When equal volumes of pH =4 and pH=6 are mixed together then th ph of the resulting solution will be [log 5 =0.7]

Options

A. 4.3
B. 4.7
C. 5
D. 5.3

Answer

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

### Step 1: Calculate the H⁺ ion concentration for each solution.
- For the solution with pH = 4:
\[ [H^+]_1 = 10^{-pH} = 10^{-4} \, \text{M} \]

- For the solution with pH = 6:
\[ [H^+]_2 = 10^{-pH} = 10^{-6} \, \text{M} \]

### Step 2: Add the H⁺ ion concentrations.
Since we are mixing equal volumes of both solutions, we can find the average concentration of H⁺ ions:
\[ [H^+]_{mix} = \frac{[H^+]_1 + [H^+]_2}{2} \]
Substituting the values:
\[ [H^+]_{mix} = \frac{10^{-4} + 10^{-6}}{2} \]

### Step 3: Simplify the expression.
To simplify, we can express \(10^{-4}\) in terms of \(10^{-6}\):
\[ 10^{-4} = 100 \times 10^{-6} \]
Thus,
\[ [H^+]_{mix} = \frac{100 \times 10^{-6} + 1 \times 10^{-6}}{2} = \frac{101 \times 10^{-6}}{2} = 50.5 \times 10^{-6} \, \text{M} \]

### Step 4: Calculate the pH of the resulting solution.
Now, we can find the pH using the formula:
\[ pH = -\log[H^+] \]
Substituting the value of \([H^+]_{mix}\):
\[ pH = -\log(50.5 \times 10^{-6}) = -\log(50.5) - \log(10^{-6}) \]
Using the properties of logarithms:
\[ pH = -\log(50.5) + 6 \]

### Step 5: Calculate \(-\log(50.5)\).
Using the approximation \(\log(5) \approx 0.7\):
\[ \log(50.5) \approx \log(5.05 \times 10^1) = \log(5.05) + 1 \approx 0.7 + 1 = 1.7 \]
Thus,
\[ pH \approx 6 - 1.7 = 4.3 \]

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

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