Chapter 17
Acid-Base Equilibria
8–1
John A. Schreifels
Chemistry 212
Chapter 17-1
Overview
• Solutions of a Weak Acid or Base
–
–
–
–
Acid ionization equilibria
Polyprotic acids
Base ionization equilbria
Acid-Base properties of Salts
• Solutions of a Weak Acid or Base with Another Solute
– Common Ion Effect
– Buffers
– Acid-Base Titration Curves
8–2
John A. Schreifels
Chemistry 212
Chapter 17-2
Acid –Ionization Equilibria
•
•
•
Weak acids and weak bases only partially dissociate; their strengths are
experimentally determined in the same way as strong acids and bases by
determining the electrical conductivity.
The reaction of a weak acid (or base) with water is the same as discussed in
previous section.
Consider the reaction:
[H3O ][ A  ]
Ka 
[HA ]
Hydronium ion concentration must be determined from the equilibrium expression.
HA(aq)  H2O(l)  H3O (aq)  A  (aq)
–
•
Relative strengths of weak acids can be determined from the value of the
equilibrium constant.
–
–
Large equilibrium constant means strong acid
Small equilibrium constant means weak acid
E.g. determine which acid is the strongest and which the weakest.
Acid
HCN
HCOOH
CH3COOH
HF
John A. Schreifels
Chemistry 212
Ka
4.9x1010
1.8x104
1.8x105
3.5x104
8–3
Chapter 17-3
Determining K from pH
• Ka determined if pH and CHA known.
– Use the equilibrium expression for the acid.
E.g. Determine the equilibrium constant of acetic acid if the pH of a
0.260 M solution was 2.68. Determine [H3O+]; [HA]; and [A].
Initial conc
 to Equil.
Equil.
HA(aq)+H2O(l)  H3O+(aq)
0.100 M
0
+x
x
+x
0.100 M  x
+ A(aq)
0
+x
+x
– Strategy
• Calculate the [H3O+] from pH; this is x in the table above.
• The rest of the quantities are obtained from the bottom row.
8–4
John A. Schreifels
Chemistry 212
Chapter 17-4
Calculating Equilibrium Concentrations in Weak–
acid Solutions
• pH determined if Ka and Ca known; for the dissociation of
acetic acid:
CH 3COOH (aq )  H 2O(l)  H3O (aq)  CH3COO  (aq)
[H3O ][CH3COO  ]
Ka 
[CH COOH ]
3
 1.8 x105
2H2O(l)  H3O (aq)  OH (aq)
K w  [H3O ][OH ]
 1.00x1014
– [H3O+]total = [H3O+]CH3COOH + [H3O+]H2O.
– [H3O+]total  [H3O+]CH3COOH.
• The total hydronium ion concentration is often equal to the
contribution from the weak acid which is usually a lot stronger
acid than water.
• The total hydronium ion concentration is needed for the
equilibrium calculation.
John A. Schreifels
Chemistry 212
Chapter 17-5
8–5
pH from Ka and Ca
E.g. Calculate the pH of 0.100M acetic acid. Given pKa = 4.76
Initial conc
 to Equil.
Equil.
CH3COOH(aq)+H2O(l)  H3O+(aq)+ CH3COO(aq)
0.100 M
0
0
+x
+x
x
+x
+x
0.100 M  x
Method I:
x2

