Chapter 16 Acid Base Equilibria

1
Chapter 16Acid Base Equilibria

• End-of-Chapter Problems See Next Page will
  discuss along with Ch 15 when through with Ch 16
  Oct 9
• Online Assignment 1, Chaps 15 16 Due Th Oct
  15 1130 PM
• Exam 1 over 15 16, Lab 1 Weak Acid Labs
  Mon Oct 19

2
End-of-Chapter Problemspp 690 - 698, Chapter 16

• Finish these on or before the last ch. 16 lecture
• 1, 2, 3, 5, 6 8 through 20 23,
  24, 26, 29,
• 30, 31, 32, 35, 37, 41, 43, 50,
  59, 60, 63,
• 65, 71, 75, 79, 85, 87, 115,
• 117a ( Ka if pH4.89 - half way to end pt),
  120

3
I. Ka Values A. Introduction

• In the preceding section we calculated the H
  if a strong acid (or base) was added to water.
  The major source of H was from the 100
  ionization of the strong acid.
• Now we will look at weak acids (HA) which only
  partially ionize. In order to determine the
  H, we will need to use the equilibrium
  expression.
• HA H A- or HA (-----) H
  A-
• Ka H A-
• HA
• - Ka weak acid equilibrium constant (from Table
  16.1, pg 656)

4
I. Ka Values A. Introduction B.
Obtaining

• Ka Large for Strong Acids, gt 1 Ka
  Small for Weak Acids, lt 1
• Examples (only need to memorize Kw for water)
• Ka 1.7 x 10-5 for HC2H3O2 Ka 6.8 x 10-4
  for HF Ka gt 102 for HCl
• B. Obtaining Ka Values 1) Table (Table 16.1,
  Page 656 or A-1314)
• 2) Conductivity Exp. Spectrophotometric Exp.
  (lab) Titration Curves (lab)
• 3) pH Measurements Example Calculate a) Ka
  b) ionization of 0.10 M phenol if the pH
  5.43.

HA A-
H
5
0.10 M Phenol (HA) pH 5.43 Determine a)
Ka b) Ionization

• Let x M of phenol (HA) that ionizes at
  equilibrium x M of H or of A-.
• x H 10-5.43 3.72 x
  10-6 M
• 
  HA H
  A-
• Initial M 0.10
  0
  0
• Equilibrium M 0.10 3.72 x 10-6
  0.10 3.72 x 10-6 3.72 x
  10-6
• H A- 3.72 x 10-6
  3.72 x 10-6
• a) Ka
  1.4 x 10-10
• HA 0.10
• b) ionization Part x 100
  3.72 x 10-6 x 100 3.7 x 10-3
• Total 0.10

6
II. Calculations Involving HA

• Introduction - two solution methods 1)
  quadratic 2) approximation
• - Once Ka is known, then can calculate
  equilibrium concentrations of HA, A-, H
• - Will frequently let X mole/L (M) of HA
  that dissociates
• - May end up with quadratic equation when
  solving for X solve two ways
• 1) Quadratic
• for a x2 b x c 0
• 2) Approximation Simplify math if causes less
  than a 5 error. Can check approximation by
  either of two methods a) make approx. check
  to see if x is less than 5 of what comparing it
  to, or use book method b) if HA / Ka gt 100,
  then approximation OK this last method allows
  you to check before doing math.
• - See next page for general problem solving
  method used for rest of chapter.

7

• 1. Write equil. rxn equil. expression with
  given chemical as reactant.
• If have to reverse, then Ka 10-14 / Kb
• 2. If a rxn takes place, then allow to occur
  calculate the new initial M of reactants and
  products. Rxn always occurs when strong acid
  mixed with any base OR strong base mixed
  with any acid.
• 3. Let X M of chemical that reacts.
• 4. Write down initial M of each then
  equilibria M of each.
• 5. Make approximation plug equilibria M into
  equilibrium expression
• 6. Solve for X.

