Acids, Bases, Salts and Buffers

1
Acids, Bases, Salts and Buffers

• Experiment 8

2
8 Acids, bases, salts and buffers

• Goals
• Understand weak acid (base) equilibria and
  conjugate acid-base pairs
• Investigate the acidity/basicity of solutions
  containing aqueous ions
• Perform calculations involving ionic equilibria
• Understand buffer solutions and how they work
• Method
• Approximate the pH of aqueous salt solutions
  using acid-base-indicators
• Make buffer solutions
• Compare the effect of added base in weak acid,
  weak base, buffer, and dilute strong acid
  solutions

3
Acid-Base Definitions

• Acids
• generate H in water
• H donors
• excess H
• Bases
• generate OH- in water
• H acceptors
• Excess OH-

4
Equilibrium in Water
Small K ? equilibrium favors reactants
5
As H rises, OH- falls
6
H and pH
H 1 ? 100 to 1 ? 10-14 in water pH
1 to 14 in water
7
Relationships
H
OH-
H2O
H gtOH-
H OH-
H ltOH-
Basic solution
Acidic solution
Neutral solution
pH gt 7
pH lt 7
pH 7
8
(No Transcript)
9
(No Transcript)
10
Strong Acids (exp. 4)

• 100 dissociation / good H donor
• equilibrium lies far to right
• HA ? H A-

Before dissociation
After dissociation
Relative moles
HCl H Cl-
HCl H Cl-
11
Weak Acids (exp. 5)

• lt100 dissociation / not-as-good H donor
• equilibrium lies far to left
• HA H A-

Before dissociation
After dissociation
Relative moles
HA H A-
HA H A-
12
Acid Dissociation Constant

• For a weak acid

?amount dissociated
?amount undissociated
10-2 lt Ka lt 10-7 2 lt pKa lt 7
13
Henderson-Hasselbach Equation
Smaller Ka? weaker acid Larger pKa? weaker acid
log(xy) log x log y log(x/y) log (y/x)
For buffer system only considerable HA,A-
14
H, pH and Ka, A-, HA

• A- lt HA ?
• A- gt HA ?
• A- HA ?

? pH lt pKa ? pH gt pKa ? pH pKa
15
Chemical Equations

• 1) Weak acid HA dissociation
• 2) Conjugate base reaction with water
• 3) Water autoionization

16
Buffer Characteristics

• Contain relatively large amounts of weak acid and
  corresponding base.
• Added H reacts to completion with weak base, A-.
• Added OH? reacts to completion with weak acid,
  HA.
• pH is determined by ratio of concentrations of
  weak acid and weak base.

17
Hydrolysis Anions of Weak Acids

• Anion of weak acids can react with proton
  sources.
• In water, the anions react with water to some
  extent to form OH- (and the conjugate acid).
• The OH- causes the solution pH to be greater than
  7.
• Ex. NO2-(aq) H2O(l) ? HNO2(aq) OH-(aq)
• The Ka of HNO2 is 4.5?10-4, so Kb for NO2- is

18
Hydrolysis Anions of Weak Acids

• Ex. OCl-(aq) H2O(l) ? HOCl(aq) OH-(aq)
• Find the pH of 0.10 M NaOCl. Ka of HOCl is
  3.0?10-8
• Kb for OCl- is
• Kb is small so x can be neglected relative to
  0.10 0.10 x ?0.10

19
Another example finding Kb from pH

• A solution of 0.10 M NaOBr has a pH of 10.85.
• Na spectator ion
• OBr- conjugate base of a weak acid (HOBr)
• Hydrolysis equation OBr-(aq) H2O(l) ?
  HOBr(aq) OH-(aq)
• pOH of the solution is 14.00 10.85 3.15
• OH- is 10-3.15 7.1?10-4

20
Hydrolysis Cations of Weak Bases

• Cations derived from weak bases react with water
  to increase the H3O concentration (acidic).
  Consider NH4 in water
• NH4(aq) H2O(l) ? NH3(aq) H3O(aq)
  
• or NH4(aq) ? NH3(aq) H (aq)
• Ka for NH4, the conjugate acid of NH3, can be
  determined using the Kb of NH3 and Kw
• Cations of the group 1A metals (Li, Na, K,
  Rb, Cs) and the group 2A metals (Ca2, Sr2,
  Ba2) do not react with water and are nonacids.
• They do not affect the pH of the solution.
• Hydrated cations of many other metals do
  hydrolyze to produce acidic solutions.
• For example Fe(H2O)63(aq) H2O(l) ?
  Fe(H2O)5(OH)2(aq) H3O(aq)
• The waters of hydration are sometimes omitted
• Fe(H2O)63(aq) Fe3(aq) Fe(H2O)5(OH)2(aq)
  Fe (OH)2(aq)

21
Summary of Hydrolysis of Salts

• The acidity, basicity, or neutrality of an
  aqueous salt solution can be predicted based on
  the strengths of the acid and base from which the
  salt was derived.
• 1. Cation from strong base anion from strong
  acid
• Ex. NaCl, KNO3
• Solution has pH 7 (neutral)
• 2. Cation from weak base anion from strong acid
• Ex. NH4Cl, Zn(NO3)2
• Solution has pH lt 7 (acidic) due to the
  hydrolysis of the cation.
• 3. Cation from strong base anion from weak
  acid
• Ex. NaF, KNO2
• Solution has pH gt 7 (basic) due to the
  hydrolysis of the anion.
• 4. Cation from weak base anion from weak acid
• Ex. NH4F, NH4C2H3O2
• Solution pH is determined by the relative Ka and
  Kb of the cation and anion.

