The dissociation constant of acetic acid at a given temperature is `1.69xx10^(-5)`.The degree of dissociation of 0.01 M acetic acid in presence of 0.01 M HCl is equal to :

Options

A. 0.41
B. 0.13
C. `1.69xx10^(-3)`
D. `0.013`

Answer

To solve the problem of finding the degree of dissociation of 0.01 M acetic acid in the presence of 0.01 M HCl, we will follow these steps:

### Step 1: Write the dissociation equation for acetic acid
The dissociation of acetic acid (CH₃COOH) can be represented as:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Define the concentrations
Let:
- The initial concentration of acetic acid = 0.01 M
- The concentration of HCl = 0.01 M (which fully dissociates into H⁺ and Cl⁻ ions)

### Step 3: Set up the equilibrium expression
The dissociation constant (K) for acetic acid is given as:
\[ K_a = 1.69 \times 10^{-5} \]

At equilibrium, if α is the degree of dissociation of acetic acid, the concentrations will be:
- [CH₃COOH] = 0.01(1 - α) M
- [CH₃COO⁻] = 0.01α M
- [H⁺] = 0.01 + 0.01α M (since the H⁺ from HCl adds to the H⁺ from acetic acid)

### Step 4: Substitute into the equilibrium expression
The expression for the dissociation constant is:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]

Substituting the equilibrium concentrations into the expression gives:
\[ 1.69 \times 10^{-5} = \frac{(0.01α)(0.01 + 0.01α)}{0.01(1 - α)} \]

### Step 5: Simplify the equation
Assuming α is very small (which is valid since K is small), we can approximate:
- 1 - α ≈ 1
- 0.01 + 0.01α ≈ 0.01 (since α is small)

Thus, the equation simplifies to:
\[ 1.69 \times 10^{-5} = \frac{(0.01α)(0.01)}{0.01} \]
\[ 1.69 \times 10^{-5} = 0.01α \]

### Step 6: Solve for α
Now, rearranging gives:
\[ α = \frac{1.69 \times 10^{-5}}{0.01} \]
\[ α = 1.69 \times 10^{-3} \]

### Conclusion
The degree of dissociation of 0.01 M acetic acid in the presence of 0.01 M HCl is approximately:
\[ α = 1.69 \times 10^{-3} \]