The Ka of a monoprotic weak acid is 0.00361. what is the precent ionization of a 0.155 M solution of this acid

Answer 1

Equation is HA → H⁺ + A^- K_A = [H⁺][A^-]/[HA]

Let's say that x = HA that ionizes in molarity

.00361 = x²/(.155-x) This can be solved for x as a quadratic equation or you can assume that x<<.155

(Usually you want K_A to be about 10^-5 to get away with that - it doesn't work here)

After solving the equation for x %ionization = (x/.155)*100