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 Initial Conditions
We are given:
- Initial volume of HCl solution, \( V_1 = 1 \, \text{L} \)
- Initial pH = 1
### Step 2: Calculate the Initial Concentration of HCl
Using the pH to find the concentration of hydrogen ions:
\[
\text{pH} = -\log[H^+]
\]
For pH = 1:
\[
[H^+] = 10^{-\text{pH}} = 10^{-1} = 0.1 \, \text{M}
\]
Thus, the initial concentration \( C_1 = 0.1 \, \text{M} \).
### Step 3: Determine the Final Conditions
We want to achieve a final pH of 2. Using the same formula:
\[
\text{pH} = 2 \implies [H^+] = 10^{-2} = 0.01 \, \text{M}
\]
So, the final concentration \( C_2 = 0.01 \, \text{M} \).
### Step 4: Use the Dilution Formula
The dilution equation is given by:
\[
C_1 V_1 = C_2 V_2
\]
Where:
- \( C_1 = 0.1 \, \text{M} \)
- \( V_1 = 1 \, \text{L} \)
- \( C_2 = 0.01 \, \text{M} \)
- \( V_2 \) is the final volume we need to find.
### Step 5: Solve for \( V_2 \)
Substituting the known values into the dilution equation:
\[
0.1 \times 1 = 0.01 \times V_2
\]
\[
V_2 = \frac{0.1}{0.01} = 10 \, \text{L}
\]
### Step 6: Calculate the Volume of Water to be Added
The total volume \( V_2 \) is 10 L, and since we started with 1 L of the 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
Therefore, the volume of water that must be added is **9 liters**.
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