pH of a lemon water bottle is 3 and that of orange juice is 4. Calculate the pH of final solution obtained by mixing above solutions in volume ratio of 1 : 2 respectively.
(Assuming that no change in volume on mixing and no chemical reaction is taking place) [Given : `log_(10)2 =0.30`]

Explanation

To calculate the pH of the final solution obtained by mixing lemon water and orange juice in a volume ratio of 1:2, we can follow these steps:

### Step 1: Determine the concentration of H⁺ ions in each solution
- The pH of lemon water is 3. Therefore, the concentration of H⁺ ions can be calculated using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{mol/L} \]

- The pH of orange juice is 4. Thus, the concentration of H⁺ ions is:
\[ [H^+] = 10^{-4} \, \text{mol/L} \]

### Step 2: Calculate the total H⁺ ions contributed by each solution
- For the lemon water (volume = 1v):
\[ \text{Total H⁺ from lemon water} = [H^+] \times \text{Volume} = 10^{-3} \times 1v = 10^{-3}v \]

- For the orange juice (volume = 2v):
\[ \text{Total H⁺ from orange juice} = [H^+] \times \text{Volume} = 10^{-4} \times 2v = 2 \times 10^{-4}v \]

### Step 3: Calculate the total H⁺ ions in the final solution
- The total H⁺ ions in the mixture is the sum of the contributions from both solutions:
\[ \text{Total H⁺} = 10^{-3}v + 2 \times 10^{-4}v = (10^{-3} + 2 \times 10^{-4})v = (10^{-3} + 0.2 \times 10^{-3})v = 1.2 \times 10^{-3}v \]

### Step 4: Calculate the total volume of the mixture
- The total volume of the mixture is:
\[ \text{Total Volume} = 1v + 2v = 3v \]

### Step 5: Calculate the final concentration of H⁺ ions in the mixture
- The final concentration of H⁺ ions in the mixture is given by:
\[ [H^+]_{\text{final}} = \frac{\text{Total H⁺}}{\text{Total Volume}} = \frac{1.2 \times 10^{-3}v}{3v} = \frac{1.2 \times 10^{-3}}{3} = 0.4 \times 10^{-3} = 4 \times 10^{-4} \, \text{mol/L} \]

### Step 6: Calculate the pH of the final solution
- Finally, we can calculate the pH using the concentration of H⁺ ions:
\[ \text{pH} = -\log[H^+] = -\log(4 \times 10^{-4}) \]
Using logarithmic properties:
\[ \text{pH} = -\log(4) - \log(10^{-4}) = -\log(4) + 4 \]
Given that \(\log(4) = 2 \cdot \log(2)\) and \(\log(2) = 0.30\):
\[ \log(4) = 2 \cdot 0.30 = 0.60 \]
Thus:
\[ \text{pH} = -0.60 + 4 = 3.40 \]

### Final Answer:
The pH of the final solution is **3.40**.

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