The weak acid HA is 2% ionized (dissociated) in a 0.20 M solution. 1. What is Ka for this acid?
2. What is the pH of this solution?

1. Ka = 8.16x10⁻⁵

2. pH = 2.40

Explanation:

1. The dissociation of a weak acid in water occurs as follows:

HA ⇄ H⁺ + A⁻

Ka = [H⁺] [A⁻] / [HA]

As 2% of the 0.20M solution is dissociated:

[H⁺] = [A⁻] = 0.20M * 2% = 0.004M -As H⁺ and A⁻ comes from the same reaction, their concentrations are the same

[HA] = 0.20M * 98% = 0.196M

Ka = (0.004)² / (0.196M) = 8.16x10⁻⁵

2. pH = -log [H⁺] = -log [0.004M]

the correct answer would be d. hope this !