The hydrogen ion concentration in 0.2 M ethanoic acid
`(K_a = 2 xx 10^(-5) mol dm^(-3))` is

Options

A. `2 xx 10^(-2)`
B. `2 xx 10^(-4)`
C. `2 xx 10^(-3)`
D. `2 xx 10^(-5)`

Answer

To find the hydrogen ion concentration in 0.2 M ethanoic acid (acetic acid), we can follow these steps:

### Step 1: Write the dissociation equation
Ethanoic acid (CH₃COOH) dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Set up the initial concentrations
Let the initial concentration of CH₃COOH be \( C = 0.2 \, \text{M} \). Initially, the concentrations of CH₃COO⁻ and H⁺ are both 0.

### Step 3: Define the change in concentration
Let \( \alpha \) be the degree of dissociation. At equilibrium, the concentrations will be:
- CH₃COOH: \( C(1 - \alpha) = 0.2(1 - \alpha) \)
- CH₃COO⁻: \( C\alpha = 0.2\alpha \)
- H⁺: \( C\alpha = 0.2\alpha \)

### Step 4: Write the expression for the acid dissociation constant \( K_a \)
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:
\[ K_a = \frac{(0.2\alpha)(0.2\alpha)}{0.2(1 - \alpha)} \]

### Step 5: Simplify the equation
Since \( \alpha \) is small, we can approximate \( 1 - \alpha \approx 1 \):
\[ K_a \approx \frac{(0.2\alpha)^2}{0.2} = 0.2\alpha^2 \]

### Step 6: Rearrange to find \( \alpha \)
From the equation above, we can rearrange to find \( \alpha \):
\[ \alpha^2 = \frac{K_a}{0.2} \]
\[ \alpha = \sqrt{\frac{K_a}{0.2}} \]

### Step 7: Substitute the known values
Given \( K_a = 2 \times 10^{-5} \, \text{mol/dm}^3 \):
\[ \alpha = \sqrt{\frac{2 \times 10^{-5}}{0.2}} \]
\[ \alpha = \sqrt{1 \times 10^{-4}} = 0.01 \]

### Step 8: Calculate the hydrogen ion concentration
The concentration of hydrogen ions \( [\text{H}^+] \) is given by:
\[ [\text{H}^+] = C\alpha = 0.2 \times 0.01 = 0.002 \, \text{M} = 2 \times 10^{-3} \, \text{M} \]

### Final Answer
The hydrogen ion concentration in 0.2 M ethanoic acid is \( 2 \times 10^{-3} \, \text{M} \).

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