A solution containing 75 mL of `0.2M` HCI and 25 mL of `0.2M` NaOH. Calculate the pH of solution.

Explanation

To calculate the pH of a solution containing 75 mL of 0.2 M HCl and 25 mL of 0.2 M NaOH, we will follow these steps:

### Step 1: Calculate the moles of HCl and NaOH
- **Moles of HCl** = Volume (L) × Concentration (M)
- Volume of HCl = 75 mL = 0.075 L
- Concentration of HCl = 0.2 M
- Moles of HCl = 0.075 L × 0.2 mol/L = 0.015 moles

- **Moles of NaOH** = Volume (L) × Concentration (M)
- Volume of NaOH = 25 mL = 0.025 L
- Concentration of NaOH = 0.2 M
- Moles of NaOH = 0.025 L × 0.2 mol/L = 0.005 moles

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

### Step 3: Calculate the remaining moles after the reaction
- Initially, we have:
- Moles of HCl = 0.015 moles
- Moles of NaOH = 0.005 moles

- After the reaction:
- Moles of HCl remaining = 0.015 - 0.005 = 0.010 moles
- Moles of NaOH remaining = 0.005 - 0.005 = 0 moles

### Step 4: Calculate the concentration of H⁺ ions in the solution
- The total volume of the solution after mixing = 75 mL + 25 mL = 100 mL = 0.1 L
- Concentration of H⁺ ions from HCl remaining:
\[ \text{Concentration of H}^+ = \frac{\text{Moles of HCl remaining}}{\text{Total Volume}} = \frac{0.010 \text{ moles}}{0.1 \text{ L}} = 0.1 \text{ M} \]

### Step 5: Calculate the pH of the solution
- The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
- Substituting the concentration:
\[ \text{pH} = -\log(0.1) = 1 \]

### Final Answer
The pH of the solution is **1**.

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