8
II. Calculations Involving HA Outline of rest
of chapter Note All problems worked by method
on previous page

• A. Weak Acids
• B. Weak Bases
• C. Salts of Weak Acids Salts of Weak Bases
• D. Common Ion (weak acid or base plus the
  salt)
• E. Buffers (same as common ion
  problem)
• F. Titrations (1. allow rxn to occur calc.
  new initial M )
• (2. determine equilibria values using one of
  above)

9
A. Overview Weak Acid (only HA)

• HA H A- Ka
  Small
• - Given Y init. M of HA let x M of HA that
  ionizes
• - Drop x if small wrt to Y - if HA / Ka gt
  100
• Ka HA- x2
  x2
• HA Y-x Y
• x v Y Ka

10
A. Weak Acids - Example

• Calculate the pH of 0.10 M Acetic Acid (HA), Ka
  1.7 x 10-5
• HA H A- Ka 1.7 x 10-5 H
  A- / HA
• Let x M of acetic acid that ionizes, then At
  equilibrium
• 0.10 x x x
• HA H
  A-
• Check to see if can drop x wrt 0.10 HA/Ka
  0.10 / 1.7 x 10-5 6000 Yes, gt100
• 1.7 x 10-5 H A-
  xx xx x2
• HA 0.10 - x
  0.10 0.10
• x v 0.10 (1.7 x 10-5) 1.3 x
  10-3 M H
• pH - log (1.3 x 10-3) 2.89

11
A. Weak Acids - Notes

• Notes 1) For 0.10 M acetic acid with a Ka
  1.7x10-5, the approximation was OK.
• - When will it not be OK? When the answer
  causes gt 5 error or when HA/Ka lt 100
• - Generally will not be OK when 1) HA is
  small /or 2) Ka is large
• Example 1.0x10-4 M acetic acid. HA/Ka
  1.0x10-4 / 1.7x10-5 5.9
• Here we cant drop x have to include
  1.0x10-4 x use quadratic eqn.
• Notes 2) May be able to make a time saving
  assumption with polyprotic acids
• Polyprotic Acid one which releases more than
  one H per molecule
• Example H3PO4 H
  H2PO41- H HPO42-
  H PO43-
• Ka1 7x10-3
  Ka2 6x10-8 Ka3 4x10-13
• Most of acid comes from 1st step so can ignore
  contributions from other steps if causes less
  than a 5 error (true when Ka2 value is 1000
  times smaller than Ka1).

12
A. Weak Acids - Polyprotic Acid Examples for
H2AKa1 1.0x10-5 Ka2 2.0x10-10 H2A
(---) H HA- HA- (---) H A-2

• Example 1 Calculate H of 2.0 M H2A
  (note Ka2 ltKa1 H2A/Ka1 2x105)
• 2.0-x x x
  Let x M of H2A that reacts M of H M
  of HA-
• H2A H HA-
• 1.0x10-5 HHA- x2 x
  v2.0x10-5 x HA-H 4.5 x 10-3 M
• H2A 2.0
• Example 2 Calculate A-2 of 2.0 M H2A (note
  Ka2 ltltlt Ka1)
• Note Starting M of H of HA- from example
  1 4.5 x 10-3 M HA-/Ka2 2x107
• 4.5x10-3 - y 4.5x10-3 y
  y Let y M of HA- that reacts
• HA- H
  A-2
• 2.0x10-10 HA-2 4.5 x 10-3y
  y A-2 2.0x10-10 M Ka2
• HA- 4.5 x 10-3

13
B. Overview Weak Base (Only Base)

• - NH3 typical weak base reacts with H2O
  producing OH-
• NH3 H2O NH4 OH- Kb Small
• Initial M Y --
  0 0
• Equil M Y-x Y
  x x
• - Given Y init. M of base. Let x M of NH3
  that reacts
• - Drop x if small wrt to Y if NH3 / Kb gt 100
• Kb NH4OH- x2 x2
• NH3 Y-x Y
• x OH- v Y Kb

