In a 0.870 M aqueous solution of a monoprotic acid, 3.51% of the acid is ionized. What is the value of its Ka?

Question

In a 0.870 M  aqueous solution of a monoprotic acid, 3.51% of the acid is ionized. What is the value of its K_a?

Answer 1

Let the monoprotic acid be represented as HA

HA <==> H⁺ + A^-

If [HA] = 0.870 M and 3.51% is ionized, then [H⁺] and [A^-] both = 3.51% x 0.870 = 0.0305 M

The Ka expression is ...

Ka = [H⁺][A^-] / [HA]

Substituting and solving for Ka, we have ...

Ka = [0.0305][0.0305] / [0.870]

Ka = 1.07x10^-3