[assaftolko]

We have the buffer Tricine 0.1M and pKa=8.15 and the pH=8.8

what are the molar conc. of tricine+ (the protonated form of tricine) and tricine0 (the deprotonated form of tricine)?

Well, if you go to henderson-hassalbach equation like they did in the solution I uploaded here, you get:
8.8=8.15+log [x/0.1-x]
where x is the molar conc. of both H3O+ and tricine0 at equillibrium, and 0.1-x is the conc. of tricine+ at equillibrium. Further solving gives x=0.082M
But I don't understand - if the pH is given to me can't I find x, which is also the H3O+ conc. at equillibrium according to the solution, from 8.8=-log[H3O+]? If I do that, I'll get a very different answer to x... why is that?
Also - if I already have the equillibria conc. of tricine+, tricine0 and H+, why can't I say:
zz  Ka=[tricine0][H+]/[tricine+] ---> 7.08*10^-9 = x*x/0.1-x ---> x=2.66*10^-5

Which is a very different result then the correct one... what's happening here? Is the reaction equation is given correctly in the solution that I uploaded?

[magician4]

If we're talking a tricine⁺ - buffer, we're not talking the dissociation of tricine⁺ "stand alone" !

for tricine⁺ "stand alone" your basic equation " tricine⁺  tricine⁰ + H⁺ " would apply  (i.e. [H⁺] = [A^-] ). on the other hand , [HA] [itex] \approx [/itex] [A-]  ( i.e. "buffering" ) doesn't apply here, as we're talking a weak acid.

for the respective buffer, the situation is completely different: here, the H⁺ is "forced" (by an external substance like additional NaOH in this case, for example), and [A^-] [itex] \neq [/itex] [H⁺] is the consequence


... and that's the whole point with your problem: whoever told you that for a buffer [A^-] equals [H⁺] was in error


regards

Ingo

[AWK]

There are two different problems - acid dissociation in water, and acid dissociation in the presence of  conjugated base (and this is your case).

[assaftolko]

Thanks to you both! So the solution is correct, only the reaction equation is written incorrectly! Would you believe that this equation and this question were given in last year's exam?!

But now another matter: Why do I necessarily have to conclude that tricine+ is 0.1-x and tricine0 is x and not let's say the opossite? (Which of course will give a different result for x)

[magician4]

Why do I necessarily have to conclude that tricine+ is 0.1-x and tricine0 is x and not let's say the opossite? (Which of course will give a different result for x)

you don't have to: the other way 'round (i.e you'd start your calculation with tricine⁰ as being of c₀ = 0.1 M) is a valid approach to the solution as well:

[tex] pH \ = \ pKa \ + log_{10} \left( \frac {0.1 \ - y }{y} \right) [/tex]

however, you would find that y = 0.1 -x will apply , and that the ratio [A-] : [HA] in equilibrium resulting therof would be the same as with the first approach

regards


Ingo