Question

What is the value of the ionization constant \( K_a \) for acetic acid (\( \text{CH}_3\text{COOH} \)) if the concentration of acetic acid is 0.1 M and the concentration of \( \text{H}^+ \) ions at equilibrium is 1.0 x \( 10^{-3} \) M?

• A. \( 1 \times 10^{-5} \)
• B. \( 1 \times 10^{-4} \)
• C. \( 1 \times 10^{-3} \)
• D. \( 1 \times 10^{-6} \)

Hint:

The ionization constant \( K_a \) represents the strength of an acid in solution. A larger \( K_a \) indicates a stronger acid.

Answer 1

The Correct Option is A

The ionization constant \( K_a \) for acetic acid is given by the expression: \[ K_a = \frac{[ \text{H}^+ ] [ \text{CH}_3\text{COO}^- ]}{[ \text{CH}_3\text{COOH} ]} \] At equilibrium, the concentration of \( \text{CH}_3\text{COOH} \) is approximately \( 0.1 - x \), where \( x \) is the concentration of \( \text{H}^+ \) ions. Given that the concentration of \( \text{H}^+ \) is \( 1.0 \times 10^{-3} \) M, we can assume \( [ \text{CH}_3\text{COO}^- ] = [ \text{H}^+ ] = 1.0 \times 10^{-3} \) M and \( [ \text{CH}_3\text{COOH} ] \approx 0.1 \) M. Now, substitute these values into the equation: \[ K_a = \frac{(1.0 \times 10^{-3}) (1.0 \times 10^{-3})}{0.1} = 1.0 \times 10^{-5} \] Thus, the value of the ionization constant \( K_a \) for acetic acid is \( 1.0 \times 10^{-5} \). Therefore, the correct answer is option (1).