5
• Substitute into equilibrium equation to get  1.75x10 
0.100  x
• x2 + 1.75x105x  1.75x106 = 0.
• Solve using quadratic equation (see book).
Method 2
• Assume x << CHA. Then x = (KaCHA)1/2.
• Check (confirm assumption to be correct)
– Analytical concentration should be: Ca = 100x[H3O+]
Method 3 method of successive approximations.
• As in Method 2; then
• x = (Ka(CHA  x1))1/2; repeat if necessary.
E.g. Calculate pH of 0.0200M lactic acid if its Ka = 8.4x104M.
John A. Schreifels
Chemistry 212
8–6
Chapter 17-6
% Dissociated (also called % Ionized) Weak Acids
•
% ionization – a useful way of expressing the strength of an acid or base.
–
–
100% ionized a strong acid.
Only partial ionization occurs with weak acids.
HA(aq)+ H2O(l)  H3O+(aq)+ A(aq)
Initial conc
CHA
0
0
+x
 to Equil.
+x
x
Equil. CHA  x
+x
+x
x
% Ionization 
 100%
CHA
E.g. determine the % ionization for 0.100 M, 0.0100 M, 0.00100M HCN if Ka =
4.9x1010.
–
Solution: determine x for each and sub into definition above. Check assumptions.
% Ionization 
CHA  K a 1/ 2
CHA
 K
  a
 CHA
•
 100%
1/ 2




 100%
8–7
Notice % ionization increases with dilution.
John A. Schreifels
Chemistry 212
Chapter 17-7
Polyprotic Acids
• Some acids can donate more
than one proton to the solution.
Thus a diprotic acid has two
protons such as H2S and
H2SO4, while a common
triprotic acid has three acidic
protons that can be donated
(H3PO4).
• First proton easily removed;
others much more difficult.
Treat Polyprotic acids
as if they were
monoprotic acids; Use
Ka1.
Formula Ka1
Ka2
Ka3
H3PO4
7.5x103 6.2x108 4.8x1011
H2SO3
1.5x102 6.3x108
•The equilibrium constant for removal of each successive proton is
about 105 times the equilibrium constant for removal of the
preceeding proton.
2
•E.g. determine the pH of 0.100 M H2SO3. Then determine [SO3. ]
8–8
John A. Schreifels
Chemistry 212
Chapter 17-8
Equilibria:Weak bases (WB)
(proton acceptor)
•
•
•
•
•
•
•
Treat bases just like we did the weak acid; except you are calculating
[OH].
The general equation that describes the behavior of a base in solution
is:
[BH ][OH ]


B(aq)  H2O(l)  BH (aq)  OH (aq)
Kb 
[B]
Set up the equilibrium table as before for the acids and substitute values
for all the quanitities in the equilibrium expression.
B(aq)+H2O(l)  BH+(aq)+ OH(aq)
Initial conc.
CB
0
0
+x
+x
 to Equil.
x
Equil. CB  x
+x
+x
Since usually CB is supplied, we have one unknown which we can
evaluate using standard equil. equation for weak base.
[BH  ][OH  ]
Kb 
Remember that x = [OH] and not [H3O+].
[B]
x2
E.g. Calculate the pH of 0.10M NH3(aq).

