`20 mL` of `0.1N HCI` is mixed with `20 ml` of `0.1N KOH`. The `pH` of the solution would be

Options

A. `0`
B. `7`
C. `2`
D. `9`

Explanation

To find the pH of the solution formed by mixing 20 mL of 0.1N HCl with 20 mL of 0.1N KOH, we can follow these steps:

### Step 1: Calculate the number of equivalents of HCl and KOH
- **HCl**:
- Normality (N) = 0.1N
- Volume (V) = 20 mL = 0.020 L
- Equivalents of HCl = Normality × Volume = 0.1 N × 0.020 L = 0.002 equivalents (or 2 milliequivalents)

- **KOH**:
- Normality (N) = 0.1N
- Volume (V) = 20 mL = 0.020 L
- Equivalents of KOH = Normality × Volume = 0.1 N × 0.020 L = 0.002 equivalents (or 2 milliequivalents)

### Step 2: Determine the reaction between HCl and KOH
- The reaction between HCl (a strong acid) and KOH (a strong base) can be represented as:
\[ \text{HCl} + \text{KOH} \rightarrow \text{KCl} + \text{H}_2\text{O} \]
- Since both HCl and KOH are present in equal amounts (2 milliequivalents each), they will completely neutralize each other.

### Step 3: Identify the resulting solution
- After the neutralization reaction, we are left with a solution of KCl and water. KCl is a neutral salt formed from a strong acid and a strong base.

### Step 4: Determine the pH of the resulting solution
- The pH of a neutral solution (formed from the reaction of a strong acid and a strong base) is 7.

### Conclusion
The pH of the solution after mixing 20 mL of 0.1N HCl with 20 mL of 0.1N KOH is **7**.

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