How much of NaOH is required to neutralize 1500 `cm^(3)` of 0.1 NHCl? (Na=23)

Options

A. 40g
B. 4g
C. 6g
D. 60g

Explanation

To solve the problem of how much NaOH is required to neutralize 1500 cm³ of 0.1 N HCl, we can follow these steps:

### Step 1: Understand the reaction
The neutralization reaction between NaOH and HCl can be written as:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
From this reaction, we can see that 1 mole of NaOH reacts with 1 mole of HCl.

### Step 2: Calculate the number of equivalents of HCl
Normality (N) is defined as the number of equivalents per liter. Therefore, we can calculate the number of equivalents of HCl in the given volume.

Given:
- Normality of HCl (N) = 0.1 N
- Volume of HCl = 1500 cm³ = 1.5 L (since 1000 cm³ = 1 L)

The number of equivalents of HCl can be calculated using the formula:
\[ \text{Equivalents of HCl} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Equivalents of HCl} = 0.1 \, \text{N} \times 1.5 \, \text{L} = 0.15 \, \text{equivalents} \]

### Step 3: Calculate the amount of NaOH required
Since the reaction between NaOH and HCl is a 1:1 ratio, the number of equivalents of NaOH required will also be 0.15 equivalents.

### Step 4: Calculate the equivalent weight of NaOH
The equivalent weight of NaOH can be calculated as follows:
- Molar mass of NaOH = Na (23) + O (16) + H (1) = 40 g/mol
- Since NaOH is a strong base and provides one hydroxide ion (OH⁻) per molecule, its equivalent weight is also 40 g.

### Step 5: Calculate the mass of NaOH required
Using the equivalent weight, we can find the mass of NaOH required:
\[ \text{Mass of NaOH} = \text{Number of equivalents} \times \text{Equivalent weight} \]
\[ \text{Mass of NaOH} = 0.15 \, \text{equivalents} \times 40 \, \text{g/equivalent} = 6 \, \text{g} \]

### Final Answer
The mass of NaOH required to neutralize 1500 cm³ of 0.1 N HCl is **6 grams**.

---