CB  x
Hint: Expect pH > 7 when with weak base.
8–9
John A. Schreifels
Chemistry 212
Chapter 17-9
Equilibria:Weak bases Structure
• Many nitrogen containing compounds
are basic –the amine most important.
• Most of the amines have a lone pair
of electrons that are available for
bonding with an acidic proton
(Brønsted-Lowry base).
• Amines usually have a carbon
residue in place of a hydrogen.
8–10
John A. Schreifels
Chemistry 212
Chapter 17-10
Relation between Ka and Kb
• Ka and Kb are always inversely related
to each other in aqueous solutions.
HA(aq)+ H2O(l)  H3O+(aq) + A(aq)
Add A(aq)+ H2O(l)  HA(aq) +OH(aq)
•
Ka 
Kb 
[ A  ][H3 O  ]
[HA ]
[HA ][OH  ]
[A  ]
2H2O(l)  H3O+(aq)+OH(aq) Kw = Ka•Kb
Inverse relationship explains why
Acid Ka
conjugate base of very weak acid is
4
HF
3.5x10
relatively strong.
HCOOH
E.g. given the Ka’s of the following acid
4
1.8x10
list their conjugate bases in terms of
8
HOCl
3.5x10
relative strength.
HCN
John A. Schreifels
Chemistry 212
4.9x1010
8–11
Chapter 17-11
Salts of WA and WB
• Salt: an ionic substance formed as a result of an acid–base
neutralization reaction.
– Salt of an acid(base) obtained by its neutralization with acid if it is a
base and base if it is an acid.
E.g. NaCl is a salt from the reaction of HCl with NaOH.
– The properties of the salt will depend upon the strengths of the acid
and base that formed the salt.
E.g.1: determine the acid–base reaction that would produce
CH3COONa, NaCN, NH4Cl, (NH4)2CO3.
• Salts are usually soluble in water because of their ionic
character.
• When they dissolve, they affect the pH of the solution. Depends
upon relative strengths of the conjugate acid and base.
8–12
John A. Schreifels
Chemistry 212
Chapter 17-12
Salt of Strong Acid and Strong Base
•
•
•
•
•
•
Neutral solution results if the salt is from the reaction of a SA + SB.
E.g. NaCl
Other cations and anions producing neutral solutions: Li+, Na+, K+,
Ca2+, Sr2+, Ba2+ and Cl, Br, I,NO 3, ClO 4 ).
E.g. what is the approximate pH of the following. NaCl, KCl, LiClO4,
etc.?
Salt of WA + SB (basic) and Salt of WB + SA (acidic).
Ignore cation (or anion) from SA (base).
Conjugate of WA is WB  basic solution.
Conjugate of WB is WA  acidic solution.
SA + SB  Neutral (very WA & WB)
SA + WB  Acidic (WA)
WA + SB  Basic (WB)
where
SA = Strong Acid; SB = Strong Base
WA = Weak Acid; WB = Weak Base
John A. Schreifels
Chemistry 212
8–13
Chapter 17-13
Calculating the pH of Salt of WA or WB (other ion
from SA(SB))
• Salt of WA: Use Kb of the conjugate base and treat it
as a weak base:
A(aq) + H2O(l)  HA(aq) + OH(aq)
E.g. determine the pH of 0.100M NaCH3COO. Ka
(CH3COOH) = 1.75x105.
E.g. determine the pH of 0.200 M NaCN. Ka(HCN) =
4.9x1010.
• Salt of WB: Use Ka of conjugate acid and treat as a
weak acid:
E.g. determine pH of 0.250M NH4Cl. Kb = 1.8x105.
E.g. determine pH of 0.100 M N2H5Br. Kb = 1.1x108. 8–14
John A. Schreifels
Chemistry 212
Chapter 17-14
Salt of WA + WB
• Determine Ka and Kb of acidic and basic portions of
salt.
• Largest K dominates to make solution either acidic
or basic.
• E.g. determine if 0.100 M NH4CN is acidic or basic.
E.g. 2 predict if 0.100 M C6H5NH3F is acidic or basic.
8–15
John A. Schreifels
Chemistry 212
Chapter 17-15
The Common Ion Effect
•
•
•
•
•
•
•
Common–Ion Effect: the change in the equilibrium that results from the
addition of an ion that is involved in the equilibrium.
E.g. NaOCl is added to 0.100 M HOCl;NH4Cl is added to NH3.
Setting up the standard equilibrium table can show the effect.
E.g. determine the pH of a solution prepared by mixing 50.0 mL of 0.100 M
HOCl with 50.0 mL of 0.100 M NaOCl (Ka = 3.5x108).
Set up equilibrium table after calculating the concentrations of each in the
final mixture.
Initial concentrations change slightly as a result of a change reaction.
HOCl + H2O H3O+ + OCl
0.0500M
0
0.0500M
+x
+x
x
+x
0.0500 + x
0.0500 x
Solve using either approximations or quadratic equation.
Shifts equilibrium towards the basic side.
8–16
John A. Schreifels
Chemistry 212
Chapter 17-16
Buffers
•
Buffer solution: a mixture of conjugate acid and base that resists pH
changes.
– Significant buffering capacity occurs when [acid] = [base], pH = pKa.
– An example of the common ion effect.
E.g. Calculate pH of solution containing 0.040M Na2HPO4 and 0.080M
KH2PO4. pKa2=7.20.
– Set up equilibrium table.
– Ignore the value of x compared to the concentrations of the common ion.
– pH in buffering region related to the relative amount of conjugate acid and
base.
•
Let r  [base ] then the equilibrium equation is:
[acid]
K a  [H3O  ]  r
K
[H3O  ]  a
r
8–17
John A. Schreifels
Chemistry 212
Chapter 17-17
Addition of Acid or Base to a Buffer
• Upon addition of a SB to the buffer we have:
– Addition of either acid or base changes ratio of acidic and basic
forms.
– Big changes in pH occur only when nearly all of one species is
consumed.
Cb Vb
r
Ca Va  Cb Vb
E.g. determine r after addition of 5.00 mL of 0.100 M NaOH to
10.00 mL of 0.100 M HOCl. Determine pH if Ka = 3.5x108.
E.g. Determine pH of 50.00 mL of phosphate buffer containing
equilmolar concentrations (0.200M) of acid/base forms, after
10.00 mL 0.100 M NaOH or 10.00 mL of HCl. pKa2 =7.20
• Changes in volume don't affect pH.
8–18
John A. Schreifels
Chemistry 212
Chapter 17-18
Henderson-Hasselbalch Equation
• The effect of r (=[A]/[HA]) on pH is better understood by taking
log of both sides of equation between K and conc. To give
 [ A ] 