14
B. Weak Bases

• - Equilibria involving a weak base can be
  treated similar to a weak acid.
• - Some weak bases contain a N with a pair of n
  electrons. NH3 NH2-CH3
• NH3 H2O (-----) NH4 OH- Kb
  NH4OH- / NH3 Water is solvent
• Example Calculate the OH- pH of 0.50 M NH3
  in water. Kb 1.8x10-5
• Let x M of NH3 that reacts Note
  0.50/ 1.8x10-5 30,000 (drop x)
• NH3 H2O (-----) NH4
  OH-
• 0.50-x 0.50
  x x
• 1.8x10-5 NH4OH- x2 x
  OH- v (1.8x10-5) x 0.50 3.0x10-3 M
• NH3 0.50
• Kw HOH- H Kw / OH-
  1.00x10-14 / 3.0x10-3 3.3x10-12 M
• pH - log 3.3x10-12 11.48

15
C. Overview Salt of Weak Base

• - NH3 (weak base) reacts with H2O to produce OH-
  NH4 Kb 1.8x10-5
• - NH4 is the salt of a weak base and is itself a
  weak acid.
• - An aqueous solution of NH4Cl (NH4 Cl-) is
  acidic.
• NH4 H NH3
• Ka HNH3
• NH4
• - Note this Ka is not in table, but Kb is and
  Ka x Kb Kw
• Ka Kw/Kb 1.0x10-14 / 1.8x10-5
  5.6x10-10
• - Let x M of NH4 that reacts solve for x
  Ka x2 / NH4
• - for 1.0 M NH4 5.6x10-10 x2 / 1.0 x
  2.4x10-5 pH 4.6

16
C. Overview Salt of Weak Acid

• - A weak acid (HA) yields H A- (weak base) in
  water.
• - If place NaA (A-) in H2O, it will pull H off to
  yield OH-
• A- H2O HA OH-
• - Salts of weak acids (NaC2H3O2, KCN) are weak
  bases in H2O. Sometimes this reaction is called
  hydrolysis.
• Kb or Kh HAOH- Note these Khs
  usually not in table
• A-
• - Get Kh from the table Ka value and Ka x Kh
  Kw
• - Let x M of A- that reacts solve for x Kh
  x2/A-

17
C. Salts of Weak Acids Bases

• - Salts of weak acids like KCN or KC2H3O2 want
  H back are bases.
• - Salts of weak bases like NH4Cl want to give up
  H are acids.
• - In water these may be called Hydrolysis
  reactions.
• Examples 1) NaCN 2) NH4Cl
  (both in water)
• 1) CN- H2O (-----) HCN OH- (Basic)
• 2) NH4 H2O (-----) NH3 H3O
  (Acidic)
• Simplified as NH4 (-----) NH3 H
  
• Note Salts of strong acids like NaCl or KNO3 do
  not influence pH.
• Examples Calculate pH of 1.0 M NaCl 1.0 M
  NH4Cl 1.0 M KCN.
• Answers NaCl 7.0 NH4Cl 4.6 KCN
  11.7 (see next page)

18
C. Salts of Weak Acids Bases

• 1) The K for hydrolysis is usually not in a
  table however
• Ka x Kb Kw or Ka x Kh Kw
• 2) Example Calculation with 1.0 M NaCN
  Calculate OH-.
• - Ka for HCN 4.9x10-10 Kb Kw/Ka
  1.0x10-14/4.9x10-10 2.0x10-5
• - Let x M of CN- that reacts.
• CN- H2O HCN
  OH-
• 1.0-x 0.10 x
  x
• 2.0x10-5 HCNOH- x2 x
  OH- 4.5x10-3 M
• CN- 1.0 pOH 2.35
• - Note pH pOH 14.00 pH 14.00 - 2.35
  11.65

19
D. Overview Common Ion Problem (Two Initial
Chemicals)

• - Past problems only had one initial component
  now figure acidity if have two initial components
  such as HA A-
• - Method a) set up equilibria
• b) let x M of HA that reacts
• c) make simplifications solve for x
• - Example 1 M HA 2 M NaA (2 M A-) Calculate
  amt H
• 1 - x 1 x 2 x 2
• HA H A-
• Ka H A- x 2 x H Ka
  / 2
• HA 1

