Interactive Buffer Simulation Lab
Model 1: Composition of a Buffer Solution
The computer simulation allows you to make solutions with varying concentrations of a weak acid and its conjugate base. Make the following solutions involving lactic acid, HC3H5O3 (pKa = 3.85), and sodium lactate, NaC3H5O3,and record the solution pH. Sodium lactate, a strong electrolyte, will completely dissociate in solution. Therefore, this component will mostly be referred to as “lactate” or C3H5O3-. Use 100.00 mL of each component.
Solution / [HC3H5O3] / [C3H5O3-] / Solution pH / pH after addition of HCl / pH after addition of NaOH1 / 0.10 / 0.90 / 4.80 / 4.61 / 5.13
2 / 0.30 / 0.70 / 4.22 / 4.12 / 4.33
3 / 0.50 / 0.50 / 3.85 / 3.76 / 3.94
4 / 0.70 / 0.30 / 3.48 / 3.37 / 3.58
5 / 0.90 / 0.10 / 2.90 / 2.57 / 3.10
Critical Thinking Questions:
- What are the concentrations of HC3H5O3 and C3H5O3-when the pH of the solution is equal to the pKa of the acid component?
- Under what conditions is the solution pH above the pKa?
- Under what conditions is the solution pH below the pKa?
Read this!
A buffered solution (or buffer) resist changes in pH when small amounts of strong acid or base are added (or when the solution is diluted).
A buffer contains a weak acid HA and its conjugate base A-. The pH of a buffer can be calculated using the Henderson-Hasselbach equation shown below.
or more generally
Where [HA] can be taken as (HA)0, assuming that dissociation of the weak acid is minimal and ignoring the dissociation of water molecules.
- Verify that the Henderson-Hasselbach equation predicts the correct pH for solutions 1 and 3.
- Use the Henderson-Hasselbach equation to calculate the pH of the solution expected for the following:
100.0 mL 0.40 M HC3H5O3(aq) / 100.0 mL 0.60 M NaC3H5O3(aq)
- Check your answer using the computer simulation. Are the values in agreement?
- Suppose you diluted solution 2 to 1/10th of its original concentration. What would [HC3H5O3] and [C3H5O3-] now be?
- Use the Henderson-Hasselbach equation calculate the pH of the diluted solution and then compare it to solution 2? Does the diluted solution have a greater, lower, or the same pH? Why?
Read this!
In general, we expect the pH of a solution to change when a strong acid or base is added. In unbuffered solutions the change in pH is drastic. However, in buffered solution, the change in pH is small and the solution is said to resist changes in pH.
Model 2: The Effective Range of a Buffer
- Using the simulation, make solution #1 again. Then go to Part II of the simulation to test the effect of adding strong acids and bases by doing the following:
- Choose HCl and adjust the amount of moles added to 0.050 (50.0 x 10-3 mol). Record the pH in Table 1.
- Choose NaOH and adjust the amount of moles added to 0.050 (50.0 x 10-3 mol). Record the pH in Table 1.
- Repeat this for solutions #2-5.
Critical Thinking Questions
- Write the ionic equation for the reaction that occurs in the buffer solution when HCl is added.
- Write the ionic equation for the reaction that occurs in the buffer solution when NaOH is added.
- Explain how these reactions help to keep the pH relatively stable when acid or base are added.
- For each solution in Table 1, calculate the change in solution pH after adding acid or base.
- Which solution undergoes the smallest change in pH (most effective buffer)? The largest (least effective buffer)?
- The buffer solutions in Table 1 show the most effective pH range for the lactic acid/lactate buffer. Over what pH range is this buffer most effective?
- Write the most effective pH range as an interval around the pKa of lactic acid (i.e. pKa + _____). Only one significant figure is needed for this expression.
Read this!
Thus, the lactic acid/lactate buffer is effective over a certain range near the pKa. Outside of this range, the amount of one of the species – lactic acid or lactate- is so small that buffering is ineffective.
- If you wanted to make a buffer of pH 5.5, would lactic acid/lactate be a good choice? Explain. Use the computer simulation to suggest another buffer pair that might be better and justify your answer.
Model 3: Buffering Capacity
- Make the following buffer and measure its pH
100.0mL 3.0 M (30 x 10-1) HC3H5O3(aq) /100.0 mL 7.0 M (70 x 10-1) NaC3H5O3(aq)
- When 50.0 x 10-3 mol HCl are added to the solution, the pH = 4.21. When 50.0 x 10-3 mol NaOH are added, the pH = 4.23.
- Compare the change in pH of this solution with that of solution 2 from Table 1. Which is a better buffer? What is the only difference between these two solutions?
Read this!
Buffering capacity is the amount of H+ or OH- a buffer can absorb without a significant change in pH. A buffer with a large capacity contains a large concentration of buffering components and can therefore absorb a relatively large amount of H+ or OH- with very little pH change. Notice that the ratio of [C3H5O3-] to [HC3H5O3] is the same, but their magnitudes are different.
Exercises:
- Calculate the pH of the following buffered solution: 0.10 M acetic acid/0.25 M sodium acetate (Ka of acetic acid = 1.8 x 10-5)
- A buffered solution contains 0.25 mol dm-3methylamine CH3NH2 (Kb = 4.38 x 10-4) and 0.40 M methylamine chloride CH3NH3Cl. Calculate the pH of this solution.
Problems:
- A buffered solution is made by adding 50.0 g NH4Cl to 1.00 L of a 0.75 M solution of NH3 (Kb = 1.8 x 10-5). Calculate the pH of the final solution. (Assume no volume change when the salt is added).
- A researcher wants to make a buffer of pH 4.60 using nitrous acid (Ka = 4.0 x 10-4) and sodium nitrite.
a)Determine the approximate ratio of the two buffer components that she should use get the desired pH.
b)Evaluate the researcher’s decision to use the nitrous acid/nitrite buffering system.