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 solve the problem of finding the pH of a solution formed by mixing equal volumes of solutions with pH 4 and pH 6, we can follow these steps:

### Step 1: Understand the pH scale
The pH scale is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. The formula for pH is given by:
\[ \text{pH} = -\log[\text{H}^+] \]
where \([\text{H}^+]\) is the concentration of hydrogen ions in moles per liter (M).

### Step 2: Calculate the concentration of \([\text{H}^+]\) for each solution
1. For the solution with pH = 4:
\[ \text{pH} = 4 \implies [\text{H}^+] = 10^{-4} \, \text{M} \]

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

### Step 3: Determine the total concentration of \([\text{H}^+]\) when mixed
When equal volumes of these two solutions are mixed, the concentrations will be averaged because the volumes are equal.

Let’s assume we mix 1 L of each solution:
- The total amount of \([\text{H}^+]\) from the first solution:
\[ n_1 = [\text{H}^+]_1 \times V_1 = 10^{-4} \, \text{mol/L} \times 1 \, \text{L} = 10^{-4} \, \text{mol} \]

- The total amount of \([\text{H}^+]\) from the second solution:
\[ n_2 = [\text{H}^+]_2 \times V_2 = 10^{-6} \, \text{mol/L} \times 1 \, \text{L} = 10^{-6} \, \text{mol} \]

### Step 4: Calculate the total moles of \([\text{H}^+]\) and the new concentration
The total moles of \([\text{H}^+]\) after mixing:
\[ n_{\text{total}} = n_1 + n_2 = 10^{-4} + 10^{-6} \approx 10^{-4} \, \text{mol} \]

The total volume after mixing is:
\[ V_{\text{total}} = V_1 + V_2 = 1 \, \text{L} + 1 \, \text{L} = 2 \, \text{L} \]

The new concentration of \([\text{H}^+]\) is:
\[ [\text{H}^+] = \frac{n_{\text{total}}}{V_{\text{total}}} = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, \text{M} \]

### Step 5: Calculate the new pH
Now, we can find the new pH:
\[ \text{pH} = -\log(5 \times 10^{-5}) = -\log(5) - \log(10^{-5}) = -\log(5) + 5 \]
Given that \(\log(5) = 0.7\):
\[ \text{pH} = -0.7 + 5 = 4.3 \]

### Final Answer
The pH of the resulting solution after mixing equal volumes of the two solutions is **4.3**.

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