When we have an acid reacting with water, we give the equilibrium constant a special symbol: Ka Consider the following reaction: HCN(aq)+H2O(I)H3O+(aq)+CN(aq)K=Ka=[HCN][H3O+][CN] The subscripted "a" just tells us that this particular K value is for the given acid reacting with water. Hence if I gave you this: Ka,HF=6.3104 I would expect you to know that this K value is for the reaction of HF reacting with water. You could write that reaction: HF(aq)+H2O(I)H3O+(aq)+F(aq) And then you could write the appropriate expression for Ka: Ka=6.3104=[HF][H3O+][F] Let's practice one: The Ka for chlorous acid, HClO2, is 1.2102. What is the equilibrium expression that accompanies this value when chlorous acid reacts with water? Just like with Ka, if you are given a Kb value, you know that is the K value for the reaction of the molecule acting as a base (accepting a proton) as it reacts with water. Try that here: The Kb for aniline, C6H5NH2, is 4.31010. What is the equilibrium expression that accompanies this value? KbKbKbKb=[C6H5NH3+][OH][C6H5NHH2]=[C6H5NH3+][OH]=[C6H5NH2][C0H5NH3+][OH]=[C6H5NH2][C6H5NH3+][H3O+] Given the two Kb values for these bases: Propylamine, C3H7NH2, with Kb=3.7104 Triethylamine, (C2H5)3N, with Kb=1.0103 would be the stronger base (the one that reacts more with water to produce more products). would have a higher concentration of [OH]. (Hint: Write out your Kb expressions for these two.) In the previous question you predicted which base, propylamine or triethylamine, would have the higher [OH]. Once we know that, we can predict which solution would have the higher [H3O+], plus their relative pH and pOH values, since all are interrelated. Review our previous equations: Kw=[H3O+][OH]pH=log[H3O+]pOH=log[OH]pH+pOH=14 Comparing propylamine (Kb=3.7104) to triethylamine (Kb=1.0103), propylamine would have a concentration of [H3O+]and a pH. (Note that we have "not" yet covered enough material to calculate these values. We are just making relative comparisons here.)