I seem to have a problem understanding this: If 1% of a weak acid dissociates in pure water. I would assume that 99% of it’s conjugate base would dissociate to form HA in pure water, but this is not the case: I tried to set up a situation below: Please help me understand what I’m misunderstanding Let’s assume we have a concentration of 1,0M acid with a Ka=10^-8 then we can discard waters Kw (due to the high concentration of acid=) and calculate the concentration of HA, A(-) and OH (-) HA + H20 = A(-) + H(+) Ka = x^2/(1,0-x) So at equilibrium we get x= 0,0001 =[A-] =[OH-] [HA] = 1,0 – 0,0001 = 0,9998 Now in situation 2, we add 1.0M of the conjugate base of the above acid to pure water and we get. The Kb= kw/Ka=10-6 A(-) + H2O = HA + OH(-) Kb=x^2/(1,0-x) x= 0.001= [HA]=[OH My question is this: How come the concentration of HA at equilibrium in the first acid + water reaction is different than the concentration of HA in the second base + water reaction? If the acid strength is inversely proportional to the conjugate base strength, should the amount of acid dissociating from from situation 1 not equal the amount of HA formed from situation 2? If 1,3% of an acid dissociates into a conjugate base, should 98,7% of it’s conjugate base not dissociate in water? What is the reason this is not the case?