Calculate the `pH` of a solution which contains `100mL` of `0.1 M HC1` and `9.9 mL` of `1.0 M NaOH`.

Explanation

To calculate the pH of a solution containing 100 mL of 0.1 M HCl and 9.9 mL of 1.0 M NaOH, we will follow these steps:

### Step 1: Calculate the moles of HCl and NaOH
- **Moles of HCl** = Molarity × Volume (in liters)
\[ \text{Moles of HCl} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles} \]

- **Moles of NaOH** = Molarity × Volume (in liters)
\[ \text{Moles of NaOH} = 1.0 \, \text{M} \times 0.0099 \, \text{L} = 0.0099 \, \text{moles} \]

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

Since we have 0.01 moles of HCl and 0.0099 moles of NaOH, NaOH is the limiting reagent.

### Step 3: Calculate the remaining moles of HCl after neutralization
- Moles of HCl remaining after reaction:
\[ \text{Remaining HCl} = \text{Initial HCl} - \text{Moles of NaOH} = 0.01 - 0.0099 = 0.0001 \, \text{moles} \]

### Step 4: Calculate the total volume of the solution
- Total volume = Volume of HCl + Volume of NaOH
\[ \text{Total Volume} = 100 \, \text{mL} + 9.9 \, \text{mL} = 109.9 \, \text{mL} = 0.1099 \, \text{L} \]

### Step 5: Calculate the concentration of H⁺ ions
- Concentration of H⁺ ions from the remaining HCl:
\[ \text{Concentration of H}^+ = \frac{\text{Remaining HCl}}{\text{Total Volume}} = \frac{0.0001 \, \text{moles}}{0.1099 \, \text{L}} \approx 0.000910 \, \text{M} \]

### Step 6: Calculate the pH
- pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
\[ \text{pH} = -\log(0.000910) \approx 3.041 \]

### Final Result
The pH of the solution is approximately **3.041**.

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