Normality of `0.1 M H_(3)PO_(3)` is

Options

A. 0.2N
B. 0.30 N
C. 0.033 N
D. 0.05 N

Explanation

To find the normality of a 0.1 M solution of H₃PO₃ (phosphorous acid), we can follow these steps:

### Step 1: Understand the relationship between Normality and Molarity
Normality (N) is related to molarity (M) through the equation:
\[ N = M \times n \]
where \( n \) is the number of equivalents or the N factor.

### Step 2: Determine the N factor for H₃PO₃
For acids, the N factor is equal to the basicity of the acid, which is defined as the number of replaceable hydrogen ions (H⁺) in the acid.

### Step 3: Analyze the structure of H₃PO₃
H₃PO₃ has the following structure:
- It contains three hydrogen atoms, but not all of them are replaceable.
- The two hydrogen atoms that are directly connected to the oxygen atoms (–OH groups) are the replaceable ones.
- The hydrogen atom connected to phosphorus is not replaceable due to the lower electronegativity difference between phosphorus and hydrogen.

### Step 4: Count the replaceable hydrogen atoms
In H₃PO₃:
- There are 2 replaceable hydrogen atoms (from the two –OH groups).
- Therefore, the basicity of H₃PO₃ is 2.

### Step 5: Calculate the N factor
Since the basicity is 2, the N factor for H₃PO₃ is:
\[ n = 2 \]

### Step 6: Substitute values into the Normality formula
Now, we can substitute the values into the normality formula:
\[ N = M \times n \]
\[ N = 0.1 \, \text{M} \times 2 \]
\[ N = 0.2 \, \text{N} \]

### Conclusion
The normality of the 0.1 M H₃PO₃ solution is:
\[ \text{Normality} = 0.2 \, \text{N} \]