 2 moles of NaOH. Therefore, the moles of NaOH required are:
\[ \text{Moles of NaOH} = 2 \times \text{Moles of H}_2\text{C}_2\text{O}_4 = 2 \times 0.001 = 0.002 \, \text{moles} \]

### Step 4: Calculate the volume of NaOH needed
Using the formula:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]
Substituting the values:
\[ \text{Volume of NaOH} = \frac{0.002 \, \text{moles}}{0.1 \, \text{M}} = 0.02 \, \text{L} \]
Convert this volume back to mL:
\[ 0.02 \, \text{L} = 20 \, \text{mL} \]

### Final Answer
The volume of 0.1 M NaOH needed for the neutralization of 20 mL of 0.05 M oxalic acid is **20 mL**.