The pH of the solution formed on mixing 20 mL of `0.05 M H_(2)SO_(4)` with 5.0 mL of 0.45 M NaOH of 298 K is

Options

A. 6
B. 2
C. 12
D. 7

Explanation

To find the pH of the solution formed by mixing 20 mL of 0.05 M \( H_2SO_4 \) with 5.0 mL of 0.45 M NaOH, we can follow these steps:

### Step 1: Calculate Normality of \( H_2SO_4 \)
- The normality (N) of an acid is given by the formula:
\[ N = M \times \text{valency factor} \]
- For \( H_2SO_4 \):
- Molarity (M) = 0.05 M
- Valency factor = 2 (since \( H_2SO_4 \) can donate 2 protons)

Therefore,
\[ N_1 = 0.05 \times 2 = 0.1 \, \text{N} \]

### Step 2: Calculate Normality of NaOH
- For NaOH:
- Molarity (M) = 0.45 M
- Valency factor = 1 (since NaOH can donate 1 proton)

Therefore,
\[ N_2 = 0.45 \times 1 = 0.45 \, \text{N} \]

### Step 3: Calculate the equivalents of \( H_2SO_4 \) and NaOH
- Calculate the equivalents of \( H_2SO_4 \) (N1V1):
\[ N_1V_1 = 0.1 \, \text{N} \times 20 \, \text{mL} = 2.0 \, \text{equivalents} \]

- Calculate the equivalents of NaOH (N2V2):
\[ N_2V_2 = 0.45 \, \text{N} \times 5 \, \text{mL} = 2.25 \, \text{equivalents} \]

### Step 4: Determine the limiting reactant
- Since \( N_2V_2 \) (2.25 equivalents of NaOH) is greater than \( N_1V_1 \) (2.0 equivalents of \( H_2SO_4 \)), NaOH is in excess.

### Step 5: Calculate the excess equivalents of NaOH
- Excess NaOH:
\[ \text{Excess} = N_2V_2 - N_1V_1 = 2.25 - 2.0 = 0.25 \, \text{equivalents} \]

### Step 6: Calculate the total volume of the solution
- Total volume:
\[ V_{\text{total}} = 20 \, \text{mL} + 5 \, \text{mL} = 25 \, \text{mL} \]

### Step 7: Calculate the concentration of excess NaOH
- Concentration of excess NaOH:
\[ \text{Concentration} = \frac{\text{Excess equivalents}}{V_{\text{total}}} = \frac{0.25 \, \text{equivalents}}{0.025 \, \text{L}} = 10 \, \text{N} \]

### Step 8: Calculate the concentration of \( OH^- \)
- Since NaOH is a strong base, the concentration of \( OH^- \) ions is equal to the normality of NaOH:
\[ [OH^-] = 10 \, \text{N} \]

### Step 9: Calculate the concentration of \( H^+ \)
- Using the relation \( K_w = [H^+][OH^-] \) where \( K_w = 1 \times 10^{-14} \):
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{10} = 1 \times 10^{-15} \, \text{M} \]

### Step 10: Calculate the pH
- Finally, calculate the pH:
\[ pH = -\log[H^+] = -\log(1 \times 10^{-15}) = 15 \]

### Final Answer
The pH of the solution is 12.