synergix
Mar31-09, 06:34 PM
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.
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.