Question

A 0.48 molar solution of a monoprotic acid (HA) in water reaches equilibrium at a concentration of 0.36 M. What is Ka for this acid? Please enter your answer rounded to two significant figures. Step by step please <3

Answer 1

The dissociation of a monoprotic acid HA can be represented as follows:

HA ⇌ H+ + A-

The equilibrium constant expression for this reaction is:

Ka = [H+][A-]/[HA]

We are given the initial concentration of the acid (HA) as 0.48 M and the equilibrium concentration as 0.36 M. At equilibrium, the concentration of H+ and A- will also be 0.36 M.

Substituting the values into the equilibrium constant expression, we get:

Ka = (0.36)^2 / 0.48 = 0.27

Therefore, the value of Ka for the acid is 0.27, rounded to two significant figures.