Question

0.05 M of a weak acid solution (HA) ionizes 0.2 percent in water. Find its acid dissociation constant (K2) and its pH value? Explain in steps, please.

Answer 1

To determine the acid dissociation constant (Ka) for the weak acid solution that ionizes 0.2 percent in water, calculate the ionized concentration, plug the values into the ionization expression, and then calculate the pH using the negative logarithm of the H+ concentration. The Ka is found to be approximately 2 × 10^-7 and the pH value is 4.

Step-by-step explanation:

To find the acid dissociation constant (Ka) and the pH value of a 0.05 M weak acid solution (HA) that ionizes 0.2 percent in water, we follow these steps:

1. Calculate the concentration of ionized acid. 0.2% of 0.05 M is 0.05 M × 0.002 = 1 × 10-4 M.
2. Since HA ionizes into H+ and A-, the concentrations of H+ and A- will also be 1 × 10-4 M.
3. Use the ionization expression: Ka = [H+][A-] / [HA] to calculate Ka. Plugging in the values, we get 1 × 10-4 × 1 × 10-4 / 0.0498 (since 0.05 M – 1 × 10-4 M is roughly 0.0498 M for HA). Ka ≈ 2 × 10-7.
4. To find the pH, take the negative logarithm of [H+]: pH = -log(1 × 10-4) = 4.

The acid dissociation constant (Ka) is approximately 2 × 10-7 and the pH value is 4.