Calculate the `pH` of a solution which contains `10 ml` of `1 M HCl` and `10 ml` of `2M NaOH`

Options

A. `11.7`
B. `12.7`
C. `13.7`
D. `10.7`

Explanation

To calculate the pH of a solution containing 10 mL of 1 M HCl and 10 mL of 2 M NaOH, we can follow these steps:

### Step 1: Calculate the moles of HCl and NaOH
- **For HCl:**
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 1 \, \text{M} \times 0.010 \, \text{L} = 0.010 \, \text{moles} \]

- **For NaOH:**
\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 2 \, \text{M} \times 0.010 \, \text{L} = 0.020 \, \text{moles} \]

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

- We have 0.010 moles of HCl and 0.020 moles of NaOH. Since HCl is in lesser amount, it is the limiting reactant.

### Step 3: Calculate the remaining moles after the reaction
- After the reaction, all of the HCl will be consumed:
\[ \text{Remaining moles of NaOH} = \text{Initial moles of NaOH} - \text{Moles of HCl} = 0.020 - 0.010 = 0.010 \, \text{moles} \]

### Step 4: Calculate the total volume of the solution
- Total volume = Volume of HCl + Volume of NaOH
\[ \text{Total volume} = 10 \, \text{mL} + 10 \, \text{mL} = 20 \, \text{mL} = 0.020 \, \text{L} \]

### Step 5: Calculate the concentration of the remaining NaOH
\[ \text{Concentration of NaOH} = \frac{\text{Remaining moles of NaOH}}{\text{Total volume}} = \frac{0.010 \, \text{moles}}{0.020 \, \text{L}} = 0.50 \, \text{M} \]

### Step 6: Calculate the pOH of the solution
\[ \text{pOH} = -\log[\text{OH}^-] = -\log[0.50] \approx 0.301 \]

### Step 7: Calculate the pH of the solution
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pH} = 14 - \text{pOH} = 14 - 0.301 \approx 13.699 \]

### Final Answer
The pH of the solution is approximately **13.70**.