How many litres of water must be added to 1 litre of an aqueous solution of an acid 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 of an acid with a pH of 4.70 to create a solution with a pH of 5.70, we can follow these steps:

### Step 1: Understand the pH and H⁺ concentration
The pH of a solution is related to the concentration of hydrogen ions (H⁺) in the solution. The formula for calculating the concentration of H⁺ from pH is:
\[ \text{[H⁺]} = 10^{-\text{pH}} \]

### Step 2: Calculate the initial H⁺ concentration
Given that the initial pH is 4.70:
\[ \text{[H⁺]}_1 = 10^{-4.70} \]
Calculating this gives:
\[ \text{[H⁺]}_1 \approx 1.99 \times 10^{-5} \, \text{mol/L} \]

### Step 3: Calculate the final H⁺ concentration
Now, we need to find the H⁺ concentration for the final pH of 5.70:
\[ \text{[H⁺]}_2 = 10^{-5.70} \]
Calculating this gives:
\[ \text{[H⁺]}_2 \approx 1.99 \times 10^{-6} \, \text{mol/L} \]

### Step 4: Set up the dilution equation
Using the dilution equation:
\[ M_1 V_1 = M_2 V_2 \]
Where:
- \( M_1 \) = initial concentration of H⁺ = \( 1.99 \times 10^{-5} \, \text{mol/L} \)
- \( V_1 \) = initial volume = 1 L
- \( M_2 \) = final concentration of H⁺ = \( 1.99 \times 10^{-6} \, \text{mol/L} \)
- \( V_2 \) = final volume (unknown)

Substituting the known values into the equation:
\[ (1.99 \times 10^{-5}) \times 1 = (1.99 \times 10^{-6}) \times V_2 \]

### Step 5: Solve for \( V_2 \)
Rearranging the equation to solve for \( V_2 \):
\[ V_2 = \frac{(1.99 \times 10^{-5}) \times 1}{(1.99 \times 10^{-6})} \]
Calculating this gives:
\[ V_2 \approx 10 \, \text{L} \]

### 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 initial solution.

---