Equal volumes of two solutions of `HCl` are mixed. One solution has a `pH = 1`, while the other has a `pH = 5`. The `pH` of the resulting solution is

Options

A. `lt 1`
B. Between `1` and `2`
C. `3`
D. Between `4` and `5`

Explanation

To solve the problem, we need to find the pH of the resulting solution when equal volumes of two HCl solutions with different pH values are mixed. Here’s a step-by-step solution:

### Step 1: Determine the H⁺ ion concentrations from the given pH values.
- For the first solution with pH = 1:
\[ [H^+]_1 = 10^{-pH} = 10^{-1} = 0.1 \, \text{mol/L} \]

- For the second solution with pH = 5:
\[ [H^+]_2 = 10^{-pH} = 10^{-5} = 0.00001 \, \text{mol/L} \]

### Step 2: Assume equal volumes of both solutions.
Let’s assume we mix 1 L of each solution. Therefore, the total volume after mixing will be:
\[ V_{total} = V_1 + V_2 = 1 \, \text{L} + 1 \, \text{L} = 2 \, \text{L} \]

### Step 3: Calculate the total moles of H⁺ ions from both solutions.
- Moles of H⁺ from the first solution:
\[ \text{Moles}_1 = [H^+]_1 \times V_1 = 0.1 \, \text{mol/L} \times 1 \, \text{L} = 0.1 \, \text{mol} \]

- Moles of H⁺ from the second solution:
\[ \text{Moles}_2 = [H^+]_2 \times V_2 = 0.00001 \, \text{mol/L} \times 1 \, \text{L} = 0.00001 \, \text{mol} \]

### Step 4: Calculate the total moles of H⁺ ions in the resulting solution.
\[ \text{Total moles of H}^+ = \text{Moles}_1 + \text{Moles}_2 = 0.1 + 0.00001 = 0.10001 \, \text{mol} \]

### Step 5: Calculate the concentration of H⁺ ions in the resulting solution.
\[ [H^+]_{total} = \frac{\text{Total moles of H}^+}{V_{total}} = \frac{0.10001 \, \text{mol}}{2 \, \text{L}} = 0.050005 \, \text{mol/L} \]

### Step 6: Calculate the pH of the resulting solution.
Using the formula for pH:
\[ pH = -\log[H^+]_{total} = -\log(0.050005) \]
Calculating this gives:
\[ pH \approx 1.301 \]

### Conclusion
The pH of the resulting solution is approximately 1.301, which lies between 1 and 2.