Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCI `K_a` of acetic acid is `1.8 xx 10^(-5)`

Options

A. 0.0018
B. 0.0001
C. 0.018
D. 0.18

Answer

To calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCl, we will follow these steps:

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

### Step 2: Set up the initial concentrations
- Initial concentration of acetic acid, [CH₃COOH] = 0.01 M
- Initial concentration of H⁺ from HCl = 0.1 M (strong acid, completely dissociates)

### Step 3: Set up the equilibrium expression
Let \( x \) be the amount of acetic acid that ionizes. At equilibrium:
- [CH₃COOH] = 0.01 - x
- [CH₃COO⁻] = x
- [H⁺] = 0.1 + x (but since x is very small compared to 0.1, we can approximate it as 0.1)

### Step 4: Write the expression for \( K_a \)
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{x(0.1)}{0.01 - x} \]

### Step 5: Substitute the known value of \( K_a \)
Given \( K_a = 1.8 \times 10^{-5} \):
\[ 1.8 \times 10^{-5} = \frac{x(0.1)}{0.01 - x} \]

### Step 6: Make approximations
Since acetic acid is a weak acid, we can assume \( x \) is very small compared to 0.01:
\[ 1.8 \times 10^{-5} \approx \frac{0.1x}{0.01} \]
This simplifies to:
\[ 1.8 \times 10^{-5} = 10x \]

### Step 7: Solve for \( x \)
Rearranging gives:
\[ x = \frac{1.8 \times 10^{-5}}{10} = 1.8 \times 10^{-6} \]

### Step 8: Calculate the percentage ionization
The percentage ionization is given by:
\[ \text{Percentage Ionization} = \left( \frac{x}{\text{Initial concentration of acetic acid}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Ionization} = \left( \frac{1.8 \times 10^{-6}}{0.01} \right) \times 100 = 0.018\% \]

### Final Answer
The percentage ionization of 0.01 M acetic acid in 0.1 M HCl is **0.018%**.

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