Note to Students: It is IMPERATIVE you have a good understanding of buffers theory and complete all pre-lab calculations before the experiment begins if you are to complete the procedure on time. Please make sure to read the discussion on buffers included in this lab and see your instructor BEFORE the lab if you have questions.
The acid ionization constant, Ka, gives useful information about an acid. For example, it tells us about the strength of the acid; the greater the Ka, the stronger the acid.
The Ka of an acid may be determined by making a buffer solution in which the concentrations of the weak acid, HA, and its conjugate base, A-, are equal. The Ka value can be determine directly by pH measurements since under these conditions, pH = pKa
pH = -log [H+] and pKa = -log Ka Equation 1
Ka = [H+] [Ac-] when [Ac-] =[HAc] then, Equation 2
Ka = [H+] and pH = pKa
Another way of determining the Ka of an acid is by its titration curve.
In this experiment, the acid ionization constant of a weak acid will be determined by using a buffer system; in next week’s experiment we will determine the acid ionization constant by the titration of the weak acid against standardized NaOH.
Materials and Equipment
· Graduated cylinder
· 2-100 mL beakers
· pH meter
· 0.10M HCl
· 0.10 M NaOH
· 0.50M unknown weak acids
On the lab bench we have 0. 10 M stock solutions that can be used to make three different common “buffer systems”. These are;
HC2H3O2-C2H3O2- NH4+-NH3 HCO3- -CO32-
acetic acid-acetate ion ammonium ion-ammonia hydrogen carbonate-carbonate
Notice the presence of a weak acid and its conjugate base. The sources of the ions will be sodium and ammonium salts containing those ions. Select one of these buffer systems for your experiment.
1. Using a graduated cylinder, measure out 15 mL of the acid component of your buffer into a 100-mL beaker. Add 15 mL of your conjugate base.
a) Measure the pH of your mixture and record it.
b) Calculate Ka for the acid.
2. Add 30 mL water to the buffer created in the previous step. Pour half of the resulting solution into another 100-mL beaker.
a) Measure the pH of the diluted buffer and record it.
b) Calculate Ka once again.
3. Add five drops of 0. 10 M NaOH to the diluted buffer and measure the pH. To the other half of the diluted buffer, add 5 drops 0. 10 M HCL, and again measure the pH. Record your results.
4. Make a buffer mixture containing 2 mL of the acid component and 20 mL of the solution containing the conjugate base.
a) Measure the pH.
b) Calculate a third value for Ka.
5. To the solution from step 4, add 3 mL 0. 10 M NaOH. This should exhaust the buffer. Measure the pH.
6. Put 25 mL distilled water into a 100-mL beaker. Measure the pH. Add 5 drops 0. 10 M HCl and measure the pH again. To that solution add 10 drops 0.10 M NaOH, mix, and measure the pH.
7. Select a pH different from any of those you observed in your experiments and calculate amounts of your conjugate acid and base that will give you the desired pH. Make up the buffer and measure its pH.
So far in this experiment we have made buffers by mixing solutions of a weak acid and its conjugate base. It is possible to prepare buffers by adding a strong base to solutions containing a weak acid. Reaction 6, as described in the buffers discussion below, will occur, quantitatively, so we will produce the same number of moles of conjugate base (B ‑) as we add of NaOH. If we add half as many moles of OH- as we have of weak acid (HB), the final solution will be half-neutralized, and will be a buffer in which [HB] equals [B-]. This solution will be completely equivalent to a solution in which we mix equal amounts of [HB] and NaB.
In this part of the experiment we will furnish you with an unknown containing a 0.50 M solution of a weak acid. Using that solution and some 0.10 M NaOH, you will be asked to design and prepare a buffer with a particular pH. The following procedure is suggested:
8. Obtain a sample of unknown acid from your instructor along with an assigned pH for the buffer to be made.
9. Dilute your unknown to 0.10 M by adding 10 mL of your sample to 40 mL distilled water and mixing thoroughly.
10. Mix 20 mL of your unknown acid with 10 mL of 0.10 M NaOH. Measure the pH of the resulting half-neutralized buffer. Calculate Ka for your unknown acid.
11. Given the pH of the buffer you need to design, and the value of Ka YOU just found, calculate the value of [HB] that is needed in the buffer. Note that [HBI = no. moles HB in the buffer.
[B-] [B-] no. moles B-
12. Find the volumes of 0.10 M NaOH and 0.10 M HB that will produce the required ratio. This is perhaps most easily done by arbitrarily deciding to add 10 mL of the NaOH to a volume of the HB solution. The number of moles of B- produced by Reaction 9 will equal the number of moles of OH- in the 10 mL of NaOH, and will also equal the number of moles of HB that will be used up in the reaction with OH-. The volume of 0.10 M HB you select must contain the number of moles of HB present in the final buffer plus the number of moles used up in producing the B- that is in the buffer. Knowing the total number of moles of HB you need to make the buffer, calculate the volume of the 0.10 M HB that is required. Mix that volume with 10 mL of the 0.10 M NaOH and measure the pH.
All waste solutions should be transferred into a waste container under the hood.
Discussion of Buffer Systems
Some solutions, called buffers, are remarkably resistant to pH changes. Water is not a buffer, since its pH is very sensitive to addition of any acidic or basic species. Even bubbling your breath through a straw into distilled water can change its pH by at least 1 unit, just due to the small amount of CO2 in exhaled air. With a good buffer solution, you could blow your exhaled air into it for half an hour and not change the pH appreciably. All living systems contain buffer solutions, since stability of pH is essential for the occurrence of many of the biochemical reactions that go on to maintain the living organism.
