How many litres of water must be added to 1 litre of an aqueous solution of and having pH 4.70 to create an aqueous solution having pH 5.70?

Options

A. 2.0 L
B. 9.0 L
C. 10 L
D. None of these

Explanation

To solve the problem of how many liters of water must be added to 1 liter of an aqueous solution with a pH of 4.70 to create an aqueous solution with a pH of 5.70, we can follow these steps:

### Step 1: Calculate the initial concentration of H⁺ ions
The pH of the initial solution is given as 4.70. We can calculate the concentration of H⁺ ions using the formula:
\[ \text{[H⁺]} = 10^{-\text{pH}} \]
Substituting the value of pH:
\[ \text{[H⁺]} = 10^{-4.70} \approx 2.00 \times 10^{-5} \, \text{mol/L} \]

### Step 2: Calculate the final concentration of H⁺ ions
The target pH is 5.70. We calculate the concentration of H⁺ ions for this pH:
\[ \text{[H⁺]} = 10^{-5.70} \approx 2.00 \times 10^{-6} \, \text{mol/L} \]

### Step 3: Use dilution formula to find the final volume
We can use the dilution equation:
\[ C_1V_1 = C_2V_2 \]
Where:
- \(C_1\) = initial concentration of H⁺ ions = \(2.00 \times 10^{-5} \, \text{mol/L}\)
- \(V_1\) = initial volume = 1 L
- \(C_2\) = final concentration of H⁺ ions = \(2.00 \times 10^{-6} \, \text{mol/L}\)
- \(V_2\) = final volume (unknown)

Substituting the known values:
\[ (2.00 \times 10^{-5})(1) = (2.00 \times 10^{-6})(V_2) \]
Solving for \(V_2\):
\[ V_2 = \frac{2.00 \times 10^{-5}}{2.00 \times 10^{-6}} = 10 \, \text{L} \]

### Step 4: Calculate the amount of water to be added
Since the final volume \(V_2\) is 10 L and the initial volume \(V_1\) is 1 L, the volume of water to be added is:
\[ \text{Volume of water} = V_2 - V_1 = 10 \, \text{L} - 1 \, \text{L} = 9 \, \text{L} \]

### Final Answer
You must add **9 liters** of water to the solution.

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