Originally posted by: da loser
Originally posted by: abeal2
Originally posted by: James3shin
fine Ka by using the relationship Ka(Kb)=Kw then use henderson-hasselbach
so
Ka x Kb = Kw
Ka x 10^-5 = 10^-14
Ka = 10^-9
pKa = 9
pH = pKa + [A]/[AH]
i know pKa and [AH] (AH being 10^-2). how do i find [A]?
i think that it's equilibrium, and since AH <-> A + H; A=H
i haven't done this in a long time, but i believe Ka or Kb is just a rate constant of the equilibrium, and since it's a weak acid the reaction does not go completely. so they give you AH, and that allows you to calculate how far the reaction goes using Kb