How many litres of water must be added to 1 L of an aqueous solution of HC 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

Answer

To solve the problem of how many liters of water must be added to 1 L of an aqueous solution of hydrochloric acid (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 hydrogen ion concentration.
The pH of a solution is related to the concentration of hydrogen ions \([H^+]\) by the formula:
\[ \text{pH} = -\log[H^+] \]
For a pH of 1:
\[ [H^+] = 10^{-1} \, \text{M} = 0.1 \, \text{M} \]
For a pH of 2:
\[ [H^+] = 10^{-2} \, \text{M} = 0.01 \, \text{M} \]

### Step 2: Use the dilution formula.
We can use the dilution equation:
\[ M_1V_1 = M_2V_2 \]
Where:
- \(M_1\) = initial concentration of HCl = 0.1 M
- \(V_1\) = initial volume of the solution = 1 L
- \(M_2\) = final concentration of HCl = 0.01 M
- \(V_2\) = final volume of the solution (unknown)

### Step 3: Substitute the known values into the dilution equation.
Substituting the known values into the equation:
\[ (0.1 \, \text{M})(1 \, \text{L}) = (0.01 \, \text{M})(V_2) \]
This simplifies to:
\[ 0.1 = 0.01 \times V_2 \]

### Step 4: Solve for \(V_2\).
Rearranging the equation to solve for \(V_2\):
\[ V_2 = \frac{0.1}{0.01} = 10 \, \text{L} \]

### Step 5: Calculate the volume of water to be added.
Since the initial volume of the solution is 1 L, the volume of water that needs 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 1 L of the HCl solution to achieve a pH of 2.

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