Find the `pH` of solution prepared by mixing `25 ml` of a `0.5 M` solution of `HCl, 10 ml` of a `0.5 M` solution of `NaOH` and `15 ml` of water

Options

A. `0.8239`
B. `1.0029`
C. `1.0239`
D. `1.8239`

Explanation

To find the pH of the solution prepared by mixing 25 ml of 0.5 M HCl, 10 ml of 0.5 M NaOH, and 15 ml of water, we can follow these steps:

### Step 1: Calculate the millimoles of HCl and NaOH
- **HCl**:
\[ \text{Millimoles of HCl} = \text{Volume (ml)} \times \text{Concentration (M)} = 25 \, \text{ml} \times 0.5 \, \text{M} = 12.5 \, \text{mmol} \]

- **NaOH**:
\[ \text{Millimoles of NaOH} = \text{Volume (ml)} \times \text{Concentration (M)} = 10 \, \text{ml} \times 0.5 \, \text{M} = 5 \, \text{mmol} \]

### Step 2: Determine the neutralization reaction
When HCl (strong acid) reacts with NaOH (strong base), they neutralize each other:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- Here, 5 mmol of NaOH will neutralize 5 mmol of HCl.

### Step 3: Calculate the remaining millimoles of HCl
After neutralization:
\[ \text{Remaining HCl} = 12.5 \, \text{mmol} - 5 \, \text{mmol} = 7.5 \, \text{mmol} \]

### Step 4: Calculate the total volume of the solution
The total volume of the mixed solution is:
\[ \text{Total Volume} = 25 \, \text{ml (HCl)} + 10 \, \text{ml (NaOH)} + 15 \, \text{ml (water)} = 50 \, \text{ml} \]

### Step 5: Calculate the concentration of H⁺ ions in the final solution
The concentration of H⁺ ions remaining in the solution is given by:
\[ \text{Concentration of H}^+ = \frac{\text{Remaining HCl (mmol)}}{\text{Total Volume (ml)}} = \frac{7.5 \, \text{mmol}}{50 \, \text{ml}} = 0.15 \, \text{M} \]

### Step 6: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the concentration:
\[ \text{pH} = -\log(0.15) \approx 0.823 \]

### Final Answer
The pH of the solution is approximately **0.82**.

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