Calculate the pH of a solution obtained by diluting 25 ml of decinormal HCl to 500 ml assuming complete ionisation of HCl.

Answer

To calculate the pH of a solution obtained by diluting 25 ml of decinormal HCl to 500 ml, we will follow these steps:

### Step 1: Understand the given data
- We have a decinormal (0.1 N) solution of HCl.
- Volume of HCl solution (V1) = 25 ml
- Final volume after dilution (V2) = 500 ml

### Step 2: Use the dilution formula
We will use the dilution formula, which states:
\[ N_1 V_1 = N_2 V_2 \]
Where:
- \( N_1 \) = initial normality (0.1 N)
- \( V_1 \) = initial volume (25 ml)
- \( N_2 \) = final normality (unknown)
- \( V_2 \) = final volume (500 ml)

### Step 3: Plug in the values
Substituting the known values into the equation:
\[ 0.1 \, \text{N} \times 25 \, \text{ml} = N_2 \times 500 \, \text{ml} \]

### Step 4: Solve for \( N_2 \)
Rearranging the equation to solve for \( N_2 \):
\[ N_2 = \frac{0.1 \times 25}{500} \]

Calculating:
\[ N_2 = \frac{2.5}{500} = 0.005 \, \text{N} \]
\[ N_2 = 5 \times 10^{-3} \, \text{N} \]

### Step 5: Relate normality to molarity
Since HCl is a strong acid and completely ionizes, the normality (N) is equal to the molarity (M) for HCl because the n-factor for HCl is 1.
Thus:
\[ \text{Molarity} = 5 \times 10^{-3} \, \text{M} \]

### Step 6: Calculate the concentration of \( H^+ \)
The concentration of \( H^+ \) ions in the solution is equal to the molarity of HCl:
\[ [H^+] = 5 \times 10^{-3} \, \text{M} \]

### Step 7: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \( [H^+] \):
\[ \text{pH} = -\log(5 \times 10^{-3}) \]

### Step 8: Solve for pH
Using logarithm properties:
\[ \text{pH} = -(\log(5) + \log(10^{-3})) \]
\[ \text{pH} = -(\log(5) - 3) \]
\[ \text{pH} = 3 - \log(5) \]
Using a calculator, \( \log(5) \approx 0.699 \):
\[ \text{pH} \approx 3 - 0.699 = 2.301 \]

### Final Answer
Thus, the pH of the diluted solution is approximately:
\[ \text{pH} \approx 2.30 \]

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