How much of NaOH is reuired to neutralise 1500 `cm^(3)` of 0.1 N HCl (Na=23)?

Options

A. (a)40 g
B. (b)4 g
C. (c )6 g
D. (d)60 g

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 Neutralization Reaction
The neutralization reaction between NaOH and HCl can be represented 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 of solute per liter of solution. Given that the normality of HCl is 0.1 N and the volume is 1500 cm³ (or 1.5 L), we can calculate the number of equivalents of HCl:
\[ \text{Number of equivalents of HCl} = \text{Normality} \times \text{Volume in liters} \]
\[ = 0.1 \, \text{N} \times 1.5 \, \text{L} = 0.15 \, \text{equivalents} \]

### Step 3: Determine the Number of Equivalents of NaOH Required
Since the reaction shows a 1:1 ratio between NaOH and HCl, the number of equivalents of NaOH required to neutralize HCl will also be 0.15 equivalents.

### Step 4: Calculate the Molecular Weight of NaOH
The molecular weight of NaOH can be calculated as follows:
- Sodium (Na) = 23 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol
\[ \text{Molecular weight of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol} \]

### Step 5: Calculate the Weight of NaOH Required
To find the weight of NaOH needed for 0.15 equivalents, we use the formula:
\[ \text{Weight} = \text{Number of equivalents} \times \text{Equivalent weight} \]
The equivalent weight of NaOH is equal to its molecular weight since it provides one hydroxide ion (OH⁻) per mole:
\[ \text{Equivalent weight of NaOH} = \text{Molecular weight of NaOH} = 40 \, \text{g/equiv} \]
Thus,
\[ \text{Weight of NaOH} = 0.15 \, \text{equivalents} \times 40 \, \text{g/equiv} = 6 \, \text{g} \]

### Conclusion
To neutralize 1500 cm³ of 0.1 N HCl, **6 grams of NaOH** is required.

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