What weight of sodium hydroxide is required to neutralise 100 ml of 0.1 N HCl?

Options

A. 4g
B. 0.4g
C. 0.04g
D. 40g

Explanation

To determine the weight of sodium hydroxide (NaOH) required to neutralize 100 ml of 0.1 N hydrochloric acid (HCl), we can follow these steps:

### Step 1: Calculate the number of equivalents of HCl
Normality (N) is defined as the number of equivalents of solute per liter of solution. For HCl, which is a strong acid, the number of equivalents is equal to the number of moles since it can donate one H⁺ ion per molecule.

Given:
- Volume of HCl = 100 ml = 0.1 L
- Normality of HCl = 0.1 N

The number of equivalents of HCl can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Number of equivalents of HCl} = 0.1 \, \text{N} \times 0.1 \, \text{L} = 0.01 \, \text{equivalents} \]

### Step 2: Determine the number of equivalents of NaOH required
Since the reaction between HCl and NaOH is a 1:1 reaction (one equivalent of NaOH neutralizes one equivalent of HCl), the number of equivalents of NaOH required will also be 0.01 equivalents.

### Step 3: Calculate the number of moles of NaOH required
To find the number of moles of NaOH, we use the relationship between equivalents and moles. For NaOH, which also has one hydroxide ion (OH⁻) that can react with one H⁺ ion, the number of equivalents is equal to the number of moles.

Thus, the number of moles of NaOH needed is:
\[ \text{Number of moles of NaOH} = \text{Number of equivalents of NaOH} = 0.01 \, \text{moles} \]

### Step 4: Calculate the weight of NaOH required
To find the weight of NaOH, we use the formula:
\[ \text{Weight} = \text{Number of moles} \times \text{Molar mass} \]
The molar mass of NaOH is approximately 40 g/mol.

Thus, the weight of NaOH required is:
\[ \text{Weight of NaOH} = 0.01 \, \text{moles} \times 40 \, \text{g/mol} = 0.4 \, \text{g} \]

### Final Answer
The weight of sodium hydroxide required to neutralize 100 ml of 0.1 N HCl is **0.4 grams**.

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