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
Equipment and Materials
Note to instructor: Click here for recipes for preparation of solutions.
Safety
Safety goggles must be worn at all times in the laboratory. Dilute solutions of acids and bases are irritating to the skin. In the event of skin contact, rinse thoroughly with plenty of water.
Experimental
Part 1: Rinsing and filling the buret
Record all data in your notebook.
Obtain a 25-mL buret and set it up using a buret clamp on a ring stand. Obtain 25-30 mL of standardized 0.10 M NaOH. Record the exact concentration of the NaOH in your notebook. Rinse the buret 3 times with 3-5 mL portions of NaOH, then fill the buret with NaOH to near the 0.00 mL mark.
Part 2: Titration
Your team will be assigned a pH meter and a particular acid-base pair from among the following to investigate:
0.1 M stock solutions of all acids and bases are available in the laboratory. Obtain supplies of your assigned acid and base in appropriately labelled beakers. You will do five pH titrations, each requiring at most 5 mL of acid solution and at most 5 mL of base solution, so take only a little more of the stock solutions than you will need. Record the exact concentrations of the stock solutions. Keep the reservoir beakers covered with watch glasses at all times to retard evaporation and exclude CO2 from the air.
(NOTE: this step may or may not be necessary; check with the instructor.) Before beginning, calibrate the pH meter using standard buffer solutions of pH 4 and pH 7, or pH 7 and pH 10. Your instructor will provide instructions for doing the calibration and will tell you which two buffers to use.
Calculate the volume of standard NaOH that will be needed to react exactly with 5.00 mL of your assigned acid.
Experiment 1: Pipet exactly 5.00 mL of acid solution to a beaker. Add exactly 5.00 mL of water and 1-2 drops of an appropriate indicator (alizarin yellow R for H2PO4-; phenolphthalein for the other acids. Titrate with standard NaOH to the indicator endpoint. This titration gives you a quick measure of how much NaOH is required to reach the equivalence point. Rinse the beaker with distilled water and dry it.
Experiment 2: Pipet exactly 5.00 mL of acid solution to the beaker. Add exactly 5.00 mL of distilled water. Rinse and dry the pH electrode and submerge it in the solution. RECORD THE INITIAL pH OF THE SOLUTION. Initiate the pH titration by adding 0.5 mL of NaOH, stirring the solution thoroughly, and reading the pH. Make sure that the meter reading is stable--that is, does not fluctuate or steadily change in one direction. After each pH reading, record the total added volume of NaOH and the pH. Continue adding NaOH 0.5 mL at a time until the total added volume is 80-90% of the volume of NaOH used in Titration 1. At this point, add the NaOH in smaller increments of first 0.2, then 0.1 mL. Continue adding NaOH incrementally until you are past the equivalence point (pH is approximately 11 or greater). When finished, discard the solution. Rinse the pH electrode and beaker with distilled water, and dry both.
Experiment 3: Pipet exactly 5.00 mL of acid solution and exactly 2.50 mL of base solution to the beaker. Add exactly 2.50 mL of distilled water. RECORD THE INITIAL pH. Titrate with NaOH in the manner described above under Experiment 2. Rinse and dry the beaker.
Experiment 4: Pipet 5.00 mL of acid solution and 5.00 mL of base solution to the beaker. RECORD THE INITIAL pH. Titrate with NaOH in the manner described above under Experiment 2. Rinse and dry the pH meter probe and beaker.
Expel the NaOH solution from the buret and rinse it thoroughly four or five times with distilled water. Rinse the buret twice more with standard HCl solution, then fill with HCl.
Experiment 5: Pipet 5.00 mL of base solution to the beaker, and add exactly 5.00 mL distilled water. RECORD THE INITIAL pH. Titrate with HCl solution (record the concentration!) until you are past the equivalence point (the pH is approximately 2).
Experiment 6: Pipet 5.00 mL of base solution, 2.50 mL of base solution, and 2.5 mL water to the plastic beaker. MEASURE THE pH. Rinse and dry the pH meter probe and beaker.
Experiment 7: Pipet 5.00 mL of base solution, 1.25 mL of acid solution, and 3.75 mL water to the beaker. MEASURE THE pH. Rinse and dry the pH meter probe and beaker.
When finished, clean all glassware; rinse the pH electrode with distilled water, dry it, cap it, and turn the meter off.
Disposal
Spent titration solutions may be flushed down the drain with plenty of water.
Part 3: pKa measurements
Based on the conclusions from Part 2, devise a method to estimate the pKa of an acid from a single pH measurement. Then apply your method to estimate the pKa's for the acids assigned to you. When finished, clean up and wait for postlab discussion.
Disposal
Solutions may be poured down the drain with plenty of water.
Clean-up. When you have finished all of your work:
Preparation
Equilibrium: Acids and Bases
Read
Problems
1. Hypochlorous acid, HOCl, reacts with water as follows:
HOCl(aq) + H2O ®
H3O+(aq) + OCl-(aq)
Ka =
[H3O+(aq)][OCl-(aq)]/[HOCl(aq)]
The equilibrium constant, Ka, for this process has the value 3.5 x 10-8.
a. What is the value of the concentration ratio, [OCl-]/[HOCl], when pH = 5.25? 9.16?
b. At what pH will [OCl-] = [HOCl]?
c. What is the significance of the pH at which [HOCl] = [OCl-]?