Equal volumes of two `HCl` solutions of `pH=3` and `pH=5` were mixed. What is the `pH` of the resulting solution ?

Options

A. `3.5`
B. `4.0`
C. `4.5`
D. `3.3`

Answer

To find the pH of the resulting solution when equal volumes of two HCl solutions with pH 3 and pH 5 are mixed, follow these steps:

### Step 1: Calculate the concentration of H⁺ ions in each solution.

1. For the first solution (pH = 3):
\[ \text{pH} = -\log[H^+] \]
\[ 3 = -\log[H^+] \]
\[ [H^+] = 10^{-3} \text{ M} \]

2. For the second solution (pH = 5):
\[ \text{pH} = -\log[H^+] \]
\[ 5 = -\log[H^+] \]
\[ [H^+] = 10^{-5} \text{ M} \]

### Step 2: Mix the two solutions.

Since equal volumes (let's say V) of both solutions are mixed, the total volume of the resulting solution will be:
\[ \text{Total Volume} = V + V = 2V \]

### Step 3: Calculate the total concentration of H⁺ ions in the mixed solution.

The total moles of H⁺ ions from both solutions can be calculated as follows:

1. Moles from the first solution:
\[ \text{Moles of } H^+ = [H^+] \times V = 10^{-3} \times V \]

2. Moles from the second solution:
\[ \text{Moles of } H^+ = [H^+] \times V = 10^{-5} \times V \]

3. Total moles of H⁺ ions in the mixed solution:
\[ \text{Total } H^+ = 10^{-3}V + 10^{-5}V = (10^{-3} + 10^{-5})V \]

### Step 4: Calculate the concentration of H⁺ ions in the resulting solution.

The concentration of H⁺ ions in the final solution (total volume = 2V):
\[ [H^+]_{final} = \frac{(10^{-3} + 10^{-5})V}{2V} = \frac{10^{-3} + 10^{-5}}{2} \]
\[ = \frac{10^{-3}(1 + 10^{-2})}{2} = \frac{10^{-3}(1 + 0.01)}{2} = \frac{10^{-3} \times 1.01}{2} = \frac{1.01 \times 10^{-3}}{2} = 0.505 \times 10^{-3} \text{ M} \]

### Step 5: Calculate the pH of the resulting solution.

Now, we can calculate the pH:
\[ \text{pH} = -\log[H^+] = -\log(0.505 \times 10^{-3}) \]
Using the logarithmic properties:
\[ = -\log(0.505) - \log(10^{-3}) = -\log(0.505) + 3 \]

### Step 6: Calculate log(0.505).

Using the approximation:
\[ \log(0.505) \approx -0.3 \quad (\text{since } 0.505 \text{ is slightly less than } 0.5) \]
Thus:
\[ \text{pH} \approx 3 - (-0.3) = 3 + 0.3 = 3.3 \]

### Final Answer:
The pH of the resulting solution is approximately **3.3**.

---