pH = pK a + log
 [HA] 
 nA 

= pK a + log
n 
 HA 
• Called Henderson-Hasselbach equation.
• Allows us to predict pH when HA/A mixed.
• When [A] /[HA] = 1 (i.e. [HA]=[A]), pH = pKa
E.g. Calculate pH of solution containing 0.040M Na2HPO4 and 0.080M
KH2PO4. pKa2=7.20.
E.g.2 determine the ratio of the concentration of the conjugate acid to
concentration of the conjugate base for a weak acid in which the pH
was 5.45 and pKa was 5.75.
E.g. determine the pH of a solution consisting of 0.100 M NH3 and
8–19
0.150 M NH4Cl.
John A. Schreifels
Chemistry 212
Chapter 17-19
Neutralization Reactions
• Neutralization Reaction: the reaction of an acid with a base to
produce water.
• Extent of reaction nearly quantitative (except if both acid and
base are weak.
• SA–SB:
– E.g. HNO3 + NaOH  NaNO3 + H2O
• SA produces: H3O+
• SB produces: OH
• Overall reaction: H3O+ + OH  2H2O K 1 = 1/Kw = 1.00x1014
w
• WA–SB: thought of as two step reaction.
– E.g. HOCl + NaOH  NaOCl + H2O K = ?
HOCl  H+ + OCl
H+ + OH H2O
Ka = 3.5x108
K w1= 1.00x10+14
HOCl + OH H2O + OCl K = KaK 1= 3.5x106
w
8–20
• Large equilibrium constant means reaction nearly quantitative.
John A. Schreifels
Chemistry 212
Chapter 17-20
Neutralization Reactions – WB + SA and WA + WB
•
WB + SA
– SA produces H3O+ ions; use base as is.
– E.g. NH3 + HCl  NH  + Cl or
4

NH3 + H2O NH4 + OH
H3O+ + OH 2H2O
•
•
•
Ka = 1.8x105
K w1= 1.00x10+14
NH3 + H3O+ NH4 + H2O
K = KaK w1= 1.8x109
HOCl + H2O  H3O+ + OCl
NH3 + H2O NH + OH
4
+