20
D. Common Ion Effect

• In the past, our calculations have involved only
  one component of an equilibria. Now predict the
  result when two components are added a weak
  acid and its salt (common ion) or a weak base and
  its salt (common ion).
• Calculate the pH of a solution which is 0.40 M in
  acetic acid (HOAc) and 0.20 M in sodium acetate
  (NaOAc). Let x M of acetic acid that
  dissociates. Ka1.7x10-5
• Initial M 0.40 0 0.20
• HOAc H OAc-
• Equilibrium M 0.40-x x
  0.20x C/Ka gt 100 for 0.4 0.2 drop x
• 1.7x10-5 HOAc- x0.20
  x H 3.4x10-5 M pH 4.47
• HOAc 0.40
• Notes 1) pH in 0.4 M HOAc 2.6 pH in
  0.4 M HOAc 0.2 M NaOAc 4.5
• 2) Le Chateliers Principle shows why
  adding 0.20 M NaOAc is more basic.

21
E. Overview Buffer Problems (Common Ion
Problem)

• - A solution which resists a change in pH is
  called a buffer.
• - Mixtures of weak acids their salts or weak
  bases and their salts are buffers.
• - Why does a weak acid (HA) its salt (NaA or
  A-) resist a change in pH when either additional
  H or OH- are added? Because a buffer consumes
  strong acids or bases.
• HA reacts with the base HA OH- -----) A-
  H2O
• A- reacts with the H A- H -----) HA
• - As long as HA and A- are available, any added
  H or OH- is consumed .

22
E. Buffers

• Example Water H2O
• If add 1 drop 10 M HCl to 10 mL water the pH
  goes from 7 to 1
• If add 1 drop 10 M NaOH to 10 mL water the pH
  goes from 7 to 13
• Water is not a buffer the ?pH was 6 in both
  cases or the ?H 106 M change!
• Example 1 M Acetic Acid and 1 M Sodium Acetate
  - HA NaA
• If add 1 drop 10 M HCl to 10 mL of the above,
  the pH goes from 4.8 to 4.7
• If add 1 drop 10 M NaOH to 10 mL of the above,
  the pH goes from 4.8 to 4.9
• The ? pH 0.1 in both cases, or the ?H
  10-1 M change!
• Notes a) the buffer capacity is a measure of
  the amount of strong acid or base consumed by
  the buffer this depends upon the initial M of
  the HA and A-. Best to have largest M of HA
  and A- possible.
• b) the effective pH range of the buffer
  pKa 1 pH unit.
• pKa of acetic acid 4.8 so, the effective
  buffer range 3.8 to 5.8 for acetic acid

23
E. Buffers

• Henderson-Hasselbalch Equation - Used for Buffers
  Common Ion
• For a weak acid HA H A-
• Ka H A- H Ka HA Now take
  neg log of
• HA A- both sides.
• pH pKa log HA pH pKa log A-
  
• A- HA
• pH pKa log A-
• HA
• Henderson-Hasselbalch Equation (H-H eqn)

24
E. Buffers Notes

• 1) A buffer work best when HA/A- 11
  1.00 When a) HA A- small, or when b)
  ratio of HA to A- is far from 11, then the
  buffer is poor. When generating a buffer, choose
  a buffer with a pKa within 1.0 of the desired
  acidity. Which acid will buffer best at pH4?
  HF (Ka 2x10-4) or HCN (Ka 5x10-10)
• 2) A quick lab way to get pKa is to measure pH of
  a mix of equal M HA A- (true at half way to the
  end point in a titration).
• pH pKa log A- /HA pH pKa log
  1 pH pKa 0 pH pKa
• 3) A weak base and its salt such as NH3 and
  NH4Cl (NH4) is also a buffer.
• 4) Buffer Problem Solving Method when adding
  strong H or OH- a) allow the added strong
  acid or base to react completely with buffer, b)
  calculate the new MOLES (or M) of HA A- and c)
  plug into the HH Equation.

