Introduction

The pH of any solution, acidic or basic, is solely dependent upon the concentration of ‘free’ protons, or hydronium ions (H⁺, H₃O⁺, H₅O²⁺, or H₉O⁴⁺) in solution. Mathematically, this relationship is taken to be the negative log of the concentration of the H₃O⁺ ion as shown:

On a side note, whenever you seen a value expressed as ‘p’ and some variable, for example pH, pOH, pK_A, or even pK_B, the lowercase ‘p’ immediately tells you to take the negative log of the number corresponding to that particular variable, [H+], [OH-], K_A, or K_B.

The concentration of hydronium ions in solution is easily manipulated by the addition of some chemical that reacts with the H₃O⁺. Specifically, this chemical alters the H₃O⁺ concentration by either consuming the acid, or supplying additional hydronium ions. There are solutions however that possess the ability to resist sudden and dramatic changes in H₃O⁺ concentration as additional acid or base is added. These solutions are often characterized as being ‘buffered’ and have therefore been coined buffers.

Buffers

Specifically, buffers attribute their resistive activity to the equilibrium between a weak acid, or base, and its conjugate, as shown in the generalized form below:

In order to thoroughly understand these so-called buffers, we will utilize a simple example buffer composed of formic acid (HCO₂H) and its conjugate sodium formate (NaHCO₂).

As you can clearly see from the equation presented below, the formate ion (HCO₂^-) is the conjugate base of the weak formic acid (HCO₂H). You should also be able to determine the equilibrium expression for this particular reaction; if you have already forgotten, all you have to remember is that the equilibrium expression for a reaction is given to be the concentration of all aqueous, or gaseous, products divided by the concentration of the same type of reactants. For our example, the following is true:

When formic acid is alone in solution, the concentration of formate at equilibrium is dependent only upon the acid’s dissociation constant (K_A)—for our example it is given to be 1.8 x 10^-4. In other words, the small value of K_A, like any other equilibrium constant, tells us that this reaction has very little tendency to progress in the forward direction and produce a significant amount of product. In fact, an initial concentration of 0.100 mol/L formic acid only dissociates only about 4%, a value you yourself should calculate to brush up on what you have been taught this term.

Common Ion Effect

Recalling that buffers resist dramatic changes in pH, what would happen if we added additional formate in the form of the specie’s soluble salt? Realizing that we have increased the concentration of formate on the product side, it seems rational that the hydronium ions formed from the dissociation of formic acid will donate H⁺ to the added HCO₂^- ions—which in fact they do. Particularly, this process occurs because of the common ion effect, a term utilized to characterize solutions containing two solutes with similar ions. In simpler terms, the ionization of our formic acid is limited by the presence of its conjugate base (HCO₂^-). Thus, the consequence of adding more formate to solution would lower the concentration of the hydronium ion, and as a result raise the pH, all of which we should have expected from Le Châtelier’s Principle.

Continuing, the most obvious consequence of adding formate is an increasing concentration of this specie. Specifically, this notion can be attributed to the miniscule amount of free H₃O⁺ in solution—remember formic acid only dissociates ~4%. In other words, an excess of formate exists because it is now at a concentration greater than that which would be normally: (1) produced from the forward reaction; and (2) consumed from the reverse reaction. As a result, the two reservoirs required for buffering are established—a reservoir for the weak acid, or base, and a reservoir for its conjugate.

The key to understanding the behavior of buffers is to remember that that the pH of the mixture is dependent upon the amount of free hydronium ion in solution. For example, if excess hydrochloric acid (HCl) is added to our buffer, it will readily donate H⁺ to the formate ion in reservoir 2. This is validated by looking at the equilibrium constant of the reverse reaction. Recall that the K_A for the forward direction was 1.8 x 10^-4, now if we take the reciprocal of this value—to obtain the equilibrium constant of the reverse direction—we arrive at a value of ~5,600. Thus, this tells us that the reverse reaction is heavily favored to form formic acid when an excess of H₃O⁺ is present. As a result, additional formic acid is formed to alleviate any potential increase in hydronium ion concentration and therefore a decrease in pH. In simpler terms, as long as there is enough formate present in solution, any excess acid added to the solution will be ‘absorbed’.

On the other hand, what happens if we were to add excess base to our buffer? In general, the behavior of buffers when a base is added can be viewed from several perspectives. The hydroxide ion (OH^-) is a much stronger base than formate; therefore the hydronium ions will readily donate H⁺ to the hydroxide ions and generate water when sodium hydroxide (NaOH) is added to the buffer. Consequentially, the formic acid in reservoir 1 then has to donate H⁺ to water to compensate for the loss of hydronium ions and restore the original pH, a process that will always occur as long as reservoir 1 possesses sufficient amounts of formic acid.

