To neutralise completely 20 mL of 0.1M aqueous solution of phosphorus acid (`H_3PO_3`), the volume of 0.1 M aqueous KOH solution required will be ?

Options

A. 10 mL
B. 20 mL
C. 40 mL
D. 60 mL

Explanation

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

### Step 1: Write the neutralization reaction
The neutralization reaction between phosphorous acid (H₃PO₃) and potassium hydroxide (KOH) can be represented as:
\[ \text{H}_3\text{PO}_3 + \text{KOH} \rightarrow \text{K}_2\text{HPO}_3 + \text{H}_2\text{O} \]

### Step 2: Determine the number of H⁺ ions from H₃PO₃
Phosphorous acid (H₃PO₃) is a diprotic acid, meaning it can donate 2 H⁺ ions per molecule. Therefore, 1 mole of H₃PO₃ will release 2 moles of H⁺ ions.

### Step 3: Calculate the moles of H₃PO₃ in the solution
We can calculate the number of moles of H₃PO₃ in the 20 mL solution using the formula:
\[ \text{Moles of H}_3\text{PO}_3 = \text{Molarity} \times \text{Volume (in L)} \]
Given that the molarity (M) is 0.1 M and the volume (V) is 20 mL (which is 0.020 L):
\[ \text{Moles of H}_3\text{PO}_3 = 0.1 \, \text{mol/L} \times 0.020 \, \text{L} = 0.002 \, \text{mol} \]

### Step 4: Calculate the total moles of H⁺ ions
Since each mole of H₃PO₃ produces 2 moles of H⁺, the total moles of H⁺ ions will be:
\[ \text{Total moles of H}^+ = 2 \times \text{Moles of H}_3\text{PO}_3 = 2 \times 0.002 \, \text{mol} = 0.004 \, \text{mol} \]

### Step 5: Determine the moles of KOH required for neutralization
For every mole of H⁺, 1 mole of KOH is required for neutralization. Therefore, the moles of KOH required will also be:
\[ \text{Moles of KOH} = 0.004 \, \text{mol} \]

### Step 6: Calculate the volume of KOH solution needed
Using the molarity of KOH (0.1 M), we can find the volume of KOH required using the formula:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.004 \, \text{mol}}{0.1 \, \text{mol/L}} = 0.04 \, \text{L} \]
Converting this to mL:
\[ \text{Volume (mL)} = 0.04 \, \text{L} \times 1000 \, \text{mL/L} = 40 \, \text{mL} \]

### Final Answer
The volume of 0.1 M KOH solution required to completely neutralize 20 mL of 0.1 M phosphorous acid is **40 mL**.

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