`100 mL` of `0.2 N NaOH` is mixed with `100` mL `0.1 NHCl` and the solution is made 1L. The pH of the solution is :

Options

A. 4
B. 8
C. 10
D. 12

Explanation

To find the pH of the solution formed by mixing 100 mL of 0.2 N NaOH with 100 mL of 0.1 N HCl and diluting it to 1 L, we can follow these steps:

### Step 1: Calculate the moles of NaOH and HCl
- **For NaOH:**
- Volume = 100 mL = 0.1 L
- Normality = 0.2 N
- Moles of NaOH = Normality × Volume = 0.2 N × 0.1 L = 0.02 moles (or 20 millimoles)

- **For HCl:**
- Volume = 100 mL = 0.1 L
- Normality = 0.1 N
- Moles of HCl = Normality × Volume = 0.1 N × 0.1 L = 0.01 moles (or 10 millimoles)

### Step 2: Determine the reaction between NaOH and HCl
The reaction between NaOH and HCl is:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 3: Calculate the remaining moles after the reaction
- NaOH reacts with HCl in a 1:1 ratio.
- Moles of NaOH remaining = Initial moles of NaOH - Moles of HCl = 20 millimoles - 10 millimoles = 10 millimoles of NaOH left.

### Step 4: Calculate the concentration of OH⁻ ions
- The total volume of the solution after dilution is 1 L.
- Concentration of OH⁻ ions = Moles of NaOH remaining / Total volume = 10 millimoles / 1 L = 0.01 moles/L = 0.01 M.

### Step 5: Calculate the pOH of the solution
- pOH = -log[OH⁻] = -log(0.01) = 2.

### Step 6: Calculate the pH of the solution
- pH + pOH = 14
- Therefore, pH = 14 - pOH = 14 - 2 = 12.

### Final Answer
The pH of the solution is **12**.

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