Calculate the normality of a solution obtained by mixing 200 mL of 1.0 N NaOH and 100 mL of pure water.

Explanation

To calculate the normality of the solution obtained by mixing 200 mL of 1.0 N NaOH and 100 mL of pure water, follow these steps:

### Step 1: Understand the Concept of Normality
Normality (N) is defined as the number of equivalents of solute per liter of solution. For NaOH, which is a strong base, the normality is equal to its molarity because it provides one hydroxide ion (OH⁻) per molecule.

### Step 2: Identify Given Values
- Normality of NaOH (N₁) = 1.0 N
- Volume of NaOH (V₁) = 200 mL
- Volume of pure water = 100 mL

### Step 3: Calculate Total Volume of the Solution
When mixing the NaOH solution with pure water, the total volume of the solution (V₂) is:
\[ V₂ = V₁ + \text{Volume of water} = 200 \text{ mL} + 100 \text{ mL} = 300 \text{ mL} \]

### Step 4: Convert Volume to Liters
Since normality is expressed in terms of liters, convert the total volume from mL to L:
\[ V₂ = 300 \text{ mL} = 0.300 \text{ L} \]

### Step 5: Use the Normality Equation
The relationship between the normality and volume before and after dilution can be expressed as:
\[ N₁ \times V₁ = N₂ \times V₂ \]
Where:
- \( N₂ \) is the normality of the final solution.

### Step 6: Substitute the Known Values
Substituting the known values into the equation:
\[ 1.0 \, \text{N} \times 200 \, \text{mL} = N₂ \times 300 \, \text{mL} \]

### Step 7: Solve for \( N₂ \)
Rearranging the equation to solve for \( N₂ \):
\[ N₂ = \frac{1.0 \, \text{N} \times 200 \, \text{mL}}{300 \, \text{mL}} \]
Calculating:
\[ N₂ = \frac{200}{300} \, \text{N} = 0.6667 \, \text{N} \]

### Step 8: Final Result
The normality of the solution after mixing is approximately:
\[ N₂ \approx 0.67 \, \text{N} \]