0.5 g of NaOH is required by 0.4 gm of a polybasic acid` H_(n)A` (Molecular weight = 96 gm) for complete neutralisation .Value of 'n' would be : (Assume all H atom are replaced )-

Answer

To solve the problem, we need to find the value of 'n' for the polybasic acid \( H_nA \) that reacts with sodium hydroxide (NaOH) for complete neutralization. Here’s the step-by-step solution:

### Step 1: Understand the Concept of Gram Equivalents
In a neutralization reaction, the number of gram equivalents of the acid is equal to the number of gram equivalents of the base. The formula for calculating gram equivalents is:

\[ \text{Gram Equivalents} = \frac{\text{Mass of solute (g)}}{\text{Equivalent mass (g/equiv)}} \]

### Step 2: Calculate the Equivalent Mass of the Acid
The equivalent mass of the acid \( H_nA \) can be expressed as:

\[ \text{Equivalent mass of } H_nA = \frac{\text{Molecular weight}}{n} \]

Given that the molecular weight of \( H_nA \) is 96 g/mol, we can write:

\[ \text{Equivalent mass of } H_nA = \frac{96}{n} \]

### Step 3: Calculate the Equivalent Mass of the Base (NaOH)
The equivalent mass of NaOH is calculated as:

\[ \text{Equivalent mass of NaOH} = \frac{\text{Molecular weight}}{\text{Basicity}} = \frac{40}{1} = 40 \text{ g/equiv} \]

### Step 4: Set Up the Equation for Gram Equivalents
Using the masses given in the problem:

- Mass of \( H_nA = 0.4 \) g
- Mass of NaOH = 0.5 g

We can write the equation for the gram equivalents of the acid and the base:

\[ \frac{0.4}{\frac{96}{n}} = \frac{0.5}{40} \]

### Step 5: Simplify the Equation
Cross-multiplying gives:

\[ 0.4 \times 40 = 0.5 \times \frac{96}{n} \]

This simplifies to:

\[ 16 = \frac{48}{n} \]

### Step 6: Solve for 'n'
Rearranging the equation to solve for \( n \):

\[ n = \frac{48}{16} = 3 \]

### Conclusion
The value of \( n \) is 3, meaning the polybasic acid \( H_nA \) can donate 3 protons (H⁺ ions) during the neutralization reaction.