Equal volumes of two soultion, one having `pH 6` and the other having`pH 4` are mixed. The `pH` of the resulting solution would be

Options

A. `5.7`
B. `4.3`
C. `5.0`
D. `5.5`

Explanation

To find the pH of the resulting solution when equal volumes of two solutions with pH 6 and pH 4 are mixed, we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions in each solution.
- For the solution with pH 6:
\[ [H^+]_1 = 10^{-pH} = 10^{-6} \, \text{M} \]

- For the solution with pH 4:
\[ [H^+]_2 = 10^{-pH} = 10^{-4} \, \text{M} \]

### Step 2: Mix the two solutions.
Since we are mixing equal volumes of both solutions, the concentrations of H⁺ ions will be halved in the resulting mixture.

### Step 3: Calculate the total concentration of H⁺ ions in the mixture.
The total concentration of H⁺ ions after mixing is:
\[ [H^+]_{total} = \frac{[H^+]_1 + [H^+]_2}{2} \]
Substituting the values we calculated:
\[ [H^+]_{total} = \frac{10^{-6} + 10^{-4}}{2} \]

### Step 4: Simplify the total concentration.
To simplify:
\[ [H^+]_{total} = \frac{10^{-6} + 10^{-4}}{2} = \frac{10^{-6} + 100 \cdot 10^{-6}}{2} = \frac{101 \cdot 10^{-6}}{2} = 50.5 \cdot 10^{-6} = 5.05 \cdot 10^{-5} \, \text{M} \]

### Step 5: Calculate the pH of the resulting solution.
Now, we can find the pH of the resulting solution using the formula:
\[ pH = -\log[H^+]_{total} \]
Substituting the value we found:
\[ pH = -\log(5.05 \cdot 10^{-5}) = -\log(5.05) - \log(10^{-5}) \]
Using the approximation \(\log(5.05) \approx 0.70\):
\[ pH \approx -0.70 + 5 = 4.30 \]

### Final Answer:
The pH of the resulting solution is approximately **4.3**.

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