22
CO2 and experimental pH

• The solutions will generally be more acidic than
  predicted primarily due to the presence of
  dissolved CO2.
• CO2 reacts with water to generate H3O (aqueous
  protons, H(aq))
• CO2(g) H2O(l) ? H2CO3(aq) ? HCO3-(aq)
  H(aq)
• The solubility of CO2 is greatest in basic
  solutions intermediate in neutral and least in
  acidic.
• Boiling can help remove the CO2

23
Part 1

• In part 1 of this experiment, the pH of water and
  several salt solutions will be tested.
• Use pH and initial concentration of each solution
  to obtain approximate value of Ka or Kb
• Approximation extent of dissociation is small
  relative to initial concentration
• A set of acid-base indicators will be used to
  determine pH

24
Testing solutions

• In each well
• 1 drop indicator and a few drops solution
• Set 1
• boiled H2O (H2O)
• unboiled H2O (H2O, CO2)
• NaCl (H2O, CO2)
• NH4Cl (NH4)
• Set 2
• NaC2H3O2 (C2H3O2-)
• ZnCl2 (Zn(H2O)62)
• KAl(SO4)2 (Al(H2O)63)
• Na2CO3 (CO32-)

25
Example colors set 1

• Approximate
• pH values
• 5.8
• 7.0
• 5.8
• 8.1

26
Example colors set 2

• Approximate
• pH values
• 4.4
• 4.4
• 3.8
• 10.4

27
Results Part 1
28
Part 1 results
29
Example Part 1

• As an example, a solution of 0.10 M NaC2H3O2 has
  a pH of 8.1. It appears
• It appears yellow in methyl orange yellow in
  methyl red blue in bromothymol blue red in
  phenol red and, colorless in phenolphthalein (no
  need to go farther).
• Since the top of the phenol red pH range is 8.0
  and the bottom of the phenophthalein range is
  8.2, an estimate of 8.1 is reasonable
• Na spectator ion
• C2H3O2- conjugate base of a weak acid (HC2H3O2)
• Hydrolysis equation C2H3O2-(aq) H2O(l) ?
  HC2H3O2 (aq) OH-(aq)
• pOH of the solution is 14.0 8.1 5.9
• OH- is 10-5.9 1.2?10-6
• Literature Ka for HC2H3O2 is 1.8?10-5 so expected
  Kb is 5.6?10-10.

30
Part 2 buffers

• Expected pH of 0.05 M HAc, Ac-, and HAc/Ac-
  solutions
• Initial concentration calculations use M1V1
  M2V2
• HAc 5 mL of 0.05 M of HAc to final volume of 100
  mL
• Ac- 5 mL of 0.05 M of Ac- to final volume of 100
  mL
• (you do)
• HAc/Ac- 5 mL each of 0.05 M of HAc and of Ac- to
  final volume of 100 mL (you do)

31
Part 2 buffers

• Expected pH of 0.05 M HAc, Ac-, and HAc/Ac-
  solutions
• Use ICE tables to find expected equilibrium H
  and then pH
• HAc HAc-(aq) ? H(aq) Ac-(aq)
• Ac- Ac-(aq) H2O ? OH-(aq) HAc (aq)
• HAc/Ac- What equilibrium expression should be
  used?
• How do your experimental results compare?

32
Part 2 - buffers

• Theoretical pH after OH- addition
• a stoichiometry problem for the neutralization
• look at moles added and resulting initial
  concentrations
• an equilibrium problem for the new concentrations
• use ICE tables to find expected values
• Which solution should show the smallest change in
  pH
• HAc, Ac-, or HAc/Ac-?

33
Part 2 - buffers

• Example, HAc after NaOH addition if initial HAc
  had pH 3.02
• Stoichimetry HAc(aq) OH-(aq) ? Ac-(aq) H2O
• Equilibrium HAc(aq) ? H(aq) Ac-(aq)

34
Report

• Abstract
• Results including
• Indicator colors of salt solutions (part 1)
• Expected and actual pH of salt solutions (part 1)
• Expected and actual pH of solutions in parts 2a
  and 2b
• Sample calculations
• Expected and actual pH of salt solutions (part 1)
• Expected and actual pH of solutions in parts 2a
  and 2b
• Percent error (generally will be large what
  are possible reasons?).
• Discussion/review questions