Calculate the pH of 0.1 M acetic acid solution if its dissociation constant is `1.8 xx 10^(-5).` If 1 litre of this solution is mixed with 0.05 mole of HCI, what will be pH of the mixture?

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the degree of dissociation of acetic acid.
Given:
- Concentration of acetic acid (C) = 0.1 M
- Dissociation constant (Ka) = \(1.8 \times 10^{-5}\)

Using the formula for the degree of dissociation (\(\alpha\)):
\[ K_a = C \cdot \alpha^2 \]
Substituting the values:
\[ 1.8 \times 10^{-5} = 0.1 \cdot \alpha^2 \]
\[ \alpha^2 = \frac{1.8 \times 10^{-5}}{0.1} = 1.8 \times 10^{-4} \]
\[ \alpha = \sqrt{1.8 \times 10^{-4}} \approx 0.0134 \]

### Step 2: Calculate the concentration of H\(^+\) ions from acetic acid.
The concentration of H\(^+\) ions produced by the dissociation of acetic acid can be calculated as:
\[ [H^+] = C \cdot \alpha = 0.1 \cdot 0.0134 \approx 0.00134 \, \text{M} \]

### Step 3: Calculate the pH of the acetic acid solution.
Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(0.00134) \approx 2.87 \]

### Step 4: Calculate the pH after mixing with HCl.
When 0.05 moles of HCl are added to 1 liter of the acetic acid solution, the total volume becomes 1 L + 1 L = 2 L. The concentration of HCl in the mixture is:
\[ \text{Concentration of HCl} = \frac{0.05 \, \text{moles}}{2 \, \text{L}} = 0.025 \, \text{M} \]

### Step 5: Calculate the total concentration of H\(^+\) ions in the mixture.
The total concentration of H\(^+\) ions in the mixture is the sum of the H\(^+\) ions from acetic acid and HCl:
\[ [H^+]_{\text{total}} = [H^+]_{\text{acetic acid}} + [H^+]_{\text{HCl}} = 0.00134 + 0.025 = 0.02634 \, \text{M} \]

### Step 6: Calculate the final pH of the mixture.
Using the pH formula again:
\[ \text{pH} = -\log(0.02634) \approx 1.58 \]

### Final Answer:
The pH of the mixture after adding HCl to the acetic acid solution is approximately **1.58**.

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