The` P^(H)` of a I lit solution is 2. It is diluted with water till its `p^(H)` becomes 4, How many litres of water is added?

Options

A. 99
B. 9
C. 999
D. 9.9

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the initial conditions
The pH of the initial 1 liter solution is given as 2.

### Step 2: Calculate the concentration of H⁺ ions before dilution
Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
we can find the concentration of H⁺ ions:
\[ [H^+] = 10^{-\text{pH}} = 10^{-2} \text{ M} \]

### Step 3: Understand the final conditions after dilution
After dilution, the pH of the solution becomes 4. We can again calculate the concentration of H⁺ ions:
\[ [H^+] = 10^{-\text{pH}} = 10^{-4} \text{ M} \]

### Step 4: Set up the dilution equation
Let \(X\) be the volume of water added in liters. The dilution equation can be expressed as:
\[ M_1V_1 = M_2V_2 \]
where:
- \(M_1 = 10^{-2} \text{ M}\) (initial concentration)
- \(V_1 = 1 \text{ L}\) (initial volume)
- \(M_2 = 10^{-4} \text{ M}\) (final concentration)
- \(V_2 = 1 + X \text{ L}\) (final volume, which is the initial volume plus the volume of water added)

### Step 5: Substitute values into the equation
Substituting the known values into the dilution equation:
\[ (10^{-2})(1) = (10^{-4})(1 + X) \]

### Step 6: Simplify the equation
This simplifies to:
\[ 10^{-2} = 10^{-4}(1 + X) \]
Dividing both sides by \(10^{-4}\):
\[ 10^{2} = 1 + X \]

### Step 7: Solve for X
Now, solving for \(X\):
\[ X = 10^{2} - 1 = 100 - 1 = 99 \text{ L} \]

### Conclusion
Thus, the volume of water added is **99 liters**.

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