25
E. Buffers - Example Problem 1

• Calculate the pH of a buffer made from adding
  0.50 mole of acetic acid and 0.30 mole of sodium
  acetate to enough water to make 2.0 L. (0.25 M
  HA 0.15 M A-)
• pH pKa log A-/HA -log 1.7x10-5 log
  0.15/0.25
• pH 4.77 log (0.60) 4.77
  (-0.22) 4.55
• Notes 1) we made OK assumption that HA x
  HA and A- x A-
• 2) volume drops out of H-H eqn so, can use
  either moles or M
• - Using moles pH -log 1.7x10-5 log
  0.30/0.50 4.55 (same value)
• Biological buffers Proteins like albumin
  their salts
• H2PO4-1 HPO4-2
• H2CO3 and HCO3-

26
E. Buffers - Example Problem 2, part 1

• 1.00 L of 0.100 M (0.100 moles) weak acid
  (HA)with a Ka1.00x10-8 is mixed with 1.00 L of
  0.200 M (0.200 moles) NaA. Calculate 1) the pH,
  and then 2) the pH after adding 0.100 L of 0.100
  M (0.0100 moles) NaOH.
• 1) Calculate the pH. Note can use moles in H-H
  eqn HA/Ka gt 100
• Initial moles 0.100 ?
  0.200
• HA H A-
• Equil. moles 0.100-x0.100
  x 0.200x0.200
• pH pKa log A- / HA -log
  (1.00x10-8) log 0.200/0.100
• pH 8.00 0.301 8.30

27
E. Buffers - Example Problem 2, part 2

• 0.100 moles HA (Ka1.00x10-8) is mixed with 0.200
  moles NaA. 2) Calculate pH after adding 0.100 L
  of 0.100 M NaOH (0.0100 mole OH-).
• Note a) allow OH- to react with HA, producing
  more A- calculate new initial moles of HA A-.
  b) Solve using Henderson Hasselbalch equation.
• a) Initial moles 0.100 0.0100
  0.200 --
• Reaction (before equil) 1HA 1OH-
  -----) 1A- H2O
• New initial moles HA 0.100 mole 0.0100 mole
  0.090 mole HA
• New initial moles A- 0.200 mole 0.0100 mole
  0.210 mole A-
• New initial moles OH- 0.0100 mole-
  0.0100 mole 0.0000 mole OH-
• b) Solve for pH
• pH pKa log A- / HA -log
  (1.00x10-8) log 0.210/0.090
• pH 8.00 0.37 8.37 (note small change
  in pH was 8.30 before adding OH-)

28
F. Overview Titrations
29
F. Overview Titration Curve (plot pH vs Vol
Std) Calculations

• pH Determination while Titrating a Strong Acid
  like HCl (or Base)
• Before Titration Starts Initial pH from M of
  HCl
• During Titration Before End Point Allow
  reaction to occur Calculate excess HCl left
  over pH from excess HCl.
• At End Point Allow rxn to occur No excess HCl
  or NaOH pH from H2O 7
• After End Point Allow reaction to occur
  Calculate Excess NaOH pH from excess NaOH.
• pH Determination while Titrating a Weak Acid like
  HA (or Base)
• Before Titration Starts Initial pH from M of HA
• During Titration Before End Point Allow
  reaction to occur Calculate excess HA A- left
  over pH from excess HA A-.
• At End Point Allow reaction to occur Only
  have A- pH from reaction of A- with H2O (get
  pH from M A- Kh.
• After End Point Allow reaction to occur
  Calculate Excess NaOH pH from excess NaOH.

30
F. Titrations

• Definition Volumetric determination of the
  amount of an acid or base by addition of a
  standard acid or base until neutralization has
  occurred.
• Notes 1) Neutralization Point End Point or
  Equivalence Point
• 2) A burette (buret) is usually used to add the
  titrant can be read to 0.02 mL
• 3) a plot of pH vs volume of titrant is called a
  pH curve or a titration curve.
• 4) End Point can be determined by either an
  indicator or by use of a pH meter. Pick an
  indicator that changes color near to the end
  point pH or if using a pH meter, the steepest
  portion of the curve is the end point (can also
  plot the 1st or 2nd derivative vs volume added).
• 5) Can easily determine Ka since pH pKa ½ way
  to the end point for a weak acid.
• 6) A quick examination of the pH curve will tell
  if titrating an acid or base as well as if they
  are strong or weak.

