`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 when `20 mL` of `0.1N HCl` is mixed with `20 mL` of `0.1N KOH`, we can follow these steps:

### Step 1: Calculate the number of equivalents of HCl and KOH
- **HCl:**
\[ \text{Normality (N)} = 0.1 \, \text{N} \]
\[ \text{Volume (V)} = 20 \, \text{mL} = 0.020 \, \text{L} \]
\[ \text{Equivalents of HCl} = N \times V = 0.1 \, \text{N} \times 0.020 \, \text{L} = 0.002 \, \text{equivalents} = 2 \, \text{milliequivalents} \]

- **KOH:**
\[ \text{Normality (N)} = 0.1 \, \text{N} \]
\[ \text{Volume (V)} = 20 \, \text{mL} = 0.020 \, \text{L} \]
\[ \text{Equivalents of KOH} = N \times V = 0.1 \, \text{N} \times 0.020 \, \text{L} = 0.002 \, \text{equivalents} = 2 \, \text{milliequivalents} \]

### 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:
\[ \text{HCl} + \text{KOH} \rightarrow \text{H}_2\text{O} + \text{KCl} \]
Since both reactants are present in equal amounts (2 milliequivalents each), they will completely neutralize each other.

### Step 3: Analyze the resulting solution
After the neutralization reaction, we are left with water (H2O) and potassium chloride (KCl). The resulting solution consists of a neutral salt (KCl) and water.

### Step 4: Determine the pH of the solution
The pH of a neutral solution (pure water) is 7. Since KCl is a neutral salt formed from a strong acid and a strong base, it does not affect the pH of the solution.

### Conclusion
Thus, the pH of the resulting solution after mixing `20 mL` of `0.1N HCl` with `20 mL` of `0.1N KOH` is:
\[ \text{pH} = 7 \]

### Final Answer
The pH of the solution is **7**.

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