Chemlab: Chemistry 3/5


Acids, Bases, and Buffers 1: Monoprotic and Polyprotic Acids

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Chemistry & Background
Acetic acid is a monoprotic acid, with one acidic hydrogen atom. It contains the acid group of carbon-containing molecules, -COOH and is called a carboxylic acid. The Lewis structure of acetic acid and its conjugate base acetate ion are shown below



For convenience, we will write CH3COOH to denote the acid formula. In water, acetic acid establishes an equilibrium between the weak acid, acetic acid, and the conjugate base, acetate ion.

CH3COOH + H2O H3O+ + CH3COO- (1)

The equilibrium constant, Ka, for this reaction is written

(2)

For acetic acid, the value of Ka equals 1.76 x 10Š5 and pKa equals 4.75. The magnitudes of the Ka and pKa values of different weak acids give us a comparison of their relative strength. A weaker acid has less dissociation to the conjugate base and the equilibrium favors the undissociated weak acid form (as in Eqn. 1). This results in a smaller Ka value (in Eqn 2). A smaller Ka value corresponds to a larger pKa, since pKa = -log Ka. In other words, the weaker the acid, the larger the pKa value. Table 1, at the end of this week's experiment compares the Ka and pKa values for several weak monoprotic and polyprotic acids. From this table you can see that acetic acid is a stronger acid than those listed with a larger pKa and a weaker acid than those with a smaller pKa value.

When a weak acid is titrated, carefully measured volumes of strong base solution are added to it. Some titrations are done for the purpose of determining the concentration of a weak acid solution. In this case, base solution is added until the equivalence point is reached. At this point, enough base has been added to react with all the weak acid present in the sample. The endpoint can be determined with a suitable indicator, as you did in the titration of KHP and your seawater ion exchange effluent solutions in Week 4's experiment. In this week's experiment, you will determine the concentration of acetic acid in a commercial vinegar solution. You will use a pH meter, rather than an indicator, to determine the equivalence point of the titration. When the equivalence point is reached, the pH of the solution will change rapidly, because all the acid has reacted with the added base. The following example illustrates the stoichiometry of a titration of vinegar.


Example: Titration of Vinegar

10.0 mL of vinegar solution is titrated with 0.1800 M NaOH solution. After 45.87 mL of base are added, the pH increases sharply, indicating that the endpoint has been reached. What is the concentration of acetic acid in the vinegar solution?

Solution: First we calculate the moles of base required to reach the equivalence point:

45.87 mL x 0.1800 mmoles/mL = 8.257 mmol OH-

Since acetic acid and sodium hydroxide react in a 1 to 1 ratio, this is also the number of moles of acetic acid that reacted with the base.

8.257 mmol OH- x 1 H3O+ / 1 OH- = 8.257 mmol H3O+

From the moles of acid and the volume of vinegar in the original sample, we can calculate the concentration of acetic acid in vinegar:

8.257 mmol H3O+ / 10.0 mL = 0.826 M


In addition to finding the concentration of acetic acid in vinegar, your titration will provide information about the behavior of this weak acid throughout the pH scale. You will measure the pH vs. volume base (mL) added to the vinegar solution, to determine the titration curve. A theoretical titration curve for acetic acid is shown below.


Figure 1
Titration curve of acetic acid.

For this titration, a 10 mL volume of 0.5 molar acetic acid is titratated with 0.5 M NaOH. Because the concentration of acid and base are the same, the equivalence point is reached when 10 mL of base are added to the solution. The equivalence point, when enough base is added to react with all the acid present, exhibits the sharp increase in pH, as discussed above. This happens when the moles of base added equals the moles of acid in the sample. It is this point that you will use to determine the concentration of acetic acid in vinegar.

The flat portion of the titration curve before the equivalence point is called the buffer region. In this part of the pH scale, the acid and conjugate base are both present in significant concentrations and the solution resists changes in pH. As base is added to a solution in this buffer region, acetic acid reacts with it to form acetate ion, without a large change in pH. If additional acid were added to a solution in the buffer region, it would react with the conjugate base, acetate ion, and, again, the pH would not change appreciably.

