Question

In a 0.2 M aqueous solution, lactic acid is 6.9% dissociated. The value of dissociation constant is

• A. \( 1.2 \times 10^{-4} \)
• B. \( 9.5 \times 10^{-4} \)
• C. \( 6.5 \times 10^{-4} \)
• D. \( 3.6 \times 10^{-2} \)

Hint:

For weak acids, the dissociation constant can be calculated using the concentration of dissociated ions and the initial concentration.

Answer 1

The Correct Option is B

To find the dissociation constant (\(K_a\)) of lactic acid, we can use the information given: the molarity of the solution is 0.2 M, and it is 6.9% dissociated. The dissociation of lactic acid (\(CH_3CH(OH)COOH\)) can be represented as:

\(CH_3CH(OH)COOH(aq) \rightleftharpoons CH_3CH(OH)COO^-(aq) + H^+(aq)\)

Initially, the concentration of \(CH_3CH(OH)COOH\) is 0.2 M. When 6.9% of the acid dissociates, the concentration of the dissociated acid is:

\[ [CH_3CH(OH)COO^-] = [H^+] = 0.2 \times \frac{6.9}{100} = 0.0138 \, \text{M} \]

The equilibrium concentration of undissociated \(CH_3CH(OH)COOH\) is:

\[ [CH_3CH(OH)COOH] = 0.2 - 0.0138 = 0.1862 \, \text{M} \]

Now, we calculate the dissociation constant \(K_a\) using the formula:

\[ K_a = \frac{[H^+][CH_3CH(OH)COO^-]}{[CH_3CH(OH)COOH]} \]

Substituting the equilibrium concentrations:

\[ K_a = \frac{(0.0138)(0.0138)}{0.1862} \]

Calculating this gives:

\[ K_a \approx \frac{0.00019044}{0.1862} \approx 0.0010236 \approx 9.5 \times 10^{-4} \]

Thus, the dissociation constant of lactic acid is \(9.5 \times 10^{-4}\).

Answer 2

The dissociation constant (\( K_a \)) of a weak acid can be calculated using the following formula: \[ K_a = \frac{[H^+][A^-]}{[HA]} \] Given that lactic acid dissociates 6.9%, we can calculate the concentration of dissociated ions. Let's denote:
- \( \text{Initial concentration of acid} = 0.2 \, M \),
- \( \text{Degree of dissociation} = 0.069 \).
Thus, the concentration of the dissociated ions (\( [H^+] \) and \( [A^-] \)) is: \[ [H^+] = [A^-] = 0.2 \times 0.069 = 0.0138 \, M \] Now, we can calculate the dissociation constant using the approximation: \[ K_a = \frac{(0.0138)(0.0138)}{0.2 - 0.0138} = 9.5 \times 10^{-4} \]
Thus, the dissociation constant is \( 9.5 \times 10^{-4} \).