Le Chatelier's Principle

Another simple way to characterize a buffer's activity during the addition of a base is to employ Le Châtelier’s Principle. In particular, the formation of additional water molecules from the reaction of OH^- with H₃O⁺ causes a shift in the equilibrium to the product side. This explanation is a little less satisfying when we consider the vast amount of water in solution. Nonetheless, the overall result of either approach remains to be only a minor change in the solution’s pH.
 
At this point in the discussion we will now dive a little deeper into the realm of these ‘reservoirs’. On certain occasions, buffers can be described as being ‘centered’, meaning that they can absorb equal amounts of acid, or base, before a significant change in pH is observed. When this occurs, there is an equal concentration of weak acid, or base, and its conjugate in the buffering solution. In detail, ‘centered’ buffers have a pH value that is equal to the pK_A of the weak acid, a rationale that will be proven now.

Our example buffer, the reaction shown again above, possesses the dissociation constant shown:

At this point it is critical to observe that the expression contains both species making up our reservoirs, one in the numerator and the other in the denominator. Recalling that a buffer which is said to be ‘centered’ has equal concentrations of a weak acid and its conjugate base, the dissociation constant expression can be simplified to reveal that the hydronium ion in such a mixture is dependent only upon the dissociation of the particular weak acid. Mathematically, the above expression can be reduced to:

Finally, taking the negative log of both sides, we see that ‘centered’ buffer do in fact have pH values that are equal to the pK_A of the weak acid.

Well sure, a reservoir ratio of 1:1 is awfully nice, but it is seldom seen, and more often than not you will be dealing with ratios that are not 1:1. To see what happens under such circumstances let’s return back to our example and add some acid. Realizing that the addition of acid to the solution increases the free H⁺, it is rather simple to understand that the formate will react with the H⁺ and shift the equilibrium left to generate more formic acid.  In particular, the concentration of formic acid would increase by however many moles of acid was added, while the concentration of formate ion would decrease by the same amount. Mathematically the dissociation constant expression would now look like this, where x is the number of moles of acid added:

Hopefully you are asking yourself the question as to how we can be dealing with moles, and not molarity. Typically, most acids and bases are added to buffers in aqueous form, and the addition of either would alter the total volume of your buffer solution. This then would require you to re-calculate the concentrations of both reservoirs—quite simply an annoying calculation. However, there is hope! Because the new total volume will remain constant, it can be canceled out of the expression, leaving you with only a ratio of the respective amounts of moles. Now looking at our K_A expression, we can clearly see that change in pH would be the cause of a change in the mole ratio of our reservoirs, and as long as x is small with respect to moles of HCO₂^- and HCO₂H, the change in pH will in fact be minimal.

A similar, but opposite, analysis applies to the case if we were to add a base to our example buffer. You already know that addition of a base will increase the dissociation of formic acid into its conjugate and shift the equilibrium to the right. Like the approach taken above, this one does not require you to memorize any formulas or learn new concepts. It only uses the general principles that govern equilibrium, simple neutralization reactions, and what you just learned. With that being said, can you, on your own, setup the generalized expression for the addition of a base to our sample buffer?

The Henderson-Hasselbalch Equation

In summary, you have learned that the pH of a buffer is regulated by the pK_A of a particular weak acid and the ratio of its reservoirs, even when they are not 1:1. The determination of the pH of a buffer solution after the addition of an acid or a base is a relatively simple process if we think of the solution as only an acid, or a base depending on what the weak electrolyte is. Further aiding us in this task is the useful Henderson-Hasselbalch equation:

For this experiment’s case, the use of this equation will help you estimate the pH of a buffer, but it is also important, as you will learn in lecture, for finding equilibrium pHs of various acid-base reactions.

Another aspect that the reservoirs come to play a role in is the buffer’s capacity. Specifically, the capacity of a buffer is the amount of acid or base it can absorb before a significant change in pH is observed, a characteristic controlled by the ‘size’ of the reservoirs (concentration). While the pH is supposed to remain reasonably constant, it does change, even before one of the reservoirs becomes exhausted, a notion that will be exploited in this experiment.

With an extensive discussion of how a buffer works behind us, we can now devote the remainder of this discussion towards the correct procedure for producing a buffer of a desirable pH. The first step in any buffer design is to decide on the pH range suitable for your experiment. With a desired pH known, we then consult a table of dissociation constants, like those presented in Appendices H and I of your textbook, to select the appropriate acid, or base. Here we then have to calculate—yes we have to do a little math—what the correct ratio of our acid to conjugate will be. Finally, with the correct molar ratio known, we obtain the corresponding salts containing our weak electrolyte, measure out the corresponding number of grams, and make dissolved and distilled water.

