[Cacoethes]

Solve this problem: Prepare a pH 8.5 buffer with a total concentration of acid + conjugate base of 0.05 M.

Ok, thus far I know HH equation is pH = pKa + log [A]/[HA]

We know the pH and the sum acid + conj. base is 0.05.

Therefore -
pH = 8.5
[HA] + [A] = 0.05

I know that if we had pKa the problem could easily be solved.  I'm not positive as to how and acquire the pKa with the given data.  Any suggestions/hints?  Couldn't i simply choose a pKa + or - 1 to the pH?  Thanks.

[Borek]

You should select a real acid with pKa close to 8.5.

[Cacoethes]

Alright, ty Borek.  Would NH4+ be acceptable with a pKa of 9.25?

If so, now the eq. pH = pKa + log [A]/[HA] can be subbed.

8.5 = 9.25 + log [A]/[HA] --> 10^(-.75) = [A]/[HA] --> .17 = [A]/[HA]

We also know that [A]+[HA]=.05

Now we can solve for either HA or A and plug it into the original equation.  Does this look ok admins?

[Borek]

Ammonia will do, but check boric acid and bicine.

[Rudi]

The equation is, incidentally, not called "Henderson-Hasselbach" but Henderson-Hasselbalch.

This should be fixed in the headline.

[Borek]

Good point, missed that.

[Cacoethes]

My apologies for rehashing this thread, but I forgot to ask one more important question.  As it seems multiple acids could have been chosen with a different pKa, all close to the pH of 8.5, wouldn't this alter the results for [A] and [HA] respectively?  Or does this not matter in the case of this problem?  Thanks again.