`pH` of a solution made by mixing `200mL` of `0.0657M NaOH, 140 mL` of `0.107M HCl` and `160mL` of `H_(2)O` is

Options

A. `3.04`
B. `2.43`
C. `2.74`
D. `2.27`

Explanation

To find the pH of the solution made by mixing 200 mL of 0.0657 M NaOH, 140 mL of 0.107 M HCl, and 160 mL of H2O, we can follow these steps:

### Step 1: Calculate the millimoles of NaOH and HCl
1. **For NaOH:**
\[ \text{Millimoles of NaOH} = \text{Molarity} \times \text{Volume (mL)} = 0.0657 \, \text{M} \times 200 \, \text{mL} = 13.14 \, \text{mmol} \]

2. **For HCl:**
\[ \text{Millimoles of HCl} = \text{Molarity} \times \text{Volume (mL)} = 0.107 \, \text{M} \times 140 \, \text{mL} = 14.98 \, \text{mmol} \]

### Step 2: Determine the limiting reactant
- The reaction between NaOH and HCl is a neutralization reaction:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- Since we have 13.14 mmol of NaOH and 14.98 mmol of HCl, NaOH is the limiting reactant.

### Step 3: Calculate the remaining moles after the reaction
- Remaining moles of HCl after reaction:
\[ \text{Remaining HCl} = \text{Initial HCl} - \text{NaOH} = 14.98 \, \text{mmol} - 13.14 \, \text{mmol} = 1.84 \, \text{mmol} \]

### Step 4: Calculate the total volume of the solution
- Total volume of the solution:
\[ \text{Total Volume} = 200 \, \text{mL (NaOH)} + 140 \, \text{mL (HCl)} + 160 \, \text{mL (H}_2\text{O)} = 500 \, \text{mL} \]

### Step 5: Calculate the concentration of H⁺ ions
- Since HCl is a strong acid, it completely dissociates in solution. The concentration of H⁺ ions is given by:
\[ [\text{H}^+] = \frac{\text{Remaining HCl (mmol)}}{\text{Total Volume (mL)}} = \frac{1.84 \, \text{mmol}}{500 \, \text{mL}} = 0.00368 \, \text{M} \]

### Step 6: Calculate the pH
- The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] = -\log(0.00368) \approx 2.43 \]

### Final Answer
The pH of the solution is approximately **2.43**.

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