Step-by-step explanation:
We have a weak acid HPI that when it is in solution it will dissociate into its ions.
HPI <---> H⁺ + PI⁻
The expression for the Ka will show us the relationship between the concentrations at the equilibrium.
Ka = [H⁺]eq * [PI⁻]eq/[HPI]eq
We are already given the concentration of H⁺ at equilibrium but we don't know the concentration of the other species. We will have to find them. To do that we will set up an ICE table.
HPI <---> H⁺ + PI⁻
I 0.67 0 0
C -x x x
E 0.67 - x x x
From the ICE table we can say that we called with the letter x to the equilibrium concentration of H+, but we already have that value. We can replace that value in the other ones to calculate them.
x = [H⁺]eq
x = 5.3 * 10^(-2) M = [H⁺]eq = [PI⁻]eq
[H⁺]eq = 5.3 * 10^(-2) M
[PI⁻]eq = 5.3 * 10^(-2) M
[HPI]eq = 0.67 M - x
[HPI]eq = 0.67 M - 5.3 * 10^(-2) M
[HPI]eq = 0.617 M
Finally we can replace these values in the Ka expression to find the answer to our problem.
Ka = [H⁺]eq * [PI⁻]eq/[HPI]eq
Ka = 5.3 * 10^(-2) M * 5.3 * 10^(-2) M/(0.617 M)
Ka = 4.6 * 10^(-3) M
Answer: Ka = 4.6 * 10^(-3) M