0.4 g of polybasic acid HnA (all the hydrogens are acidic) requries 0.5 g of NaOH for complete neutralisation. The number of replaceable hydrogen atoms and the molecular weight of A would be (Mw of `acid=96)`

Options

A. `2,94`
B. `1,95`
C. `3,93`
D. `4,92`

Explanation

To solve the problem step by step, we need to determine the number of replaceable hydrogen atoms (N) in the polybasic acid \( H_nA \) and the molecular weight of the component \( A \).

### Step 1: Understand the neutralization reaction
In a neutralization reaction, the moles of acid will equal the moles of base at the equivalence point. We have:
- Mass of polybasic acid \( H_nA = 0.4 \, \text{g} \)
- Mass of NaOH = \( 0.5 \, \text{g} \)

### Step 2: Calculate the normality of the acid
The normality (N) of an acid can be calculated using the formula:
\[ N = \frac{\text{Weight}}{\text{Molar Mass} \times n} \]
Where:
- Weight = mass of the acid
- Molar Mass = 96 g/mol (given)
- \( n \) = number of replaceable hydrogen atoms (which we need to find)

Thus, for the acid:
\[ N_{acid} = \frac{0.4}{96 \times n} \]

### Step 3: Calculate the normality of the base
For NaOH, the normality can be calculated similarly:
\[ N_{base} = \frac{\text{Weight}}{\text{Molar Mass} \times n} \]
Where:
- Weight = mass of NaOH = 0.5 g
- Molar Mass of NaOH = 40 g/mol
- \( n \) = 1 (since NaOH can release one OH⁻ ion)

Thus, for the base:
\[ N_{base} = \frac{0.5}{40 \times 1} = \frac{0.5}{40} = 0.0125 \, \text{N} \]

### Step 4: Set up the equivalence equation
Since the moles of acid equal the moles of base at neutralization:
\[ N_{acid} \times V_{acid} = N_{base} \times V_{base} \]
Assuming both volumes are 1 L, we can set:
\[ \frac{0.4}{96 \times n} = \frac{0.5}{40} \]

### Step 5: Solve for \( n \)
Cross-multiplying gives:
\[ 0.4 \times 40 = 0.5 \times 96 \times n \]
\[ 16 = 48n \]
\[ n = \frac{16}{48} = \frac{1}{3} \]

### Step 6: Calculate the molecular weight of \( A \)
The total molecular weight of the acid \( H_nA \) is given as 96 g/mol. Since \( n = 3 \), the contribution of the hydrogen atoms is \( 3 \times 1 = 3 \) g/mol. Therefore, the molecular weight of \( A \) is:
\[ \text{Molecular weight of } A = 96 - 3 = 93 \, \text{g/mol} \]

### Final Answers
- Number of replaceable hydrogen atoms \( n = 3 \)
- Molecular weight of \( A = 93 \, \text{g/mol} \)