31
F. Titration 1 - 25.0 mL of 0.100 M HCl with
0.100 M NaOH (Strong Acid with a Strong Base).
Note Large change in pH near end point and pH
at equivalence point 7.00
32
F. Titration 2 - 25.0 mL of 0.100 M Nicotinic
Acid (Ka1.4x10-5) with 0.100 M NaOH (Weak Acid
with a Strong Base). Note Smaller change in pH
near end point pH at equival. point 8.78
33
F. Titration of a Strong Acid with NaOH
Calculations

• Calculate pH for titration of 10. mL of 0.80 M
  HCl with 1.0 M NaOH at 0.0 mL, 3.0 mL, 8.0 mL
  12.0 mL 1.0 M NaOH added.
• Notes allow reaction to occur then calculate
  pH. The pH is due to excess strong acid or base
  except at end point.
• Titration Rxn
• 1HCl 1NaOH
  1NaCl 1H2O
• 0.0 mL NaOH added Only have HCl. pH - log
  0.80 0.097
• 3.0 mL NaOH added 0.0080 moles HCl 0.0030
  moles NaOH yields 0.0050 moles of HCl 0.0030
  moles of NaCl left over in a total volume of 13.0
  mL. 0.0050 m / 0.0130 L 0.384 M HCl
• pH - log 0.384 0.41

34
F. Titration of a Strong Acid with NaOH
Calculations - Continued

• Calculate pH for titration of 10. mL of 0.80 M
  HCl with 1.0 M NaOH at 0.0 mL, 3.0 mL, 8.0 mL
  12.0 mL 1.0 M NaOH added.
• 1HCl 1NaOH
  1NaCl 1H2O
• 8.0 mL NaOH added have 0.0080 moles of HCl
  0.0080 moles of NaOH. All HCl NaOH reacted and
  have 0.0080 moles of NaCl (salt of a strong acid)
  in 16.0 mL H2O pH pH of neutral water 7.0
• 12.0 mL NaOH added 0.0080 moles HCl 0.012
  moles NaOH yields 0.000 moles of HCl, 0.004 moles
  of NaOH, 0.0080 moles of NaCl left over in a
  total volume of 22.0 mL. 0.004 m / 0.022 L
  0.18 M NaOH.
• pOH - log 0.18 0.74 pH Kw - pOH
  14.00 0.74 13.3

35
F. Titration of a weak Acid with NaOH
Calculations

• Calculate pH for titration of 10. mL of 0.80 M HA
  with 1.0 M NaOH at 0.0 mL, 3.0 mL, 8.0 mL 12.0
  mL 1.0 M NaOH added. Ka1.0x10-6
• Notes allow reaction to occur then calculate
  pH.
• Titration Rxn 1HA 1NaOH
  1NaA 1H2O
• Equil. Rxn HA H A-
  1.0x10-6 HA- / HA
• 0.0 mL added Only have weak acid. Let x M of
  HA that reacts.
• At equilibrium H A- x HA
  0.80-x 0.80 since x small
• 1.0x10-6 HA- x2 x
  8.9x10-4 H
• HA 0.80
• pH - log 8.9x10-4 3.05 (Note
  start less acidic than HCl titration)

36
F. Titration of a weak Acid with NaOH
Calculations - continued

• Calculate pH for titration of 10. mL of 0.80 M HA
  with 1.0 M NaOH at 0.0 mL, 3.0 mL, 8.0 mL 12.0
  mL 1.0 M NaOH added. Ka1.0x10-6
• Notes allow reaction to occur then calculate
  pH.
• 1HA 1NaOH
  1NaA 1H2O
• HA H A- 1.0x10-6
  HA- / HA
• 3.0 mL added Allow the NaOH to react with the
  HA.
• 0.0080 moles HA 0.0030 moles NaOH yields
• 0.0050 moles of HA 0.0030 moles of NaA left
  over in 13.0 mL.
• pH pKa log A- / HA -log 1.0x10-6
  log 0.0030 / 0.0050
• pH 6.00 0.222 5.78