H3O + OH 2H2O
Ka = 3.5x108
Kb = 1.8x105

NH3 + HOCl NH4 + OCl
K = KaKbK w1 = 63
Conclusion: Quantitatively generate product (nearly).
WA + WB: initially undissociated species dominates.
K w1= 1.00x10+14
Conclusion: Reaction will sometimes, but not always, be quantitative.
E.g. determine the extent of reaction when di methyl amine (Kb =
5.4x104) reacts with either HF (Ka = 3.5x104) or HOCl (Ka = 3.5x108). 8–21
John A. Schreifels
Chemistry 212
Chapter 17-21
pH Titration Curves
pH
Titration of 0.100 M HA with
• Titration curve: plot of pH of
0.100 M NaOH
the solution as a function of
14
12
the volume of base (acid)
10
added to an acid (base).
8
6
WA
• Sharp rise in curve is
4
SA
equivalence point.
2
0
• pH at equivalence point is 7.0
0
10
20
30
40
for SA but higher for WA.
Volume Base Added, mL
• Equivalence point can be used to determine the
concentration of the titrant.
E.g. the equivalence point for 15.00 mL of an acid
8–22
occurred when 25.00 mL of 0.075 M NaOH was
added. What was the molarity of the acid?
John A. Schreifels
Chemistry 212
Chapter 17-22
SA–SB Titrations
• Base removes some acid and pH increases.
• Let nb = moles of base added
na,r = moles of acid remaining
na,r = na  nb = CaVa  CbVb
• Moles of hydronium ion same as moles of acid remaining. nH3O+
= na,r;
nH3 O+

[H3O ] 
V a + Vb
C V  Cb Vb
 a a
V a + Vb
• Valid until very close to equivalence point.
• Equivalence point(EP): pH = 7.00
• Beyond EP: pH due only to base added (i.e. excess base). Use
total volume.
• E.g. Determine pH of 10.0 mL of 0.100M HCl after addition of 8–23
5.00, 10.0 and 15.0mL of 0.100M NaOH.
John A. Schreifels
Chemistry 212
Chapter 17-23
Titration of SB with SA
•
•
Acid removes some of the base and pH is changed by amount of base
removed.
Let na = moles of acid added
nb,r = moles of base remaining
nb,r = CbVb  CaVa
Moles of hydroxide ion same as moles of base remaining.
nOH = nb,r;

[OH ] 
n
OH
V a + VB
C V  Ca Va
 b b
V a + VB
– Valid until EP.
•
•
EP: pH = 7.00
Beyond EP: pH due only to excess acid. Use total volume.
E.g. Determine pH of 10.0 mL of 0.100M NaOH after addition of 5.00,
10.0 and 15.0mL of 0.100M HCl.
John A. Schreifels
Chemistry 212
8–24
Chapter 17-24
WA with SB Titration
•
As above base removes some of the acid and pH is changed by
amount of acid removed.
Let nb = moles of base added
nHA = moles of acid remaining
nHA = CHAVHA  CbVb
nA = nb = CbVb
n 
pH  pK a  log A
nHA
Cb Vb
 pK a  log
Ca Va  Cb Vb
•
Up to equivalence point moles of hydronium ions must be determined
from equilibrium expression.
Equivalence point: pH = pH of salt of WA
Beyond Equivalence point: Use amount of excess base to determine
pH.
E.g. determine pH of 10.0 mL of 0.100M HA after addition of 5.00, 10.0
and 15.0mL of 0.100M NaOH. Ka = 1.75x105.
•
•
•
John A. Schreifels
Chemistry 212
8–25
Chapter 17-25
WB–SA Titrations
•
•
•
•
Acid removes some of the base and decreases the pH.
Let na = moles of acid added
nb,r = moles of base remaining
nb,r = CbVb  CaVa
nBH+ = na = CaVa
n 
pOH  pK b  log BH
nB
Ca Va
 pK b  log
Cb Vb  Ca Va
Moles of hydroxide ions must be determined from equilibrium
expression. Valid until EP.
EP: pH = pH of salt of weak base.
Beyond EP: pH due only to presence of acid added after endpoint (i.e.
excess acid) as seen for strong base. Volume correction needed as
above (total volume).
E.g. Determine pH of 10.0 mL of 0.100M B after addition of 5.00, 10.0 8–26
and 15.0mL of 0.100M HCl. Kb = 1.75x105.
John A. Schreifels
Chemistry 212
Chapter 17-26