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

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: Understand the relationship between pH and H⁺ concentration
The pH of a solution is defined as:
\[ \text{pH} = -\log[\text{H}^+] \]
Given that the pH of the initial solution is 1, we can calculate the concentration of hydrogen ions \([\text{H}^+]\) in the solution.

### Step 2: Calculate the initial concentration of H⁺ ions
From the pH of 1:
\[ [\text{H}^+] = 10^{-1} \text{ M} = 0.1 \text{ M} \]

### Step 3: Determine the concentration of H⁺ ions for the final pH
For the final solution with a pH of 2:
\[ [\text{H}^+] = 10^{-2} \text{ M} = 0.01 \text{ M} \]

### Step 4: Use the dilution equation
We can use the dilution equation \(M_1V_1 = M_2V_2\) to find the final volume after dilution. Here:
- \(M_1 = 0.1 \text{ M}\) (initial concentration)
- \(V_1 = 1 \text{ L}\) (initial volume)
- \(M_2 = 0.01 \text{ M}\) (final concentration)
- \(V_2\) is the final volume we need to find.

Substituting the values into the equation:
\[ 0.1 \times 1 = 0.01 \times V_2 \]

### Step 5: Solve for \(V_2\)
\[ 0.1 = 0.01 \times V_2 \]
\[ V_2 = \frac{0.1}{0.01} = 10 \text{ L} \]

### Step 6: Calculate the volume of water to be added
Since the initial volume 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
Therefore, **9 liters of water must be added** to the solution.

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