York University CHEM 1001 3.0 Acids and Bases - 48 Buffers and Acid-Base Titrations Reading: from chapter 18 of Petrucci, Harwood and Herring (8th edition): Required : Sections 18-2 through 18-6. Recommended : Section 18-1. Examples: 18-3 through 18-9. Problem Set: Chapter 18 questions: 5, 6, 7, 8, 9, 26 , 27a-e, 32. Additional problems from Chapter 18: 54, 55.
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York University CHEM 1001 3.0 Acids and Bases - 48
Buffers and Acid-Base Titrations
Reading: from chapter 18 of Petrucci, Harwood and Herring(8th edition):
York University CHEM 1001 3.0 Acids and Bases - 49
Adding Base to a Weak Acid SolutionHF is a weak acid; it partially ionizes in water:
HF + H2O W H3O+ + F- pKa = 3.18
What effect does adding NaOH have on this equilibrium?
NaOH 6 Na+ + OH- complete ionization
H3O+ + OH- 6 H2O neutralization
Consequences:
C Adding NaOH consumes H3O+.
C The HF equilibrium shifts to the right to partiallyoffset the change (replaces some H3O
+).
C The increase in pH is less than if the same amount ofNaOH was added to pure water.
York University CHEM 1001 3.0 Acids and Bases - 50
Adding Acid to a Salt of a Weak AcidNaF is a salt of a weak acid (HF). It hydrolyzes in water:
F- + H2O W OH- + HF pKb = 10.82
What effect does adding HCl have on this equilibrium?
HCl + H2O 6 H3O+ + Cl- complete ionization
H3O+ + OH- 6 H2O neutralization
Consequences:
C Adding HCl consumes OH-.
C The hydrolysis equilibrium shifts to the right to partiallyoffset the change (replaces some OH-).
C The decrease in pH is less than if the same amount ofHCl was added to pure water.
York University CHEM 1001 3.0 Acids and Bases - 51
Solution of a Weak Acid and its Salt
Example: Solution of HF and NaF in water.
HF + H2O W H3O+ + F- pKa = 3.18
C Adding acid shifts the equilibrium to the left. Thispartially offsets the reduction in pH.
C Adding base shifts the equilibrium to the right. Thispartially offsets the reduction in pH.
C Significant amounts of acid or base can be neutralizedsince the solution contains substantial [HF] and [F-].
C The pH is kept nearly constant. This solution is a buffer.
York University CHEM 1001 3.0 Acids and Bases - 52
Applications of Buffer Solutions
Many chemical processes require control of pH.
C The rates and mechanisms of many chemical reactionsare sensitive to pH.
C pH affects the structure of and charges on proteins.
C The activity of enzymes is very sensitive to pH.
C Human blood is buffered to a pH of 7.4
C Many industrial chemical processes require acontrolled pH.
York University CHEM 1001 3.0 Acids and Bases - 53
Acid-Base Buffers
An acid-base buffer is a solution that is resistant tochanges in pH. In other words, when an acid or base isadded to a buffer, the pH changes only slightly.
For a solution to be a buffer, it must contain:
C comparable concentrations of a weak acid its conjugatebase:
0.1×[acid] < [conjugate base] < 10×[acid]
C a concentration of the conjugate base at least 100 timesthe value of Ka for the acid:
[conjugate base] > 100 Ka
York University CHEM 1001 3.0 Acids and Bases - 54
pH of a Buffer SolutionReaction HA + H2O W H3O
+ + A-
Initial [HA]0 - 0 [A-]0 M
Change -x - x x M
Equil. [HA]0-x - x [A-]0+x M
Simplifying assumptions: x n [HA]0 and x n [A-]0
Y Ka = [H3O+][A-]/[HA] . x [A-]0 / [HA]0
Y [H3O+] = x = Ka [HA]0 / [A
-]0
Check that the assumptions are valid:
C Since [A-]0 > 100Ka, x < 0.01[HA]0.
C Since [A-]0 > 100Ka and [HA]0 < 10[A-]0, x < 0.1[A-]0.
York University CHEM 1001 3.0 Acids and Bases - 55
pH of a Buffer Solution - continued
In a buffer, we have:
[H3O+] = Ka [HA] / [A-]
Take the negative logarithm of both sides:
-log[H3O+] = -logKa - log([HA]/[A-])
pHbuffer = pKa + log([A-]/[HA])
This is called the Henderson-Hasselbalch equation.
C Relates the pH of a buffer to its composition.
C Uses initial concentrations of acid and conjugate base.
C Don't memorize - it is easy to mess up but easy to derive.
York University CHEM 1001 3.0 Acids and Bases - 56
Buffer RangeTo have a buffer, we must have:
0.1 < [A-] / [HA] < 10
Taking logarithms:
-1.0 < log([A-]/[HA]) < 1.0
Substitute into the Henderson-Hasselbalch equation:
pHbuffer = pKa + log([A-]/[HA])
pHbuffer = pKa ± 1.0
Conclusion: A particular weak acid can used to preparebuffers with pH values within 1 unit of the acid's pKa.
York University CHEM 1001 3.0 Acids and Bases - 57
Preparing Buffer Solutions
Chemists and biologists often need to prepare solutionswith a specific, constant pH.
To make a buffer with a specific pH:
C Find an acid with a pKa within 1 unit of the required pH.
C Use the acid ionization constant to determine therequired ratio of acid to conjugate base.
C Make the buffer solution.
C Use a pH meter to check the pH. Add acid or base asneeded to get the correct pH.
York University CHEM 1001 3.0 Acids and Bases - 58
Acid pKa
Citric Acid 3.13
Benzoic acid 4.20
Acetic acid 4.77
Carbonic acid 6.36
Ammonium ion 9.25
Phenol 9.89
Preparing a Buffer - Example
To prepare a buffer with pH = 3.50:
» Closest to desired pH.
Find ratio of acid to base:
[HA]/[A-] = [H3O+]/Ka = 10pKA-pH
[HA]/[A-] = 103.13-3.50 = 0.43
if [A-] = 0.10 M (>100Ka)then [HA] = 0.043 M.
For one liter, use 0.143 mol H3C6H5O7 plus 0.100 mol NaOH.
York University CHEM 1001 3.0 Acids and Bases - 59
Adding Acids or Bases to a Buffer
Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
C Adding an acid decreases [A-] and increases [HA]. So itlowers the pH.
C Adding a base decreases [A-] and increases [HA]. So itraises the pH.
C Qualitatively, this is just like any other solution. But ina buffer, the pH changes are much smaller.
Q: How much does the pH change when acid or base is added?
A: Calculate the changes in [HA] and [A-].
York University CHEM 1001 3.0 Acids and Bases - 60
pH Change in a Buffer
Given a citric acid-citrate buffer with [A-]0 = 0.100 M,[HA]0 = 0.043 M, and pH = 3.50 ([H3O
+] = 3.16×10-4 M).
Find the pH after adding 3.0×10-3 mol HCl to 1.0 liter ofthis buffer.
Solution: Calculate the changes in [HA] and [A-].
[A-] = [A-]0 - 3.0×10-3 M = (0.100-0.003) M = 0.097 M
[HA] = [HA]0 + 3.0×10-3 M = (0.043+0.003) M = 0.046 M