pH calculations

Last time we described how strong and weak acids and bases behave in water. We basically saw that strong acids or bases disolve practically all the way, giving a [H⁺] or [OH^-] concentration equal to the initial concentration of acid or base, respectivelly. In a strong acid - strong base titration, both will react fully and the pH will be determined by the excess of either acid or base.

In weak acids and bases, the situation is different. The acid (or base) will not dissolve completely in water because the reaction is reversible: The conjugated base reacts with water and re-generates acid. The system reaches equilibrium and we have some protonated acid and some conjugated base in solution, such that they satisfy the equlibrium condition for that particular weak acid.

Say that we add a certain ammount of a weak acid AH to pure water to a final concentration AH_o. What would the pH be? The reaction we have to work with is the equilibrium of the acid:

Now, some of the acid will disolve to give us a final concentration x of conjugated base, giving us an equal concentration of H⁺, and some will remain behind, (AH_o - x). The easiest thing is to build a table with the initial concentrations, and those after dissolution takes place:
 

[AH]	[H+]	[A-]
AHo	~ 0	0
AHo - x	~ x	x

Note that the [H⁺] is ~ 0 initialy and ~ x after dissolution because we are not taking into account the H⁺ liberated by water through self-dissociation. Now we have to write out the equation with these numbers:

Now, we transofrm this into a quadratic equation and obtain x:

Solving it for x we get:

Say that we do it with and initial concentration of 0.1 M of CH₃COOH (keq = 1.74 x 10^-5). We get

Fine. Did we have to do all this math? How does [H⁺] compares to [AH]_o? Its two orders of magnitude smaller, so we could have just droped it from the denominator of the original equilibrium equation. If we try it that way:

For the same initial concentration of acetic acid (0.1 M), we get:

We see that we can simplify these calculations if we had recognized that the [H+] was going to be a lot smaller than the original concentration of acid. This will usually be the case when we are dealing with weak acids and bases.

Now lets try another case. Lets say we make a solution by combining a certain ammount of AH, AH_o, and some of the conjugate base as the salt, NaA. This is basically a buffer solution. Remember that NaA will dissociate in solution to give a certain inital concentration of A^-, A^-_o. We have:

Again, we build up the little table:
 

[AH]	[H+]	[A-]
AHo	~ 0	A-o
AHo - x	~ x	A-o + x

In this case, we have to remember that we have an initial concentration of congujate base because we dissolved a salt in water. Some of it will react with the liberated H+ from the acid, and some will appear from the the dissociation of the acid. If we write out the equilibrium equation:

Now, as we had done before, x will be very small compared to the initial concentrations of acid and base, because AH is a weak acid - We can eliminate them from the picture:

If we do some math with the logarithms we get:

This equation, as we will see later, is a very comvenient way of writing the pH as a function of the pK_a and the ratios of the acid and its conjugated salt. Despite that it is nothing more than re-writing the equilibrium equation and working the logarithms, a couple of guys figured it out and gave it their name: The Henderson-Hasselbach equation. PLEASE!!! Do not get 'married' to this equation. It works ONLY for the weak acid - conjugated base (or weak base - conjugated acid) pair in a range of pH values. Furthermore, working out the math is not that awful, and may save you from your own foly. In any event, the darn thing is pretty useful to do buffer solution problems.

Talking about buffer problems, lets fill in some numbers. Say that the AH/A- combination we used above was the infamous CH₃COOH/CH₃COO^- pair, and each had initial concentrations of 0.5 M. We get:

Good. First 'rule' of the buffer pairs: If the concentration of acid and conjugated base are the same, the pH will equal the pKa of the acid (the same thing with for base and its conjugated acid).

If we think of it in a slightly different way, we can see that the concentrations of the CH₃COOH and CH₃COO^- are the same if the pH equals the pK_a. This means that at a pH equal to the pK_a of a weak acid or weak base, half of it will be dissociated (make this one the second rule...). We will have to keep this one in mind when we discuss the ionization states of amino acid groups at different pHs.