In the middle of the buffer region lies the half-equivalence point. Here the volume of base added is half that required to reach the equivalence point and half the acetic acid has been converted to the conjugate base, acetate ion. This means that the concentrations of acetic acid and acetate ion are equal. If we examine the equilibrium expression at the half-equivalence point, we find something interesting:



At the half-equivalence point, [CH3COOH] = [CH3COO-], so

Ka = [H3O+]

And taking the log and multplying both sides by -1 yields

pKa = pH

So at the half-equivalence point, half the acetic acid has been converted to the conjugate base, acetate ion, and the pH will be equal to the pKa of the acid. This gives us an experimental way to determine the pKa of a weak acid. In this week's experiment, you will determine the pKa of acetic acid and compare it to the literature value. Next week, you may find this method useful for identifying unknown solutions of weak acids, bases, or buffers.

More About Buffers
In the buffer region of a titration, little pH change occurs upon addition of OH- or H3O+. In this region, the acid and its conjugate base are present in similar (within a factor of 10) concentrations. Such a solution has the important property of being able to resist changes in pH and is called a buffer solution.

Buffers and buffering ability play a central role in much of biology and medicine. Living systems require that their physiological fluids be maintained at certain pH values for the proper biochemical reactions to occur. Blood, because of its role in nutrient transport, must be maintained at pH 7.4, in spite of moving large quantities of CO2 from the cells to the lungs. This pH is maintained primarily by the buffering action of carbon dioxide, hydrogen carbonate, and carbonate ions. Two major components involved in the buffering of intracellular fluids are the dihydrogen phosphate and hydrogen phosphate ions involved in one of the acid-base equilibria that you will study in this week's experiment.

For experimental work in aqueous solutions, it is fundamentally important to be able to prepare a buffer solution at a desired pH. Based on what you know about the buffer region of a titration curve, a buffer solution capable of moderating H3O+ and OH- additions equally well would have equal concentrations of an acid and its conjugate base. The dissociation equilibrium relationship for the general weak acid HA can be written:

(3)

Rearranging (3) gives an expression for the hydronium ion concentration:

(4)

We then take the logarithm of both sides to obtain:

(5)

Multiplying both sides by -1 and rearranging gives an expression for the pH in terms of the log of the ratio [A-]/[HA] and the pKa value of the acid: (remember that log(a/b) = -log(b/a) and pKa = -log Ka)

(6)

Equation (6) is called the Henderson-Hasselbalch relationship. Using it, we can predict the pH of a solution from the concentration of an acid and its conjugate base. It can also be used as a guide to making a buffer at a given pH. There are a large number of acids with pKa's that span the common pH range, and generally one can be chosen with a pKa near the pH value which one wishes to buffer.

It is useful to consider the half-equivalence point again, using the Henderson-Hasselbalch relation. As we saw above, at the half equivalence point for a weak acid, [A-] = [HA] and pH = pKa (because log [A-] / [HA] = log 1 = 0). This solution will be equally effective as a buffer towards either H3O+ or OH- because equal amounts of acid and conjugate base are present. If you consider the titration curve, in Figure 1, you can see that the pH equals the pKa at a point midway between equivalence points, at the center of the buffer region.

The Henderson-Hasselbalch relationship is very useful in preparing buffer solutions. It certainly is possible to prepare a buffer solution starting from the pure acid (HA) and adding a strong base (i.e. NaOH) until the desired pH (in the buffer region of the acid) is achieved; conversely one could start with the pure base (A-) and add the appropriate amount of strong acid (i.e. HCl) until the same point is reached. The Henderson-Hasselbalch equation, however, allows us to calculate the correct ratio of basic form to acidic form which can be mixed to achieve the desired buffered pH. An example is acetic acid and its conjugate base, acetate ion, from sodium acetate.


Example: Acetic acid/acetate buffer

A buffer solution with pH 5 is to be prepared by adding sodium acetate and acetic acid to enough water to make 1.00 L of solution. The pKa of acetic acid is 4.75. Given 0.600 mol of sodium acetate, what amount of acetic acid should be added to produce 1.00 L of a buffer solution at pH = 5.00?

