The pH of a solution prepared by mixing 2.0 mL of HCI solution of pH 3.0 and 3.0 mL of NaOH of pH 10.0 is

Options

A. 2.5
B. 3.5
C. 5.5
D. 6.5

Answer

To find the pH of the solution prepared by mixing 2.0 mL of HCl solution with a pH of 3.0 and 3.0 mL of NaOH solution with a pH of 10.0, we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions from HCl
The pH of the HCl solution is given as 3.0. We can use the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
From this, we can find the concentration of H⁺ ions:
\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

### Step 2: Calculate the concentration of OH⁻ ions from NaOH
The pH of the NaOH solution is given as 10.0. First, we calculate the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 10 = 4 \]
Now, we can find the concentration of OH⁻ ions:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-4} \, \text{M} \]

### Step 3: Calculate the milliequivalents of H⁺ and OH⁻
Next, we calculate the milliequivalents of H⁺ and OH⁻ ions in the respective solutions:
- For HCl:
\[ \text{Milliequivalents of H}^+ = \text{n-factor} \times \text{Volume (L)} \times [\text{H}^+] \]
\[ = 1 \times 0.002 \, \text{L} \times 10^{-3} \, \text{M} = 2 \times 10^{-6} \, \text{equivalents} \]

- For NaOH:
\[ \text{Milliequivalents of OH}^- = \text{n-factor} \times \text{Volume (L)} \times [\text{OH}^-] \]
\[ = 1 \times 0.003 \, \text{L} \times 10^{-4} \, \text{M} = 3 \times 10^{-7} \, \text{equivalents} \]

### Step 4: Determine the excess H⁺ ions
Now we find the net milliequivalents of H⁺ ions after neutralization:
\[ \text{Excess H}^+ = \text{Milliequivalents of H}^+ - \text{Milliequivalents of OH}^- \]
\[ = 2 \times 10^{-6} - 3 \times 10^{-7} = 1.7 \times 10^{-6} \, \text{equivalents} \]

### Step 5: Calculate the concentration of excess H⁺ ions in the total volume
The total volume of the mixed solution is:
\[ V_{\text{total}} = 2 \, \text{mL} + 3 \, \text{mL} = 5 \, \text{mL} = 0.005 \, \text{L} \]
Now we can calculate the concentration of excess H⁺ ions:
\[ [\text{H}^+] = \frac{\text{Excess H}^+}{V_{\text{total}}} = \frac{1.7 \times 10^{-6}}{0.005} = 3.4 \times 10^{-4} \, \text{M} \]

### Step 6: Calculate the pH of the resultant solution
Finally, we can find the pH of the resultant solution using the concentration of H⁺ ions:
\[ \text{pH} = -\log[\text{H}^+] = -\log(3.4 \times 10^{-4}) \approx 3.5 \]

### Conclusion
The pH of the solution prepared by mixing the two solutions is approximately **3.5**.

---