What will be the percentage of dissociation in 1.0 M `CH_(3)CO OH` at equilibrium having dissociation constant of `1.8xx10^(-5)`?

Explanation

To find the percentage of dissociation of 1.0 M acetic acid (CH₃COOH) at equilibrium, given the dissociation constant (Kₐ) of 1.8 × 10⁻⁵, we can follow these steps:

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

### Step 2: Define the equilibrium expression
The equilibrium constant expression for the dissociation of acetic acid is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]

### Step 3: Set up the equilibrium concentrations
Let the degree of dissociation be represented by \( \alpha \). At equilibrium:
- The concentration of CH₃COOH will be \( 1 - \alpha \) M.
- The concentration of CH₃COO⁻ and H⁺ will each be \( \alpha \) M.

### Step 4: Substitute into the equilibrium expression
Substituting these values into the equilibrium expression gives:
\[ K_a = \frac{\alpha \cdot \alpha}{1 - \alpha} = \frac{\alpha^2}{1 - \alpha} \]

### Step 5: Substitute the known values
We know that \( K_a = 1.8 \times 10^{-5} \) and the initial concentration \( C = 1.0 \) M. Thus:
\[ 1.8 \times 10^{-5} = \frac{\alpha^2}{1 - \alpha} \]

### Step 6: Make the assumption for small \( \alpha \)
Since \( K_a \) is small, we can assume that \( \alpha \) is also small, so \( 1 - \alpha \approx 1 \). Therefore, the equation simplifies to:
\[ 1.8 \times 10^{-5} \approx \alpha^2 \]

### Step 7: Solve for \( \alpha \)
Taking the square root of both sides:
\[ \alpha = \sqrt{1.8 \times 10^{-5}} \]
Calculating this gives:
\[ \alpha \approx 4.24 \times 10^{-3} \]

### Step 8: Calculate the percentage of dissociation
The percentage of dissociation is given by:
\[ \text{Percentage of dissociation} = \alpha \times 100\% \]
Substituting the value of \( \alpha \):
\[ \text{Percentage of dissociation} = 4.24 \times 10^{-3} \times 100\% = 0.424\% \]

### Final Answer
The percentage of dissociation of 1.0 M acetic acid at equilibrium is approximately **0.424%**.

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