What is the pKa of the new acid?

Question

When you prepare a 1.139M solution of the monoprotic acid, the measured pH of the solution is 4.619. What is the pKa of the new acid?

Answer 1

Let the monoprotic acid be HA

HA <==> H⁺ + A^-

From the pH, we can calculate the [H⁺] which is also the [A-]

pH = -log [H⁺]

[H⁺] = 1x10^-pH

[H⁺] = 1x10^-4.619

[H^+] = [A-] = 4.16x10^-5 M

Now we can write the expression for the ionization of the acid...

Ka = [H⁺][A-] / [HA]

Plugging in values and solving for Ka, we have...

Ka = (4.16x10^-5)(4.16x10^-5) / 1.139

Ka = 1.52x10^-9