Solution: We first rewrite the Henderson-Hasselbalch equation as



In the buffer solution, virtually all 0.60 mol of sodium acetate, CH3COONa, will be completely dissociated to acetate ion, CH3COO-, and the acetic acid we add will be present as undissociated CH3COOH (to a very good approximation). Thus we solve the equation above for [CH3COOH], finding

[CH3COOH] = 10-0.25[CH3COO-] = (0.56) (0.60 mol/L of buffer solution)
= 0.34 mol acetic acid per L of buffer solution

We can check the calculation with the original Henderson-Hasselbalch equation:






Note that the Henderson-Hasselbalch relationship indicates that the pH of a buffer solution does not depend on the total concentration of the buffering acid and conjugate base but only on the pKa and the ratio of the concentration of these two species. On the other hand, the buffering capacity of a solution quantifies the amount of H3O+ or OH- the solution is capable of neutralizing before the acid or conjugate base form is saturated and the pH begins to fall or rise precipitously. This will depend on the total concentration of the acid and conjugate base buffer ions. For example, a solution 0.2 M in total acetate, distributed between acetic acid, CH3COOH, and acetate ion, CH3COO-, will be able to neutralize more H3O+ or OH- than a solution at the same pH that is 0.1 M in total acetate. Also, the buffering capacity may be different towards addition of acid than towards base. This will be true unless the pH of the buffer solution is identical to the pKa of the buffering acid-base equilibrium.

We will use the following definition, applicable to any solution, for the buffering capacity:

• the acidic buffering capacity of a solution is the number of moles of H3O+ per liter of buffer which are required to lower the pH by one unit

• the basic buffering capacity of a solution is the number of moles of OH- per liter of buffer which will raise the pH by one unit.

The figure below indicates how the acidic and basic buffering capacity of a buffer solution can be graphically determined from a titration curve, using the pH = 5.00 acetate buffer solution from the Example above.

Figure 3
Figure 2
Buffer Capacity

We take the measured volume of base added to raise the pH by one unit, multiply by the base concentration, and divide by the initial volume of buffer. This gives the base buffer capacity as moles of base per liter of buffer.

Polyprotic Acids
In contrast to a simple monoprotic acid like acetic acid, with only one equilibrium between the acid and conjugate base, a polyprotic acid contains more than one acidic hydrogen. For a polyprotic acid, n acidic hydrogens will exist in solution in equilibrium with n conjugate base forms (for a total of n+1 species). For example, when phosphoric acid (n = 3, a triprotic acid) is dissolved in solution, the following equilibria are established among the four species H3PO4 (phosphoric acid itself), H2PO4-1 (dihydrogen phosphate anion), HPO4-2 (hydrogen phosphate anion), and PO4-3 (phosphate anion):

(7-9)

Note that each successive pair of species is linked in an independent equilibrium with H3O+ and that each link has a unique equilibrium constant. If we want to understand how polyprotic acid solutions respond to chemical changes (such as addition of OH- or H3O+), we need to learn how to predict the concentrations of the four phosphate species, as well as H3O+ and OH-, under a variety of conditions. To account for all six species, we will use relationships such as chemical equilibrium reactions and associated equilibrium constant equations (Eq. (7)-(9)), the mass balance, electroneutrality and water auto-ionization equilibrium relationships (Eq.'s (10) - (12)):

mass balance (total phosphate):
[ H3PO4]initial = [ H3PO4] + [ H2PO4-1] + [ HPO4-2] + [ PO4-3] (10)

electroneutrality (charge balance):
[Na+] + [H3O+] = [ H2PO4-1] + 2[ HPO4-2] + 3[ PO4-3] + [OH-] (11)

The concentrations are multiplied by species' charges here. For example, one phosphate ion carries a -3 charge; thus, its concentration is multiplied by 3 in the charge balance equation.

water dissociation (auto-ionization) equilibrium:
2 H2O OH- + H3O+
Kw = [OH-][H3O+] = 1.0 x 10-14(12)

Therefore, we have six (not very simple) equations with six unknowns that describe our triprotic acid solution. Because these equations are non-linear (i.e. they are algebraic equations in which the variables--our species' concentrations--may be raised to powers other than 1), solving these equations for the concentrations is most easily done by a computer program. Simple elementary algebra of the "six equations in six unknowns" variety wonÕt work.

Despite this complicated set of equations, we can still understand the relationship between the concentration of acid and conjugate base pairs. Note that the ratio of concentrations of any two species in a polyprotic acid solution can be determined by using only the dissociation equilibrium constant (such as Eq.'s (7)-(9)) and [H3O+]. For example, from Eq. (7), we can write

concentration ratio

and find the phosphoric acid to dihydrogen phosphate concentration ratio at any pH.

