Equilibrium: Conjugate Acids and Bases

1 lab period; work in pairs. Complete the Preparation page before coming to lab.

Goals

Background

According to the Bronsted view, an acid is a substance capable of donating a proton, H+, to a base. Thus an acid is a proton donor, and a base is a proton acceptor. Reaction between a generic acid, HA (note that we represent the donatable proton explicitly) and a generic base, B, is shown in equation 1:

(1)

HA +

:B ®

BH+ +

A-

acid

base

acid

base

 

B uses a non-bonding pair of electrons to attach the proton; this pair is shown in the equation. The reverse of reaction 1 also involves proton transfer, with BH+ serving as the acid and A- as the base. The pair of species, HA and A- are related by gain or loss of H+. As such they are related and are therefore called a conjugate acid-base pair. B and BH+ are also a conjugate acid-base pair.

Our primary interest in acids in this experiment involves their reaction with water, shown in equation 2.

(2) HA(aq) + H2O ® H3O+(aq) + A-(aq)

In this reaction, water functions as a Bronsted base. The extent of reaction 2 is indicated quantitatively using the equilibrium constant, Keq. The equilibrium constant expression for 2 is given in 3:

(3) Keq = Ka = [H3O+(aq)][A-(aq)]/[HA(aq)]

The equilibrium constant for reaction of an acid with water is usually symbolized Ka, to remind us of the type of reaction being dealt with. The reactant water, since it is present in huge concentration and is thus essentially a pure liquid, is not included in the Ka expression. The strength of an acid in aqueous solution is defined in terms of the magnitude of Ka for reaction 2. Strong acids have Ka values larger than 1; weak acids have Ka values less than 1. The common strong acids are HCl (hydrochloric), HBr (hydrobromic), HI (hydroiodic), HNO3 (nitric), H2SO4 (sulfuric), HClO3 (chloric) and HClO4 (perchloric). All other acids that we will commonly encounter are weak; i.e., their reaction with water according to 2 occurs to only a minor extent.

The equilibrium established when a weak acid reacts with water according to 2 can be explored using a procedure called titration, involving the following steps:

Example. 32.16 mL of 0.1042 M NaOH is needed to titrate a 25.00-mL aliquot of a solution of H3PO4 of unknown concentration, according to

3NaOH(aq) + H3PO4(aq) ® Na3PO4(aq) + 3H2O

Calculate the concentration of H3PO4 in the original solution.

Solution. We can calculate the moles of NaOH used; divide it by 3 to obtain the moles H3PO4 present; and divide by 25.00 mL, the volume in which this number of moles was contained:

moles NaOH = 0.03216 L x 0.1042 moles/L = 3.3511 x 10-3 moles (we retain more significant figures than justified until the end).

moles H3PO4 = moles NaOH x 1mole H3PO4/3 moles NaOH

= 3.3511 x 10-3 x 1/3 = 1.1170 x 10-3 moles H3PO4

[H3PO4] = moles/volume = 1.1170 x 10-3/0.025 L = 0.04468 M (4 significant figures are justified).

Titration involves a very important practical problem: how is the equivalence point detected? Unless we can detect the equivalence point, we will not know when to stop adding titrant. Clearly, the pH of the solution must change as the titration proceeds. At the beginning of the process, before base is added, the pH of the solution is fairly low because it contains acid. As titration proceeds, acid is neutralized by the added base, and pH rises. Addition of base after all of the acid has been neutralized produces a basic solution, with a high pH. During the titration, then, pH runs the gamut from low to high. We can detect the equivalence point from the manner in which the pH change occurs.

Two methods are commonly used to detect the equivalence point in a titration. In the first, an appropriate acid-base indicator is used to signal the equivalence point via a color change. In the second, an instrument called a pH meter is used to monitor the pH of the solution as base is added during the titration. A pH meter responds to the electrical potential of an electrode immersed in the solution being titrated. This potential is a function of [H3O+] in the solution. A plot of pH versus the volume of titrant added to the solution gives the so-called titration curve. The experimental curve for titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH is shown in Figure 1. We observe that

This provides us with the first method for determining the equivalence point: we successively add small volumes of base, measure pH after each addition, and plot the titration curve, from which we may find Vbase at the inflection point (the equivalence point). Moles of acid in the original aliquot is calculated as follows:

(4) moles acid = Vbase at inflection point x M base

The titration curve for a weak acid, for example acetic acid, looks a bit different from that for a strong acid such as HCl. In this experiment we will explore the shape of the weak acid titration curve, and its relationship to the equilibrium constant expression in (3). We will carry out titrations on a very small scale, using a very small buret (a microburet), which may be constructed from inexpensive and readily available materials.

Focus Questions

  1. What effect does the presence of conjugate base have on the initial pH of the acid?
  2. What effect does the presence of conjugate base have on the equivalence point pH in the titration of acid with NaOH?
  3. For each titration, determine the point during the titration at which the concentrations of conjugate acid and base are the same. What is the pH at this point? Are the 4 pH values the same or different?
  4. What is the significance of the pH at which the concentrations of conjugate acid and base are the same?
  5. How are the titration curves of pure conjugate acid and pure conjugate base related?
  6. Using data from your titrations, construct a plot of pH versus log R (R = [A-]/[HA]) for your conjugate pair, and explain how Ka for the acid can be determined from this plot. (HINT: Can you use pH data from experiments 3, 4, 6, and 7?)

Equipment and Materials