When a buffer (for example, a mixture of 100 mM acetic acid and 100 mM acetate), is diluted 10-fold, all the concentrations change drastically. Surprisingly, after re-establishment of equilibrium, the pH is not much different from the starting pH.

How can you visualize how diluting a buffer disturbs the equilibrium of a buffer and how the system returns to equilibrium? To make things simple in a thought experiment, we will posit that dilution is faster than acid/base chemistry, i.e. the first step is dilution without any reactions, and the second step is acid base chemistry without further dilution.

The explanation/visualization could be at the molecular or bulk level, and could use kinetic or K vs. Q or Gibbs free energy arguments.


2 Answers 2


Consider an acid buffer with $pK_a=5$. Ignore the hydroxyl concentration, assuming it is negligible.

Dilution leads to an identical (proportional) reduction in the concentration of all solutes: undissociated acid, complementary base (anion), and protium. Because the concentration of anion and protium is reduced, encounters between the two will occur less frequently. On the other hand, dissociation of the acid into component ions continues unabated, since this process is (we are ignoring activities) a first order reaction. The result is an increase in the concentrations of anion and protium, thus reestablishing a value of the protium concentration close to that prior dilution.

Other important factors should be remembered in the context of ideal buffers and the Henderson-Hasselbach equation: (1) the concentrations of $\ce{A^-}$ and $\ce{HA}$ are set equivalent in an ideal buffer (2) the $\ce{H^+}$ concentration is set equal to the $K_a$ and (3) the concentrations of $\ce{A^-}$ and $\ce{HA}$ are much higher than those of $\ce{H^+}$ . That means that in both absolute and relative amounts, the concentrations of $\ce{A^-}$ and $\ce{HA}$ need to change by only a small amount in order to bring the pH back near the desired set point. This is very clear from the following plots (concentrations are molar and $pK_a = 5$). Every 0.1 time units the concentration is diluted by 17%. The pH spikes up but then settles back again as the acid dissociates. You practically don't see the changes in the undissociated acid and anion on the plot during the recovery because they are tiny.

enter image description here

You can compare this to the case when your buffer is not as good (same $pK_a$, but lower capacity, 100 $\mu M$ concentration). I altered the kinetics too so the pH would stabilize after each dilution: enter image description here

I should add: ignoring $[OH^-]$ is admittedly bad near neutral pH/low buffer capacity, but then I was attempting to provide a visualization. Adding terms to cope with the hydroxyl equilibrium etc would complicate things without adding much value or accuracy (since the kinetics are not meant to be accurate anyway, just illustrative).

Addendum: Here's the code:

K = 1e-5;  % M
kd=10;     % arbitrary, set to alter convergence speed, numerical stability
dt = 1e-7; % arbitrary, alter to change convergence speed, numerical stability 
v0 = [0.2; 0.2 ; K]; % M [HA], [A], [H],

%% for dilute conditions
%dt = 1e-5;
%v0 = [0.0001; 0.0001 ; K]; % [HA], [A], [H]

Nstep = 10000;
t = [0:dt:(Nstep-1)*dt];
tdilute = t(round(Nstep*[0.1:0.1:0.9]));

dv_dt = @(kd,K,v) [ -kd*v(1)+kd/K*v(2)*v(3); kd*v(1)-kd/K*v(2)*v(3)] ;
dv_ = @(v,dt) dv_dt(kd,K,v)*dt;
dv = @(v) dv_(v,dt);
v = v0;
conc = v0;
for ii=2:Nstep
    vstep = dv(v);
    v = v + [vstep(1); vstep(2); vstep(2)];
    conc(:,ii) = v;
    if any(t(ii) == tdilute)
        v = v/1.2;  % <-- change factor to dilute by here

hold on
xlabel('time (au)')
  • $\begingroup$ Very cool, which tool are you using to simulate the kinetics? $\endgroup$
    – Karsten
    Feb 23, 2019 at 17:24
  • 1
    $\begingroup$ Well, yes and no. With a pKa =5 you are getting close to where the autoionization of water makes a difference. So if the ratios change depends on how many significant figures you're using and how dilute the buffer becomes. The limitations of the Henderson-Hasselbach equation seem to be often overlooked. $\endgroup$
    – MaxW
    Feb 23, 2019 at 18:09
  • $\begingroup$ @MaxW Yes and no? I did not use H-H explicitly in the calculations, but pointed out that they inform on restrictions for use of a buffer. $\endgroup$
    – Buck Thorn
    Feb 23, 2019 at 18:15
  • $\begingroup$ @KarstenTheis Sims can be done with matlab or octave. $\endgroup$
    – Buck Thorn
    Feb 23, 2019 at 18:17
  • $\begingroup$ I should add: ignoring $[OH^-]$ is admittedly bad near neutral pH/low buffer capacity, but then I was attempting to provide a visualization. Adding terms to cope with the hydroxyl equilibrium etc would complicate things without adding much value or accuracy (since the kinetics are not meant to be accurate, just illustrative). $\endgroup$
    – Buck Thorn
    Feb 23, 2019 at 18:31

Abbreviating the concentration of acetic acid with A, that of acetate with B, and that of hydronium ions with H, we can write the equilibrium constant as:

$$ K_a = \frac{B \times H}{A} $$

Multiplying by A and taking the negative base 10 logarithm, we get:

$$ pKa + pA = pB + pH $$

Now, we can visualize pKa (blue), pA (orange), pB (yellow) and pH (green). When the blue orange pillar has the same height as the green yellow pillar, the system is at equilibrium.

enter image description here

In the first part of the visualization (dilution), all concentrations drop by the same factor 10, so pA, pB, and pH increase by one, and the reaction is no longer at equilibrium (mismatch of pillar heights). In the second part of the visualization (net acid dissociation to re-establish equilibrium), the pH changes a lot while pA and pB change very little (imperceptibly) because acetic acid and acetate are present at much higher concentration (major species) compared to hydronium ions (minor species).

I'm interested in the calculation and numbers

Before dilution, you can estimate the pH using Henderson-Hasselbalch:

$$ \mathrm{pH} = \mathrm{pK_a} + \log([\ce{A-}]/[\ce{AH}]) $$

The pH is equal to the pKa (4.75), and the concentrations of acetate and acetic acid are 100 mM each if you neglect the small amount of acetic acid that has to dissociate to bring up the hydronium ion concentration (about 0.01 mM). For the thought experiment concerning 10-fold dilution, you just have to divide all the concentrations by 10. For the equilibrium after dilution, the simplest method is to use Henderson-Hasselbalch again. The pH will be equal to the pKa, and the concentrations of acetate and acetic acids are 10 mM each (again neglecting the 0.01 mM of additional dissociation of acetic acid).

  • 1
    $\begingroup$ I am interested in the answer, If possible to explain it by using numbers and calculation. $\endgroup$ Feb 23, 2019 at 16:58
  • 1
    $\begingroup$ How did you make the neat graph? $\endgroup$
    – Buck Thorn
    Feb 23, 2019 at 18:39
  • 1
    $\begingroup$ I used excel to do the calculations and then turned the frames into a gif with camtasia. $\endgroup$
    – Karsten
    Feb 23, 2019 at 21:02

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