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- Lab 8 - Buffers
Lab 8 - Buffers
Purpose
To prepare buffers and measure the pH of each, and to prepare a buffer at a specific pH. Goals
- 1 To learn to prepare buffers by both the direct and indirect methods.
- 2 To learn to identify solutions that are buffers.
- 3 To understand the how a buffer resists changes in pH upon addition of acid or base solutions.
Introduction
In dilute aqueous solutions, weak acids are slightly dissociated. They produce a small concentration of hydronium ion (H3O+) and an equal concentration of the conjugate base of the acid. Such dissociation reactions are equilibria, and equilibrium mathematics can be used to calculate concentrations of the species present in solution. Consider formic acid (CH2O2), which is what red ants inject when they bite. The acid dissociation constant Ka for formic acid is 1.7 x 10-4. The concentration of H3O+ present in a 0.010 M solution of formic acid can be calculated from the equilibrium expression and a reaction table. ( 1 )
| HCOOH(aq) | + | H2O(l) |  | H3O+(aq) | + | HCOO-(aq) |
| initial | 0.010 | 0 | 0 |
| Δ | -x | +x | +x |
| equilibrium | 0.010 - x | x | x |
( 2 )
Ka = 1.7 x 10−4 = = Solving for x, we find that the solution is 0.0012 M in H3O+ and HCOO-.* Expressing [H3O+] as pH ( 3 )
pH = −log[H3O+ ] = log(0.0012) = 2.92 * For this calculation, the quadratic formula was used. If we make the simplifying assumption that x is small relative to [HCOOH], the calculated value of [H3O+] is 0.0013 M. Since this is >5% ionization of the acid, using the quadratic equation will help us calculate more accurate values for [H3O+] and [HCOO-]. Although a weak acid dissociates only slightly in water, the soluble salt of a weak acid (for example, sodium formate) is a strong electrolyte and dissociates completely. ( 4 )
HCOONa(s) + H2O(l) → Na+(aq) + HCOO−(aq) The salt, if added to the weak acid solution, produces a large amount of formate ion in comparison to that produced by the acid dissociation. Adding formate ion to the equilibrium of equation 1 stresses the system by adding a product. According to Le Châtelier's principle, the equilibrium will shift to the left (toward reactants) and the concentration of H3O+ will decrease (and the pH will increase). The suppressed dissociation caused by adding an ion already present in the solution is called the common ion effect. One can prepare a solution using a weak acid and its conjugate base (the common ion). The resulting solution will resist major changes in pH when an acid or base is added to buffer solutions. Consider what would happen in a solution containing both formic acid and sodium formate when acid or base are added. Adding acid (a source of H3O+) stresses the system by adding a product. The equilibrium of equation 1 shifts toward the reactants, consuming formate ion and some of the added H3O+. The result: a small decrease in the pH. Adding a base causes the hydronium ion to neutralize the base. This stresses the system by removing a product. Some formic acid dissociates to replace the H3O+ and the equilibrium of equation 1 shifts toward the products. The result: a small increase in the pH. The concentrations of the acid and its conjugate base in a buffer will determine how much additional acid or base can be added to the solution before its buffering ability is exhausted. This is called the buffer capacity of the solution. The higher the concentrations of acid and conjugate base, the larger the buffer capacity. The preceding discussion will also apply if a buffer is prepared using a weak base and its conjugate acid. However, buffers cannot be made with strong acids or strong bases and their conjugates. No buffer capacity exists in such solutions because there is no equilibrium; everything has completely dissociated into ions. The pH of a buffer solution may be calculated with the Henderson-Hasselbalch equation: ( 5 )
pH = pKa + log or pH = pKa + log| moles of base |
| moles of acid |
