I should probably put this in the EK errors thread, but I thought that the Ka of water was simply 7. I also thought that the whole reason for the Henderson-Hasselbach equation (pH=pKa - Log (base/acid) was to find the pH of a buffer solution.
Sorry for the vague question but my computer is about to crash so I have to post fast!!!
lol yeah it is kind of vague. the pKa of water is 15.7ish. So the Ka of water is 10^-15.7 which is about 2*10^-16.
to determine this, you can use what you know about water:
Kw=[H+][OH-]=10^-14. Now, the concentration of water can be determined by determining the number of moles in one liter of water. So one liter of water is 1000 cm^3 of water and the density of water is 1 g/cm^3 so that means that you have 1000 g of water in one liter of water. The molecular weight of water is 18.02 g/mol so 1000g/(18.02 g/mol) is 55.49 mol and that's in one liter so the concentration of pure water is 55.49 M.
So since Ka=[H+][A-]/[HA], for water that's Ka=Kw/[H2O] soooo...
Ka=(10^-14)/55.49= 1.8*10^-16
The reason a lot of people think the pKa of water is 7 is because everyone knows the pH of pure water is 7 and people think, oh, well, the pKa is the pH at which the acidic and basic forms are in the same concentration. Well, that's true, but at pH 7 the acidic and basic forms of water are not in the same concentration. At pH 7, the concentration of H+ and OH- is the same. But OH- is the basic form (A-), and H20 is the acidic form, not H+. Remember, pH deals with proton concentration, and pure water has a pH of 7 at 25 deg C. That's all it means.
55.5 M is actually the concentration of water if none of it dissociated. So, from pH 0-13, the idea that [H2O]=55.5 M holds true. That's because [OH-] is much lower than the 55.5 M amount. Now at pH 14, [OH-]=1 M, right? But now the [OH-] is approaching the 55.5 M amount. So that means that the [H2O]=54.5 M approximately. So you can see now how as pH approaches 15.7ish, the [H2O] starts to drop significantly and the [OH-] starts to rise until at pH 15.7, [H2O]=[OH-].
So that brings me to the next part, the Henderson Hasselbach equation. All the Henderson-Hasselbach equation is is a reorganized form of the Ka expression. It is derived by simply manipulating the expression slightly. Anyway, the point is, you use it for many purposes. In reality, the most common purpose is to determine how much of a species is protonated and deprotonated. So, look at ethylamine. pKa of ethylamine is 10.7. When the pH of a solution of ethylamine is 10.7, that means that the concentration of the deprotonated form is equal to the concentration of the protonated form. (for ethylamine this means the neutral amine versus the ammonium form). This is what the Henderson-Hasselbach equation tells us:
pH=pKa+log([A-]/[HA])
so for our example, pH=10.7, pKa=10.7:
10.7=10.7+log([EtNH2]/[EtNH3+])
so 0=log([EtNH2]/[EtNH3+])
1=[EtNH2]/[EtNH3+]
so [EtNH3+]=[EtNH2]
now, if pH is one unit above the pKa, so pH=11.7:
10.7=10.7+log([EtNH2]/[EtNH3+])
so 1=log([EtNH2]/[EtNH3+])
10=[EtNH2]/[EtNH3+]
so 10[EtNH3+]=[EtNH2]
so that means that approximately 90% of the amine is in the neutral, deprotonated form, right?
this makes sense, because at pH 11.7, the solution is now more basic, so more protons are pulled off the ammonium into solution.
at one pH unit below the pKa, so at pH=9.7, 90% is in the cationic, protonated form, for the same reasons. the solution is more acidic now, so there are more protons in solution so that drives the equilibrium towards formation of HA which in this case is the ammonium.
quickly, at 2 pH units above pKa, 99% is deprotonated, and at 2 pH units below the pKa, 99% is protonated.
so that's the primary application of the henderson hasselbach equation. Now, if you happened to have both A- and HA and you knew the pKa, you could calculate the pH of your buffer. So like, if I had sodium acetate and acetic acid, and I added 1 M of each, then my pH would be 4.76 because the pKa is 4.76 and i have an equal amount of the protonated and deprotonated form. However, making a buffer this way is extremely rare, since all you really have to do is add some acetic acid to water, and then raise the pH with NaOH until you got to 4.76 and you'd have the same thing. Or you could start with sodium acetate and drop the pH with HCl.
Anyway, so I hope that helps.
