The `pH` of `10^(-5) M HCI` solution if `1mol` of it is diluted to `1000 ml` is :

Options

A. 5
B. 8
C. 7.02
D. 6.98

Explanation

To find the pH of a \(10^{-5} \, M\) HCl solution that has been diluted to \(1000 \, ml\), follow these steps:

### Step 1: Understand the Initial Concentration
The initial concentration of HCl is given as \(10^{-5} \, M\). HCl is a strong acid and dissociates completely in solution to give \(H^+\) ions.

### Step 2: Use Dilution Formula
When 1 mole of \(10^{-5} \, M\) HCl is diluted to \(1000 \, ml\), we can use the dilution formula \(M_1V_1 = M_2V_2\) to find the new concentration (\(M_2\)).

- Here, \(M_1 = 10^{-5} \, M\)
- \(V_1 = 1 \, ml = 0.001 \, L\)
- \(V_2 = 1000 \, ml = 1 \, L\)

Using the dilution formula:
\[ M_2 = \frac{M_1 \cdot V_1}{V_2} = \frac{10^{-5} \cdot 0.001}{1} = 10^{-8} \, M \]

### Step 3: Consider Autoionization of Water
In addition to the \(H^+\) ions from HCl, water also contributes \(H^+\) ions due to its autoionization. The concentration of \(H^+\) ions from water at neutral conditions is \(10^{-7} \, M\).

### Step 4: Calculate Total \(H^+\) Concentration
The total concentration of \(H^+\) ions in the solution is the sum of the contributions from HCl and water:
\[ [H^+] = [H^+]_{HCl} + [H^+]_{water} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \, M \]

### Step 5: Calculate pH
The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
Substituting the total \(H^+\) concentration:
\[ pH = -\log(1.1 \times 10^{-7}) \]
Using logarithmic properties:
\[ pH \approx 6.96 \text{ (approximately 6.98)} \]

### Final Answer
The pH of the diluted \(10^{-5} \, M\) HCl solution is approximately **6.98**.

---