The derivation of this equation follows from the general dissociation equilibrium expression for a weak acid, and includes the assumption that [H3O+] is small relative to [HA]: ( 6 )
HA + H2O H3O+ + A- ( 7 )
Ka = Solving for [H3O+] gives: ( 8 )
[H3O+] = Taking the negative logarithm of both sides puts the equation in terms of pH: ( 9 )
pH = − log[H3O+ ] = − log Ka − log([HA]/[A−]) By definition, pKa = - log Ka and - log ([HA]/[A- ]) = log ([A- ]/[ HA]). Substituting these terms into equation 9: ( 10 )
pH = pKa + log([A− ]/[HA]) Equation 5 shows that pH can be found using either concentrations of acid and base, or the number of moles of each. This follows from the fact that the volume term is the same for the acid and its conjugate base, and cancels in the calculation. The Henderson-Hasselbalch equation shows that the pH of a buffer is close to the pKa of the weak acid from which it is made. The exact pH is dependent on the ratio [A- ]/[HA]. If [HA] = [A- ], log ([A- ]/[HA]) = 0 and the pH of the buffer will be exactly the pKa of the acid. If one wishes to make a buffer of a specific pH, one selects an acid with a pKa near that value and adjusts the ratio of [A- ]/[HA] to obtain the desired buffer. Suppose we want a pH 4.00 formate buffer. The pKa of formic acid is -log (1.7 x 10-4) = 3.77. ( 11 )
| pH | = | pKa + log([A− ]/[HA]) |
| 4.00 | = | 3.77 + log([A− ]/[HA]) |
| [A− ]/[HA] = 1.70 |
A ratio of 1.70 moles of formate anion per mole of formic acid should produce a buffer of the desired pH. Buffer capacity can be controlled by the concentrations of each. A buffer prepared with 0.17 mole of formate and 0.1 mole of formic acid per liter would have ten times the capacity of a buffer containing 0.017 mole of formate and 0.010 mole of formic acid, but the initial pH of both should be the same. In Part A of this experiment, you will prepare various acetate buffer solutions by the direct method and measure the pH of each solution. In the direct method, the conjugate acid and base are added together in solution to get the desired base to acid ratio. For example, acetic acid and sodium acetate will be combined in solution. In Part B, you will prepare acetate buffers by the indirect method and measure the pH of each solution. In the indirect method, strong base is added to the weak acid OR strong acid is added to the conjugate weak base. For example, acetic acid will be mixed with sodium hydroxide solution. Upon mixing, the weak acid and hydroxide ion react to produce acetate ions and water. As long as hydroxide ion is the limiting reagent, and some acetic acid remains, the solution contains both species of the conjugate pair (acetic acid and acetate ions), and a buffer exists. In Part C of the experiment, you will be given a target pH and you will need to prepare a phosphate buffer with the given pH. Phosphoric acid is a polyprotic acid and there are several buffer regions. You will use Na2HPO4 · 7 H2O and another phosphate salt, either NaH2PO4 · H2O or Na3PO4 · 12 H2O. The Henderson-Hasselbalch equation suggests that even at a 10:1 acid:base ratio (or a 1:10 ratio) the pH of the resulting solution will only differ from the pKa by one pH unit. Therefore, you need to select a conjugate pair whose pKa is close to your target pH. The acid/base table shows that the H2PO4-/HPO42- conjugate pair has a pKa of about 7.2, so it should be a good system to use for buffers in the pH range of about 6.5 to 8.0. The HPO42-/PO43- conjugate pair has a pKa of about 12.3, so it should be a good system to use for buffers in the pH range of about 11.5 to 13.0. Equipment
- 2 30 mL beakers
- 2 100 mL volumetric flasks
- 1 10.0 mL volumetric pipet
- 1 pipet bulb
- 1 50 mL graduated cylinder
- 1 10 mL graduated cylinder
- 1 50 mL beaker
- 1 100 mL beaker
- 1 glass stir rod
- 1 MicroLab Interface
- 1 MicroLab pH Measurement Instruction Sheet
- 1 pH electrode in pH 7.00 buffer
- 1 ring stand
- 1 clamp
- 1 250 mL beaker for electrode rinsings
- 1 deionized water squirt bottle
- 1 box Kimwipes
Reagents
- ~10 mL 6.0 M acetic acid (HC2H3O2)
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