In order to better understand the process, let’s now go through the detailed step by step process of correctly generating a buffered solution with our example buffer. Specifically, let’s assume we have been assigned the task of preparing 2.0-L of a 0.75 mol/L solution of this buffer at a pH of 4.10. The first step we will take is to determine the principal components of our buffer system. Generally this is a relatively easy task for monoprotic systems, but does get a tad more daunting when we have to work with diprotic and polyprotic solutions. From what we already know about our example, we can easily identify the conjugate acid-base pair as HCO₂H and HCO2-. After determine the components of our system, the next step is to calculate the desired ratio of the conjugate acid-base pair using the Henderson-Hasselbalch equation. For our example, the equation is as shown:

Since we are concerned only with the ratio inside the log, we need to rearrange the equation in order to make it easier to deal with. By subtracting pK_A from both sides of the equation and taking the anti-log of both sides we arrive at:

After calculating the pK_A for formic acid and subtracting it from our desired pH, our expression becomes:

which is then simplified further leaving an expression extremely easy to work with.

Now at this point of the process things start to get a little tricky since we are left with an equation with two unknowns. However, from what we know about the buffer acid and base concentrations—the sum of the having to equal 0.75 mol/L—we can solve the equation. Specifically, the first expression we can arrive it is just a simple rearrangement of our previous equation. As for the second expression, we know that the total concentration of our formate buffer cannot exceed 0.75 mol/L, thus we are left with the following expression:

Remembering all the way back to Algebra II, you should recall that when you have two equations with two unknowns, all you have to do is solve both of them for the same variable. For our example, let’s solve for the concentration of HCO₂^-:

Setting these two equal to each other, our example just becomes a simple algebra problem solved as shown:

Adding the [HCO₂H] term to both sides and solving we finally arrive at the concentration of formic acid we need in our buffer.

[HCO₂H] = 0.228 M

With a concentration for formic acid now known, we can substitute it back into one of our original two equations and solve for the necessary formate concentration. For our example, if we do the math right, we get a formate concentration of 0.52 mol/L.

With the correct concentrations of each component of our buffer, the next step is to determine the most feasible means of obtaining the desired components. In the lab you will be provided with a series of salts containing the necessary components, but for our example we will assume we have ample amounts of anhydrous formic acid and sodium formate. Further, upon dissolution, we will finally possess both components of the acid-base pair leaving us left to only figure out the correct mass of each component.
 
Realizing that we need 2.0-L of buffer, the next step is to take the concentrations just calculated and multiply them by two. In particular, this converts our concentrations into the number of moles of formate and formic acid, a number much easier to ‘measure’. With the number of moles and the corresponding molecular weights of both substances, we can obtain the number of grams required.

 

All that we have to do now is actually make the solution. Specifically, we would weight out 21.2 grams of anhydrous HCO₂H and 70.7 grams of NaHCO₂ and dissolve them in ~1.9-L of distilled water. We then check the pH, adjusting if necessary, and then bring the total volume to 2.0-L to complete our buffers preparation.

It is important to realize that when this is accomplished in the lab, that the results are only approximate as the solution pH is dependent upon the number of free hydronium ions in solution, which in turn is determined by both the solution’s temperature and the pH of the distilled water utilized to make the buffer—two factors not accounted for in our original buffer calculation. In fact, even if all of the above factors were carefully considered, we still would not observe the pH we thought. This is due to a particular topic which if you go further in chemistry you will learn far more about. Briefly, the ionic strength of particular species in solution varies from concentration to concentration. Because of this, the solubility of a particular specie can also be affected, and therefore alter the pH. Thus, when you initially check the pH of your buffer in this experiment, you will most likely get a pH several tenths off from what you expected. Do not chastise yourself, if you followed the procedure described above carefully you will make your buffer the right way, there are just some factors in chemistry that we cannot control. In order to ease your mind, think of this aspect like you think of the Ideal Gas Law. You know that gases do not behave ideally, yet you still use the PV = nRT equation to solve most questions dealing with gases. Whether you know it or not, there are still several aspects of chemistry where assumptions like this still take precedence.
 
As for your particular experiment, you will be investigating the hydrogen phosphate buffer system, an equilibrium that consists solely of ionic species, as shown:

Phosphoric acid (H₃PO₄) is capable of quickly changing into dihydrogen phosphate (H₂PO₄^-), an excellent buffer. Specifically, the strength of H₂PO₄^- as a buffer is due to its ability to either absorb H⁺, becoming phosphoric acid, or dissociate H⁺ and form hydrogen phosphate (HPO₄^2-). In further detail, both ions are readily obtained from potassium and sodium salts, and with a pK_A value of 7.20 the hydrogen phosphate buffer system can be used to generate buffers with a pH anywhere from ~6.90 to ~7.50.

On a final note, real laboratory chemists do not do these kinds of calculations before preparing buffer solutions. Buffer solutions with integer pH values are commercially available, and when buffers with non-integer pH values are required, approximate solutions are prepared and additional strong acid or base is carefully added. However, even with buffers commercially available, the concepts learned in the construction are vital to each and every chemist alike.