50 mL of 0.1 M HCl and 50 mL of 2.0 M NaOH are mixed. The pH of the resulting solution is

Options

A. `1.30`
B. `4.2`
C. `12.70`
D. `11.70`

Explanation

To find the pH of the resulting solution when 50 mL of 0.1 M HCl is mixed with 50 mL of 2.0 M NaOH, we can follow these steps:

### Step 1: Calculate the moles of HCl and NaOH
- **Moles of HCl**:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \, (5 \, \text{mmol}) \]

- **Moles of NaOH**:
\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 2.0 \, \text{M} \times 0.050 \, \text{L} = 0.1 \, \text{mol} \, (10 \, \text{mmol}) \]

### Step 2: Determine the neutralization reaction
Since HCl is a strong acid and NaOH is a strong base, they will completely neutralize each other:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- **Moles of HCl neutralized**: 5 mmol
- **Moles of NaOH neutralized**: 5 mmol

### Step 3: Calculate the remaining moles of NaOH
After neutralization:
- Remaining moles of NaOH = Initial moles of NaOH - Moles of NaOH neutralized
\[ \text{Remaining moles of NaOH} = 10 \, \text{mmol} - 5 \, \text{mmol} = 5 \, \text{mmol} \]

### Step 4: Calculate the concentration of OH⁻ ions
The total volume of the solution after mixing is:
\[ \text{Total Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]
The concentration of OH⁻ ions is:
\[ \text{Concentration of OH}^- = \frac{\text{Remaining moles of NaOH}}{\text{Total Volume}} = \frac{5 \, \text{mmol}}{0.1 \, \text{L}} = 0.05 \, \text{M} \]

### Step 5: Calculate the concentration of H⁺ ions
Using the ion product of water (\(K_w\)):
\[ K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \]
We can find the concentration of H⁺:
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{0.05} = 2.0 \times 10^{-13} \, \text{M} \]

### Step 6: Calculate the pH
Finally, we can calculate the pH:
\[ \text{pH} = -\log[H^+] = -\log(2.0 \times 10^{-13}) \approx 12.7 \]

### Conclusion
The pH of the resulting solution is approximately **12.7**.

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