37
F. Titration of a weak Acid with NaOH
Calculations - continued

• Calculate pH for titration of 10. mL of 0.80 M HA
  with 1.0 M NaOH at 0.0 mL, 3.0 mL, 8.0 mL 12.0
  mL 1.0 M NaOH added. Ka1.0x10-6
• Notes allow reaction to occur then calculate
  pH.
• 1HA 1NaOH
  1NaA 1H2O
• 8.0 mL added 0.0080 moles HA 0.0080 moles
  NaOH yields 0.00 moles of HA 0.0080 moles of
  NaA left over in 18.0 mL.
• A- 0.0080 m / 0.0180 L 0.444 M A-
• A- H2O (------) HA OH- Kh Kw/Ka
  1.0x10-14 / 1.0x10-6 1.0x10-8
• Let x M of A- that reacts A- 0.44-x
  0.44 HA OH- x
• 1.0x10-8 x2 / 0.444 x OH- 6.66x10-5
• H 1.0x10-14 / 6.66x10-5 1.50x10-10 pH
  9.82 (end point)

38
F. Titration of a weak Acid with NaOH
Calculations - continued

• Calculate pH for titration of 10. mL of 0.80 M HA
  with 1.0 M NaOH at 0.0 mL, 3.0 mL, 8.0 mL 12.0
  mL 1.0 M NaOH added. Ka1.0x10-6
• Notes allow reaction to occur then calculate
  pH.
• 1HA 1NaOH
  1NaA 1H2O
• 12.0 mL added 0.0080 moles HA 0.0120 moles
  NaOH yields 0.00 moles of HA 0.0080 moles of
  NaA 0.0040 moles of NaOH left over in 22.0 mL.
• A- 0.0080 m / 0.0220 L 0.36 M A-
• OH- 0.0040 m / 0.0220 L 0.182 M OH-
• Note In a mixture of a weak strong base, pH
  is due to strong base
• H Kw / OH- 1.0x10-14 / 0.182
  5.5x10-14 pH 13.26

39
III. Summary Let X M of reactant that reacts
to reach equilibria

• Weak Acids HA (---) HA-
  KaHA-/HA
• Salts of Weak Acids A-H2O (---)
  HAOH- KhHAOH-/A-
• Common Ion/Buffer HA (---) HA-
  KaHA-/HA
• Notes 1. Can use H-H Eqn pH pKa Log
  A-/HA
• 2. If add strong H or OH- a) allow to react,
  b) calculate new moles or M of HA A-, c)
  plug into H-H eqn.
• Titrations Strong H with OH-. a) initially
  pH due to strong H. b) before end pt. allow
  to react calculate excess strong H. c) at end
  pt. pH 7.0. d) after end pt. allow to
  react calculate excess OH-, pH due to strong
  OH-.
• Titrations Weak HA with OH-. a) initially pH
  due to HA. b) before end pt. allow to react
  calculate excess HA A- . c) at end pt. pH
  due to A-. d) after end pt. allow to react
  calculate excess OH-, pH due to strong OH-.

40
III. Summary - One method for all weak
acid-base problems

• 1. Write equil. rxn equil. expression with
  given chemical as reactant.
• If have to reverse, then Ka 1.00 x10-14
  / Kb
• 2. If a rxn takes place, then allow to occur
  calculate the new initial M of reactants and
  products. Rxn always occurs when strong acid
  mixed with any base or strong base mixed with any
  acid.
• 3. Let X M of chemical that reacts.
• 4. Write down initial M of each equilibria M
  of each.
• 5. Make approximations plug equilibria M into
  equil. expression.
• 6. Solve for X.