What is the pH of solution in which `25.0` mL of `0.1` M NaOH is added to 25 mL of `0.08`M HCl and final solution is diluted to 500 mL?

Options

A. 3
B. 11
C. 12
D. 13

Answer

To find the pH of the solution when 25.0 mL of 0.1 M NaOH is added to 25.0 mL of 0.08 M HCl and the final solution is diluted to 500 mL, we can follow these steps:

### Step 1: Calculate the number of milliequivalents of HCl
The number of milliequivalents of HCl can be calculated using the formula:
\[ \text{milliequivalents of HCl} = \text{concentration (M)} \times \text{volume (mL)} \]
For HCl:
\[ \text{milliequivalents of HCl} = 0.08 \, \text{M} \times 25 \, \text{mL} = 2 \, \text{milliequivalents} \]

### Step 2: Calculate the number of milliequivalents of NaOH
Similarly, for NaOH:
\[ \text{milliequivalents of NaOH} = 0.1 \, \text{M} \times 25 \, \text{mL} = 2.5 \, \text{milliequivalents} \]

### Step 3: Determine the excess base
Since NaOH is in excess, we can find the remaining milliequivalents of NaOH after neutralizing HCl:
\[ \text{Excess NaOH} = \text{milliequivalents of NaOH} - \text{milliequivalents of HCl} = 2.5 - 2 = 0.5 \, \text{milliequivalents} \]

### Step 4: Calculate the concentration of OH⁻ in the final solution
The total volume of the final solution is 500 mL. To find the concentration of OH⁻ ions:
\[ \text{Concentration of OH}^- = \frac{\text{Excess NaOH (milliequivalents)}}{\text{Total volume (mL)}} = \frac{0.5}{500} = 0.001 \, \text{M} = 10^{-3} \, \text{M} \]

### Step 5: Calculate the concentration of H⁺ ions
Using the relationship between H⁺ and OH⁻ concentrations:
\[ [\text{H}^+] \times [\text{OH}^-] = 10^{-14} \]
We can find [H⁺]:
\[ [\text{H}^+] = \frac{10^{-14}}{[\text{OH}^-]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \, \text{M} \]

### Step 6: Calculate the pH of the solution
Finally, we can calculate the pH using the formula:
\[ \text{pH} = -\log[\text{H}^+] = -\log(10^{-11}) = 11 \]

### Final Answer
The pH of the solution is **11**.

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