1 lab period; work in pairs. Complete the Preparation page before laboratory.
Goals
Background
When dissolved in water, acids transfer the acidic proton to water to varying extents according to the following generic reaction:
(1) HA(aq) + H2O ®
H3O+(aq) + A-(aq)
Ka =
[H3O+][A-]/[HA]
The strength of the acid relative to H3O+ is indicated by the size of the equilibrium constant, Ka. Acids with Ka >>1 are strong acids, for which reaction (1) proceeds essentially completely to the right. Perchloric acid, HClO4, is an example of a strong acid. Acids with Ka < 1 are weak acids. Formic acid, HCOOH, is an example of a weak acid. For weak acids, relatively few HA molecules transfer a proton to water, resulting in relatively small concentrations of H3O+ and A- in solution. Acids with Ka < 10-14 are very weak acids; they have no measurable tendency to transfer a proton to water. Examples of very weak acids are NH3, H2, and CH4.
Reaction (1) is called a proton transfer reaction, because it depicts the transfer of a proton, H+, from HA to H2O. Proton transfer processes are the focus of the Bronsted-Lowry theory of acids and bases, according to which an acid is a proton donor, and a base is a proton acceptor. Note in (1) that loss of a proton from the acid, HA, results in a species, A-, that is a proton acceptor. Thus when an acid reacts, it becomes a base! The closely related pair of species, HA and A-, is referred to as a conjugate acid-base pair. They are related via addition or loss of a proton, as in (2):
(2) HA(aq) = H+(aq) + A-(aq)
HA =
acid; A- = base
Water and hydronium ion, H3O+, are also a conjugate pair. Any proton transfer process involves two conjugate acid-base pairs, as shown below:
(3) HA + B- ® HB + A-
If HA is a stronger acid than HB, the reaction will proceed primarily to the right. If HB is the stronger acid, reaction will proceed primarily to the left.
We can achieve a couple of interesting results by rearranging and manipulating the equilibrium constant expression for the proton transfer process, (1).
(4) Ka/[H3O+] = [A-]/[HA] = R (for ratio)
This arrangement of the expression shows us that the ratio of the base ("deprotonated" form, A-) to acid ("protonated" form, HA) forms of HA depends on, first, how strong the acid is (Ka); and on what the concentration of H3O+ happens to be. If [H3O+] is large, there are lots of protons around, which tends to shift (2) to the left in favor of the conjugate acid form. Thus the value of R (= [A-]/[HA]) is small. If [H3O+] is small, there is a paucity of protons, shifting (2) to the right and favoring the conjugate base, A-. Thus R is large. A very interesting situation is that in which the hydronium ion concentration is equal to the value of Ka for the acid. In this circumstance, the value of R is precisely 1; that is, the concentrations of conjugate acid and base are equal in the solution.
A commonly seen version of (4) can be developed by, first, taking the common logarithm of both sides of (4):
log Ka - log [H3O+] = log [A-]/[HA]
rearranging to isolate -log[H3O+] on the left side:
-log[H3O+] = -log Ka + log R
and, finally, recognizing that -log[H3O+] is pH:
(5) pH = pKa + log [A-]/[HA]
Here we have generalized the concept of pH to any quantity, X, so that pX = -log X. Thus pKa is simply the negative of the log of Ka for the acid.
Equation (5) is called the Henderson-Hasselbalch equation. It leads directly to the following very useful general statements:
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:
The buret is a graduated (i.e., marked) glass tube fitted with a stopcock (a valve) at the bottom. The tube, or barrel, has a capacity of 25 mL. Graduation marks completely encircling the barrel correspond to whole numbers of milliliters. Between any 2 of these marks are 9 additional marks corresponding to tenths of a milliliter (0.10 mL). These marks are separated by a distance of approximately 1 mm. To read the level of liquid in a buret, locate the bottom of the liquid meniscus curve, which normally will fall between two of the 0.1-mL marks. To estimate hundredths of a milliliter, assess the position of the meniscus between the two 0.1-mL marks. For example, suppose the meniscus falls about 40% of the way between the marks corresponding to 2.2 and 2.3 mL. The liquid level would then be read as 2.24 mL. The volume of liquid delivered from the buret in a titration is the difference between the liquid levels at the end and at the start of the titration.
