Mixing Acids and Bases Practice III – BUFFERS – Chemistry 121A Hanson
When mixing acids and bases, what are the possibilities? Strong with strong, strong with weak, weak with weak,
more acid than base, more base than acid, an “equivalent” amount of each… It’s quite the matrix of situations.
Believe it or not, you can now solve just about any question that starts with “What’s the pH of…” for relatively
simple systems. We have a few more we need to consider, and we need to establish what the limitations are. Most
“weak + weak” situations are quite beyond us, and polyprotic acids such as H3PO4, which lead to multiple
concurrent equilibria, those are going to be treated in the junior-level course Analytical Chemistry, which is
mostly for chemistry and environmental studies majors. So we won’t be going there this time around.
So what’s left? A very famous situation is one where in solution there are substantial concentrations of both a
weak acid and its conjugate base. For example, HF and F− or NH4+ and NH3. This might come from adding a
limiting amount of strong acid to a weak base, adding a limiting amount of strong base to a weak acid, or from
adding both a weak acid, such as HF, and a salt of its conjugate base, such as NaF, to water.
This situation, called a buffer, is incredibly useful in all areas of chemistry, from organic chemistry to analytical
chemistry, to biochemistry, to molecular biology, to environmental chemistry, in cooking, everywhere. If you can
“get” what a buffer is and how it works now, it will serve you for life. I kid you not.
A solution is prepared by mixing 25 mL of 0.20 M NaOH and 100 mL of 0.10 M HF. What is the expected pH?
(a) OH− (5 mmol) and HF (10 mmol) are the principal species.
(b) stoichiometric reaction is HF + OH−  F− + H2O
(c) 10 mmol of HF reacts with 5 mmol of OH− to give 5 mmol of F− with 5 mmol of HF remaining
(d) equilibrium is HF + H2O
F− + H3O+ with K = Ka = 3.5 x 10-4
−
(e) [HF]o = [F ]o = 5 mmol / 125 mL = 0.040 M; we approximate [H3O+] o = 0.
(f) algebra gives (0.040 + x)x/(0.040 – x) = 3.5 x 10-4 where [H3O+] = x = 3.44031E-4, and pH = 3.46
That’s the long way. There is nothing new here! Feel free to use that. But we could have used the HendersonHasselbalch approximation. Note the little “o”s indicating “initial”:
pH ≈ pKa + log([F−]o/[HF]o) = pKa + log((mmol F−)o/(mmol HF)o)
and we have in this case (the first three steps are STILL NEEDED):
(d) pH ≈ -log(3.5 x 10-4) + log(1.0) = 3.46
This approximation assumes that x in (0.040 + x) and (0.040 – x) is small. It’s a pretty good approximation
provided the two concentrations are both significantly larger than the Ka. For the following problems, predict the
pH both ways, using algebra and using Henderson-Hasselbalch. Compare your results. How good is this
approximation?
1. A solution prepared by adding 8.2 mmol of NaF and 3.5 mmol of HF to enough water to make 750 mL.
2. A solution prepared by adding 20 mmol of NH4Cl and 10.0 mmol of NH3 to enough water to make 500 mL.
3. A solution prepared by adding 15 mL of 0.12 M NaOH solution to 40 mL of 0.10 M HF solution. [Don’t forget
to do Steps (b) and (c)!]
The next three problems illustrate why buffers are so useful. Compare your answers with those above to see why
this is so cool.
4. Solution (1) diluted to 3.0 L.
5. Solution (2) mixed with 10 mL of 0.10 M NaOH.
6. Solution (2) mixed with 10 mL of 0.10 M HCl.
Answers:
algebra
Henderson-Hasselbalch
1.
3.84
3.83
no significant difference
2.
8.95
8.95
no difference
3.
3.38
3.37
no significant difference
4.
3.90
3.83
HH is about 2% off (note that [H3O+] was about 10% off)
5.
9.01
9.01
no difference
6.
8.88
8.88
no difference