Ka = 9.60 * 10⁻¹⁰
The dissociation constant of an acid, Ka = [H⁺][A⁻] / [HA]
pH = -log[H⁺]
pH of acid = 5.12
log[H⁺] = -5.12
[H⁺] = 10⁻⁵°¹²
[H⁺] = 7.59 * 10⁻⁶ M
Assuming the acid is a monoprotic acid, the dissociation of the acid is given below:
HA ⇄ H⁺ + A⁻
Moles of acid produced = moles of acid conjugate produced
Therefore, at equilibrium, [H⁺] = [A⁻]
Also, at equilibrium, the final concentration of the acid, concentration of acid, [HA] = [HA] - [H⁺]
[HA] - [H⁺] = (0.06 - 7.59 * 10⁻⁶) M
Using the equation Ka = [H⁺][A⁻] / [HA]
Ka = (7.59 * 10⁻⁶ M) * (7.59 * 10⁻⁶ M) / (0.06 - 7.59 * 10⁻⁶ ) M
Ka = 9.60 * 10⁻¹⁰
Learn more at: https://brainly.com/question/15076429
