posted by Amandeep on .
What is the percent ionization of a 1.38 mol/L weak acid if its Ka = 2.7 x 10^-6?
Write the ionization equation. I would call the weak acid HA.
Write the Ka expression.
Set up an ICE chart if that will help.
Let y = (H^+), then (A^-) will be y and the unionized HA will be 1.38 - y.
Substitute these variables into the Ka expression and solve for y = (H^+).
Then percent ionization =
[(H^+)/(HA)]*100 = ??
Post your work if you need further help.