To neutralize completely 20 mL of 0.1 M aqueous solution of phosphorus acid, the volume of 0.1 M aqueous `NaOH` solution required is

Options

A. 20 mL
B. 30 mL
C. 40 mL
D. 60 mL

Answer

To find the volume of 0.1 M NaOH solution required to completely neutralize 20 mL of 0.1 M phosphorous acid (H₃PO₃), we can follow these steps:

### Step 1: Understand the Neutralization Reaction
Phosphorous acid (H₃PO₃) is a diprotic acid, meaning it can donate two protons (H⁺ ions) in a reaction. The balanced neutralization reaction with sodium hydroxide (NaOH) can be represented as:
\[ \text{H}_3\text{PO}_3 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{HPO}_3 + 2 \text{H}_2\text{O} \]

### Step 2: Calculate the Moles of H₃PO₃
To find the moles of H₃PO₃ in the solution, we use the formula:
\[ \text{Moles of H}_3\text{PO}_3 = \text{Volume (L)} \times \text{Concentration (M)} \]
Given:
- Volume = 20 mL = 0.020 L
- Concentration = 0.1 M

Calculating the moles:
\[ \text{Moles of H}_3\text{PO}_3 = 0.020 \, \text{L} \times 0.1 \, \text{mol/L} = 0.002 \, \text{mol} \]

### Step 3: Determine the Moles of NaOH Required
Since 1 mole of H₃PO₃ reacts with 2 moles of NaOH, the moles of NaOH required will be:
\[ \text{Moles of NaOH} = 2 \times \text{Moles of H}_3\text{PO}_3 = 2 \times 0.002 \, \text{mol} = 0.004 \, \text{mol} \]

### Step 4: Calculate the Volume of NaOH Required
Now, we can calculate the volume of NaOH solution needed to provide 0.004 moles. We use the formula:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Concentration (M)}} \]
Given:
- Moles of NaOH = 0.004 mol
- Concentration of NaOH = 0.1 M

Calculating the volume:
\[ \text{Volume of NaOH} = \frac{0.004 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.04 \, \text{L} = 40 \, \text{mL} \]

### Conclusion
The volume of 0.1 M NaOH solution required to completely neutralize 20 mL of 0.1 M phosphorous acid is **40 mL**.

---