Example. 32.16 mL of 0.1042 M NaOH is needed to titrate a 25.00-mL aliquot of a solution of HCl of unknown concentration, according to
NaOH(aq) + HCl(aq) ® NaCl(aq) + H2O
Calculate the concentration of HCl in the original solution.
Solution. We can calculate the moles of NaOH used; realize that 1 mole of NaOH will react with 1 mole of HCl; 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 HCl = moles NaOH x 1mole HCl/1 mole NaOH
= 3.3511 x 10-3 moles HCl
[HCl] = moles/volume = 3.3511 x 10-3/0.025 L = 0.1489 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 instrument called a pH meter is used to monitor the pH of the solution as base is added during the titration. In this method, 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:
moles acid = Vbase at inflection point x M base (4)
The second method for detecting the equivalence point is to use an appropriate indicator, a weak acid that exhibits different colors in its protonated and deprotonated forms. The indicator is chosen so that it changes color at the pH of the equivalence point for the acid to be titrated. In this experiment we will make use of 2 common indicators, phenolphthalein and methyl red.
Focus Questions
Equipment and Materials
Note to instructor: Click here for recipes for preparation of solutions.
Safety
Safety glasses must be worn at all times in the laboratory. You will work with solutions of acids and bases. Exercise due caution. If you should spill solution on your skin, rinse with copious volumes of water.
Experimental
Record all data in your notebook.
Initial Activities. Obtain a 25-mL buret from the instructor and set it up in a buret clamp on a ring stand. Obtain about 25-30 mL of 0.10 M NaOH. Rinse the buret 3 times with 3-5 mL portions of NaOH solution, then fill the buret to near the 0.00 mark. IT IS NOT NECESSARY TO ADJUST THE LIQUID LEVEL TO EXACTLY THE 0.00 mL MARK!
The instructor will assign you an acid or base to investigate.
Determination of Concentration of NaOH Solution. Using a 5-mL graduated pipet, transfer a 5-mL aliquot of the standard solution of potassium hydrogen phthalate to a small, clean Erlenmeyer flask or beaker. Read and record the initial volume level in the buret. Titrate the KHP with the approximately 0.10 M NaOH solution in the buret, using phenolphthalein as indicator (1 or 2 drops). Read the final volume level in the buret. Repeat the titration twice more. Your 3 delivered volumes of NaOH should agree within 0.05 mL. Determine the concentration of the NaOH solution to a precision warranted by your titration results. You have now standardized the NaOH solution.
Determination of Concentration of HCl Solution. Titrate a 5-mL aliquot of the approximately 0.1 M HCl solution with the standard NaOH solution, again using phenolphthalein as indicator. Repeat the titration twice more, and determine the concentration of the HCl solution to a precision warranted by your titration results.
Determination of the Unknown Concentration of a Solution of Your Assigned Acid or Base. Titrate 3 5-mL aliquots of your assigned acid (or base) solution with standard NaOH (HCl) using the indicator specified below.
| Acid or base | pH at equivalence point | Indicator |
|---|---|---|
| HNO3 | 7 | phenolphthalein |
| HCl | 7 | phenolphthalein |
| HCOOH | 8.2 | phenolphthalein |
| CH3COOH | 8.7 | phenolphthalein |
| NaHCO3 | 3.8 | methyl orange |
| NaHSO4 | 7.3 | phenolphthalein |
| NaF | 2.2 | thymol blue |
| NaHCOO | 2.5 | thymol blue |
| NaCH3COO | 3.0 | erythrosin B |
Determine the concentration of acid (base) in your solution to a precision warranted by your titration results.
Determination of Ka of Your Assigned Acid or Base. To a 5-mL aliquot of your assigned acid (base) solution, add exactly one-half of the volume of standard NaOH (HCl) solution needed to react with the acid (base) in the solution. To the resulting solution, add 2 drops of universal indicator solution. Determine the pH of the solution by comparing the color with the colors of the standard solutions provided in the laboratory. Estimate the value of Ka for your acid.
Pool your results with those of other groups and compile a list of acids in decreasing order of strength (Ka).
Determination of the Acidity of Commercial Products. If time permits, select one of the commercial substances provided and determine its acid content as precisely as you can.
Clean-up. When you have finished all of your work:
Disposal Methods
Return all equipment to the instructor. Solutions of acids and bases may be flushed down the drain with plenty of water.
Preparation
Equilibrium: Bronsted-Lowry Acids and Bases