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`

Explanation

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

### Step 1: Calculate the H⁺ ion concentrations for both solutions.
The pH of a solution is related to the concentration of hydrogen ions (H⁺) by the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
From this, we can derive the concentrations of H⁺ ions:

- For the first solution (pH = 3):
\[ [\text{H}^+]_1 = 10^{-3} \, \text{M} \]

- For the second solution (pH = 5):
\[ [\text{H}^+]_2 = 10^{-5} \, \text{M} \]

### Step 2: Determine the total H⁺ ion concentration when the two solutions are mixed.
Since equal volumes of both solutions are mixed, the total H⁺ ion concentration in the resulting solution can be calculated by adding the contributions from both solutions:

\[ [\text{H}^+]_{\text{total}} = [\text{H}^+]_1 + [\text{H}^+]_2 \]

Substituting the values:
\[ [\text{H}^+]_{\text{total}} = 10^{-3} + 10^{-5} \]

### Step 3: Simplify the total H⁺ concentration.
Since \(10^{-3}\) is much larger than \(10^{-5}\), we can approximate:
\[ [\text{H}^+]_{\text{total}} \approx 10^{-3} \, \text{M} \]

### Step 4: Calculate the pH of the resulting solution.
Now, we can find the pH of the resulting solution using the H⁺ concentration:
\[ \text{pH} = -\log([\text{H}^+]_{\text{total}}) \]
Substituting the approximate value:
\[ \text{pH} \approx -\log(10^{-3}) = 3 \]

### Step 5: Adjust for dilution due to mixing.
Since we mixed equal volumes of both solutions, the concentration of H⁺ ions will be halved:
\[ [\text{H}^+]_{\text{result}} = \frac{10^{-3}}{2} = 5 \times 10^{-4} \, \text{M} \]

Now, calculate the new pH:
\[ \text{pH} = -\log(5 \times 10^{-4}) \approx 3.3 \]

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