Calculate the normality of a solution obtained by mixing 100 mL of 0.2 N KOH and 100 mL of 0.1 `MH_(2)SO_(4)`.

Explanation

To calculate the normality of the solution obtained by mixing 100 mL of 0.2 N KOH and 100 mL of 0.1 M H₂SO₄, we can follow these steps:

### Step 1: Understand the reaction
KOH is a strong base that dissociates completely in solution to give K⁺ and OH⁻ ions. H₂SO₄ is a strong acid that dissociates to give 2 H⁺ ions per molecule. The neutralization reaction can be represented as:

\[ \text{H}_2\text{SO}_4 + 2 \text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

### Step 2: Calculate equivalents of KOH and H₂SO₄
- **For KOH:**
- Normality (N) = 0.2 N
- Volume (V) = 100 mL = 0.1 L
- Equivalents of KOH = Normality × Volume = 0.2 N × 0.1 L = 0.02 equivalents

- **For H₂SO₄:**
- Molarity (M) = 0.1 M
- Volume (V) = 100 mL = 0.1 L
- H₂SO₄ gives 2 H⁺ ions, so its normality is:
\[ \text{Normality} = \text{Molarity} \times \text{number of H}^+ \text{ ions} = 0.1 M \times 2 = 0.2 N \]
- Equivalents of H₂SO₄ = Normality × Volume = 0.2 N × 0.1 L = 0.02 equivalents

### Step 3: Determine the total equivalents after mixing
When we mix the two solutions, the equivalents of KOH and H₂SO₄ will neutralize each other:

- Equivalents of KOH = 0.02
- Equivalents of H₂SO₄ = 0.02

Since they are equal, they will completely neutralize each other.

### Step 4: Calculate the normality of the resulting solution
After neutralization, we have no excess H⁺ or OH⁻ ions, which means the resulting solution is neutral. Therefore, the normality of the resulting solution is:

\[ \text{Normality} = 0 \, \text{N} \]

### Final Answer:
The normality of the solution obtained by mixing 100 mL of 0.2 N KOH and 100 mL of 0.1 M H₂SO₄ is **0 N**.

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