What will be the pH of a solution formed by mixing 50 mL of 0.5 M HCl solution and 150 mL of 0.5 M NaOH solution and 300 mL `H_(2) O` ?

Options

A. 13
B. 12.7
C. 7
D. 11

Explanation

To find the pH of the solution formed by mixing 50 mL of 0.5 M HCl and 150 mL of 0.5 M NaOH, along with 300 mL of water, we can follow these steps:

### Step 1: Calculate the moles of HCl and NaOH
- **Moles of HCl**:
\[ \text{Moles of HCl} = \text{Volume (L)} \times \text{Concentration (M)} = 0.050 \, \text{L} \times 0.5 \, \text{mol/L} = 0.025 \, \text{mol} \]

- **Moles of NaOH**:
\[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.150 \, \text{L} \times 0.5 \, \text{mol/L} = 0.075 \, \text{mol} \]

### Step 2: Determine the reaction between HCl and NaOH
- HCl and NaOH react in a 1:1 ratio:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 3: Calculate the remaining moles after the reaction
- **Moles of HCl remaining**:
\[ \text{Remaining moles of HCl} = 0.025 \, \text{mol} - 0.025 \, \text{mol} = 0 \, \text{mol} \]

- **Moles of NaOH remaining**:
\[ \text{Remaining moles of NaOH} = 0.075 \, \text{mol} - 0.025 \, \text{mol} = 0.050 \, \text{mol} \]

### Step 4: Calculate the concentration of NaOH in the final solution
- **Total volume of the solution**:
\[ \text{Total volume} = 50 \, \text{mL (HCl)} + 150 \, \text{mL (NaOH)} + 300 \, \text{mL (water)} = 500 \, \text{mL} = 0.500 \, \text{L} \]

- **Concentration of NaOH**:
\[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Total Volume}} = \frac{0.050 \, \text{mol}}{0.500 \, \text{L}} = 0.1 \, \text{M} \]

### Step 5: Calculate the pOH of the solution
- **pOH**:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log[0.1] = 1 \]

### Step 6: Calculate the pH of the solution
- Using the relationship \( \text{pH} + \text{pOH} = 14 \):
\[ \text{pH} = 14 - \text{pOH} = 14 - 1 = 13 \]

### Final Answer:
The pH of the solution is **13**.

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