1. The problem statement, all variables and given/known data Write two equations that illustrate that an aqeous solution of NaHCO3 can act either as an acid or a base. In pure water show quantitatively which of the two reactions predominate. ka1= 4.2x10-7, Ka2=4.8x10-11 3. The attempt at a solution as an acid: NaHCO3(aq) + H2O(l)---> H3O+(aq) + NaCO3-(aq) Base: NaHCO3(aq) + H2O(l) ------> OH-(aq) + NaH2CO3+(aq) The second part is where I am having some trouble. I am not sure if I am doing this correctly so i will try to show and explain what I think will happen as best I can. I think Ka1 will predominate the reason being that kb1= (1x10^-14)/(4.2x10^-7)=2.38x10^-8 so Ka1>Kb1 therefore this reaction will proceed. Meaning most of the sodium bicarbonate will act as an acid and release protons. whereas Kb2=(1x10^-14)/(4.8x10^11)=2.08x10^-4. so Kb2>>Ka2 so releasing a 2nd proton is not very favorable the species is much more likely to pick up another proton then it is to release another proton.