Final Answer:
The Kₐ for the monoprotic acid is 4.00 x 10⁻⁴.
Step-by-step explanation:
In order to calculate the Kₐ for the monoprotic acid, we first need to determine the concentration of hydronium ions ([H₃O⁺]) in the solution. The pH of the solution is given as 2.80, which means the concentration of [H₃O⁺] is 10⁻².⁸⁰ M. Since the acid is monoprotic, this also represents the concentration of the acid ion (HA).
The equilibrium expression for the dissociation of a weak monoprotic acid (HA) in water is given by Kₐ = [H₃O⁺][A⁻] / [HA]. Since the acid is monoprotic, [A⁻] is equal to the concentration of [H₃O⁺], which is 10⁻².⁸⁰ M. The concentration of HA is given as 1.28 M. Substituting these values into the equilibrium expression, we get:
Kₐ = (10⁻².⁸⁰)² / (1.28 - 10⁻².⁸⁰)
Simplifying this expression gives the final answer for Kₐ as 4.00 x 10⁻⁴.
This result indicates that the acid is a weak acid, as Kₐ is small. The smaller the Kₐ, the weaker the acid. It's important to note that the pH and Kₐ are related in a logarithmic manner – a small change in pH corresponds to a large change in acidity. Therefore, the given pH of 2.80 implies a relatively high concentration of hydronium ions, resulting in a low Kₐ value for the weak acid.