How many litres of water must be added to `1 L` of an aqueous solution of `HCl` with a `pH` of `1` to create an aqueous solution with `pH` of `2`?

Options

A. `0.1 L`
B. `0.9 L`
C. `2.0 L`
D. `9.0 L`

Explanation

To solve the problem of how many liters of water must be added to 1 L of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with a pH of 2, we can follow these steps:

### Step 1: Determine the initial concentration of H⁺ ions
Given that the pH of the solution is 1, we can calculate the concentration of H⁺ ions using the formula:
\[ \text{[H⁺]} = 10^{-\text{pH}} = 10^{-1} = 0.1 \, \text{mol/L} \]

### Step 2: Calculate the number of moles of HCl in the initial solution
Since we have 1 L of the solution, the number of moles of HCl can be calculated as:
\[ \text{Number of moles of HCl} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 1 \, \text{L} = 0.1 \, \text{mol} \]

### Step 3: Determine the final concentration of H⁺ ions for the desired pH
To achieve a pH of 2, we calculate the new concentration of H⁺ ions:
\[ \text{[H⁺]} = 10^{-\text{pH}} = 10^{-2} = 0.01 \, \text{mol/L} \]

### Step 4: Set up the dilution equation
Using the dilution equation \( M_1V_1 = M_2V_2 \), where:
- \( M_1 = 0.1 \, \text{mol/L} \) (initial concentration)
- \( V_1 = 1 \, \text{L} \) (initial volume)
- \( M_2 = 0.01 \, \text{mol/L} \) (final concentration)
- \( V_2 \) is the final volume we need to find.

Substituting the known values into the equation:
\[ 0.1 \, \text{mol/L} \times 1 \, \text{L} = 0.01 \, \text{mol/L} \times V_2 \]

### Step 5: Solve for \( V_2 \)
Rearranging the equation to find \( V_2 \):
\[ V_2 = \frac{0.1 \, \text{mol/L} \times 1 \, \text{L}}{0.01 \, \text{mol/L}} = 10 \, \text{L} \]

### Step 6: Calculate the volume of water to be added
Since the final volume \( V_2 \) is 10 L and we started with 1 L of HCl solution, 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 need to add **9 liters** of water to achieve a pH of 2.

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