For polyprotic acids with equilibrium constants well separated in magnitude (i.e. different by factors of 103 to 104 or so), approximations can be made which allow us to consider only those species involved in one of the deprotonation equilibria. For example, Ka1 for phosphoric acid is so much larger than either Ka2 or Ka3 that the initial part of a titration of H3PO4 against standardized NaOH involves almost entirely equilibrium (7) between H3PO4 and its conjugate base H2PO4-1. At pH values just past the first equivalence point, H2PO4-1 will be the dominant species in solution, and the next region of the base titration will involve the second deprotonation equilibrium, (8). Thus, a titration curve for a polyprotic acid which meets these criteria will appear to be a sequence of individual titration curves, with the post-equivalence region of one equilibrium overlapped with the buffer region of the next. In polyprotic acid systems such as these, at any pH value, one acidic form and its conjugate base will be present in much higher concentration than all other polyprotic acid forms. The concentrations of these minor forms can usually be ignored.

A theoretical titration curve for the titration of 10 mL of 0.1 M phosphoric acid with 0.1 M NaOH is shown below. Because the initial acid and the titrating base concentrations are equal and the initial acid volume was 10 mL, the first equivalence point should appear at 10 mL added base. The graph shows this point clearly, along with the clear second equivalence point at 20 mL added base. But the third equivalence point (at 30 mL added base) is not distinct. This point is washed out because Ka3 is close to Kw, the water dissociation constant. Phosphate ion is a fairly strong base and competes with OH- for hydrogen ions, at high pH. Note that the half-equivalence points are found in each buffer region, midway between adjacent equivalence points.

Figure 1
Figure 3
Titration Curve for Phosphoric Acid

This type of plot can be produced by acid/base software available in the General Chemistry folder on the Public File Server. This program allows you to select mono-, di-, or tri-protic acids by name and to explore the theoretical titration curves for such acids with any strong base. You may want to run this program with a series of "artificial" Ka values to see how the shape of the titration curve changes with the absolute magnitude and the relative closeness of Ka's. You can do this by creating your own acid or base. Instructions for using the software are at the end of this section of the manual.

The equilibrium expressions and the total phosphate mass balance expression can be manipulated to give equations for the fraction of total phosphate in any one form as a function of pH. The program also uses these equations to produce what are referred to as alpha plots, which look like the Figure 4.
Figure 4
Figure 4
Alpha Plot for Phosphoric Acid

This figure shows that at any pH, only one acidic form and its conjugate base dominate the concentration of all phosphate species. At, for instance, pH = 2, the concentrations of H3PO4 and H2PO4-1 are roughly equal while the concentrations of HPO4-2 and PO4-3 are negligibly small. We can use this graph to make the following table, showing the dominant acid-conjugate base pair at various pH ranges:

pH range Dominant species
0 - 4.7 H3PO4, H2PO4-1
4.7 - 9.7 H2PO4-1, HPO4-2
9.7 - 14 HPO4-2, PO4-3

This situation is fairly common, not only among polyprotic inorganic acids such as H3PO4, but also among biologically important acids, such as amino acids.

Table 1: Weak Acids, Ka, and pKa values

Acid HA A- Ka pKa
Acetic CH3COOH CH3COO- 1.76 x 10-5 4.75
Ammonium NH4+ NH3 5.6 x 10-10 9.25
Benzoic C6H5COOH C6H5COO- 6.46 x 10-5 4.19
Carbonic H2CO3 HCO3- 4.3 x 10-7 6.37
HCO3- CO32- 4.8 x 10-11 10.32
Chloroacetic CH2ClCOOH CH2ClCOO- 1.4 x 10-3 2.85
Citric C6O7H8 C6O7H7- 7.41 x 10-4 3.13
C6O7H7- C6O7H62- 1.74 x 10-5 4.76
C6O7H62- C6O7H53- 3.98 x 10-7 6.40
Formic HCOOH HCOO- 1.77 x 10-4 3.75
Phosphoric H3PO4 H2PO4- 7.52 x 10-3 2.12
H2PO4- HPO4-2 6.23 x 10-8 7.21
HPO42- PO43- 2.2 x 10-13 12.67
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