To make a buffer a solution containing a weak acid and its conjugate base is required. An example of such a solution is one containing acetic acid, HAc, and acetate ion, Ac-, its conjugate base obtained from the dissolution of the salt NaAc in water. The pH of such a buffer is established by the relative concentrations of weak acid and base in solution. In such a solution [H+] can be calculated by manipulating Equation 2:
[H+] = Ka [HAc] Equation 3
Taking the negative logarithm on both sides of the equation and by the definition of pH given in Equation 1 we get:
-log [H+] = -log Ka -log [HAc] ; pH = pKa + log [Ac-] Equation 4
In general terms: pH = pKa + log [conjugate base] Equation 4a
This equation is known as the Henderson Hasselbalch equation.
If, for example, we mix 500 mL of 0.10 M HAc with 500 mL 0. 10 M NaAc, we will have a typical buffer, containing an acid and its conjugate base. Since we have equal amounts of HAc and Ac, their concentrations are equal and by Equation 3, the [H+] in this solution equals Ka; and, by Equation 4, pH is equal to pKa.
You may wonder why the concentrations of the acid and conjugate base do not change when the species are mixed. Actually, they do, very very slightly, just enough to generate enough H+ ion to satisfy the equilibrium conditions imposed by Equation 2. Ordinarily, however, Ka is small, so [H+] is also small. If Ka is 1x 10-5, [H+] will be 1 x 10-5 M, and so, in our example, where we have 1 L of solution, we will have 1 x 10-5 moles of H+. This means that 1 x 10-5 moles of HAc will dissociate out of 0.050 moles initially present, only a negligible decrease in [HAc] occurs, and only a tiny increase in [Ac-] as a result of the reaction to form the equilibrium system. We can conclude then that the acid and conjugate base in a buffer do not react appreciably when mixed and that the relative concentrations can be calculated just from the way the buffer was put together.
Another interesting property of a buffer is that its pH does not change appreciably on dilution. Looking at the buffer in our example, if we increased the volume from I L to 5 L by adding water, the ratio of [HAc] to [Ac] would not change, and since that ratio fixes [H+], the pH would not change.
We can adjust the pH of the buffer, within limits, to bring it to some desired value. In our example, if Ka for the buffer is 1 x 10 -5, the pH of the buffer solution would be 5.0. If we wish to make a buffer of pH equal to 4.5, we need to simply select volumes of the acid and conjugate base such that the resultant ratio
of [HAc] to [Ac-] would make [H+] equal to 10-4.5 or 3.2 x 10-5 M. Then by Equation 3,
3.2 x 10-5 = 1.0 x 10-5 x [HAc] [HAc]
[Ac -] and [Ac -] = 3.2
So, to make the desired buffer we could use 320 mL of 0.10 M HAc and 100.0 mL 0.10 M NaAc.
Or, if our stock solutions were of different concentrations, we would select reagent volumes such that the number of moles of HAc used would be 3.2 times as large as the number of moles of Ac-.
Buffer’s Resistance to pH changes
The reason that a buffer has a stable pH is that its two components can "soak up" added H+ or OH- ions. If we add a little HCl, a strong acid, to our buffer, the extra H+, will react with the base in the buffer according to the following reaction:
H+(aq) + Ac - ® HAc(aq) + H2O Reaction 5
If we added a little NaOH, a strong base, it will react with the acid present:
OH- (aq) + HAc(aq) ® Ac-(aq) + H2O Reaction 6
As a result of these reactions, [HAc] and [Ac-] will change slightly, but if the amounts of H+ and OH- ions that are added are small as compared to the amounts of HAc and Ac- present in the buffer, the effect on the pH will be small since the ratio of [HAc] to [Ac-] will not change much.
The range over which a buffer is useful is limited to about 2 pH units. In the example we used earlier, if we mixed 500 mL 0.10 M HAc with 50 mL 0.10 M NaAc, in the buffer [HAc]/[Ac-] would be 10, and so [H+] would be 1 x 10-4 M and the pH would be 4.0. This buffer could deal with added NaOH much better than with added HCl, since the amount of available HAc is much greater than that of Ac - . Indeed, if we add enough HCl to react with all of the Ac - present, the buffer would be “exhausted," since it would contain only HAc, and any excess HCl would produce a pH with just about the same value as if the HCl were added to water. Similar behavior would occur if we made the buffer in such a way that [HB]/[B-] were equal to 0. 1. Then the pH would be 6, and the buffer would have very little capacity for added NaOH.
Prelaboratory Questions Ka Determination by Buffer Systems
Answer these in your notebook.
1. The pH of a 0.10 M solution of HCN is 5.2.
What is the [H+] in the solution?
What is [CN-]? What is [HCN]?
What is the Ka for HCN?
2. Which of the following solutions would yield a buffer system? Explain.
a) HCl and NaOH
b) HCl and NaC2H3O2
c) HC2H3O2 and NaC2H3O2
d) HC2H3O2 and NaOH
3. Formic acid has a Ka equal to 1.7 x 10-4. A student is asked to prepare two buffers with pH 3.50 using 10 mL of 0.10 M solutions of formic acid (HFor) and the following bases:
a) 0.10 M sodium formate (NaFor)
b) 0.10 M NaOH
How many milliliters of sodium formate should he to make buffer a?
How many milliliters of sodium hydroxide should he add to make buffer b?
In your notebook, create a well-organized table to collect all data and perform all required calculations.
Buffer system selected: ___________________ Unknown Weak Acid Number ___________
Post Laboratory Questions-
1) Suggest reasons why you might want to buffer a system. In particular, comment on the fact that biological systems are usually buffered.