Now, lets see about the buffering power of this solution, or how much it will resist to changes in pH. Say that we add a certain ammount of solid NaOH (no volume change to make life easy) to the solution above to make the final concentration of NaOH 0.1 M. The OH- liberated by the strong base will react with the protonated acid almost completely, giving us an equal concentration of conjugated base:

If we make the little table again, this time including OH to include the initial concentration of this ion:
 

[AH]	[H+]	[OH-]	[A-]
AHo	~ 0	Y	A-o
AHo - x - Y	~ x	-	A-o + x + Y

After the system reach equilibrium, the OH^- will have reacted with the protonated acid completely to give us the same concentration of conjugated base. Thus, the final concentrations of AH will be (AH_o - Y), and that of conjugated base (A^-_o + Y). Again, x is very small compared to (AH_o - Y) or (A^-_o + Y), so we don't consider it. Re-writing the buffer system equation for pH (I'll resist using the H-H word):

Using the concentrations for the acid, conjugated base, and the initial [OH-]. we get:

We see that there is a relatively small change in pH for a relatively big change in [OH^-]. Now lets see what would have happened if the pH had been kept at 4.76 only with HCl, a strong acid. The concentration of HCl would have been exactly that of [H⁺], or 10^-pH. This is 1.74 x 10^-5 M. Now, after adding NaOH to this HCl solution at pH 4.76 to a final NaOH concentration of 0.1 M, we would have (0.1 - 1.74 x 10^-5) M of [OH^-], which is approximately 0.1 M NaOH. Remember that strong acids react completely with strong bases. The final pH would have been 13.00!!!

So far, so good. We see that with a buffer system the pH of a system has small variations upon release of bases into solution. How about acids? The same thing, but in this case the conjugated base will be consumed, and more protonated acid will appear. For a Y final contentration of acid:

Using the concentrations for the acid, conjugated base, and an initial [H+] of 0.1 M. we get:

Again, if we had a pH of 4.76 based only on HCl, the final pH would have been 1!!!

Now, how much can we buffer a solution. Say that the Y concentration of NaOH in the example above was 0.4 M. If we do the math:

If we had added HCl instead of NaOH to the system to a final concentration of 0.4, the final pH would have been 3.81. We start seeing that the buffer system will only function in a range (pK_a - 1) to (pK_a + 1), which correspond to concentrations of added base or added acid slightly smaller than the initial concentration of AH/A- we had in the system. If we go either way now, all the acid will be entirelly deprotonated (A^-) or fully protonated (AH).

After this point, if we add more acid or base, all the conjugated base will react or all the weak acid will be deptrotonated, because the equilibrium will be pushed all the way to one side or the other. The buffer system breaks down, and the Henderson-Hasselbach equation is not worth the paper you are writing it on to solve the problem!!! This, I guess, will make you realize that you have to first think, then use packaged formulas...

Anyway, is there any way we can increase the power of our buffer system? We saw from the examples above that the ratio that is important here is (A^-_o + Y) / (AH_o - Y) in the case of an added base, or (A^-_o - Y) / (AH_o + Y) in the case of an added acid. From this, it is pretty obvious that is we increase the initial [A^-_o] and [AH_o] the ratio will vary a lot less with addition of acid or bases. Lets use the last example of base and acid additiona (0.4 M) in different setting:

First, lets make [CH₃COOH] and [CH₃COO^-] 1.0 M. For an addition of to an initial [NaOH] of 0.4 M, we will have:

For an initial [HCl] = 0.4 M:

Now, if we make [CH₃COOH] and [CH₃COO^-] 5.0 M, we will have

We see that the more concentrated our buffer system, the better it resists to changes in pH. In the lab, we have to decide on the buffer system composition and concentration following certain rules. First, we have to know the pH range we want our reaction/bugs to work. Second, we need a rough idea of how big the pH changes could be (and therefore how big the [H+] variation can be) so that we choose the right concentration of weak acid and conjugated base salt. There is an 'upper limit' imposed by the solubility of the acids and salts in pure water.

Next time we will start talking about diprotic acid (base) systems, and we will evaluate how to calculate group ionization states as a function of their pK and the pH of the solution.

Prepared by Guillermo Moyna, 1999.