The 𝐾 a Ka of a monoprotic weak acid is 0.00584. 0.00584. What is the percent ionization of a 0.181 M 0.181 M solution of this acid?

Answer 1

I'm assuming you meant the Ka = 0.00584 for this monoprotic acid, which we'll refer to as HA.

HA <==> H⁺ + A^-

Ka = 0.00584 = [H⁺][A^-]/[HA]

0.00584 = (x)(x)/0.181 - x and assuming x is small relative to 0.181, we will ignore it for now.

x² = 0.001057

x = [H⁺] = [A^-] = 0.0325 (this value is actually around 18% of 0.180 so our assumption was incorrect, and we must now use the quadratic equation to solve for x)

x² = 0.001057 - 0.00584x

x = 0.0297

% ionization = 0.0297/0.181 (x100%) = 16.4%