1. Chapter 8 Monoprotic Acid-Base Eqlilibria

2. 8-1 Strong Acids and Bases
Completely dissociates
: [H3O+] or [OH-] equals concentration of strong acid or base
(8-1)
Ex. What is the pH of a 0.1M solution of HCl?
Ex. What is the pH of a 0.1M solution of KOH?
Ka of strong acids
Acid Ka
HCl
103.9
HBr
105.8 HI
1010.4
HNO3
101.4
[OH-], M
[H+], M pH
11.00
10.00
9.00

3. The Dilemma What is the pH of 1.0x10-8 M KOH?
How can a base produce an acidic solution?
⇒ Wrong Assumption!!
For large concentration of strong acid or base,
[H+] = [acid] or [OH-] = [base]
For small concentration, must account for water dissociation
⇒ In pure water [OH-] = 1.0×10-7M, which is greater than [KOH] = 1×10-8M
⇒ Must Use Systematic Treatment of Equilibrium

4. The Cure Step 1: Pertinent reactions:
(Completely dissociates, not pertinent)
Step 2: Charge Balance:
Step 3: Mass Balance:
(All K+ comes from KOH)
Step 4: Equilibrium constant expression (one for each reaction):
Step 5: Count equations and unknowns:
Three equations: Kw, MB, CB
Three unknowns: K+, H+, OH-
Step 6: Solve (Seeking pH, [H+]):
From CB, MB:
For Kw:
Use quadratic equation
(pH slightly basic, consistent with low [KOH])

5. - Three Regions depending on acid/base concentrations
pHs of Strong Acids and Bases for different concentrations of Strong Acids and Bases
- Three Regions depending on acid/base concentrations
Low concentrations, (10-6~10-8M), H2O ionization ≈ H+,OH- of acid/base  systematic equilibrium calculation necessary
High concentrations (≥10-6M), pH considered just from the added H+,OH-
Very low concentrations (≤10-8M), pH=7  not enough H+, OH- added to change pH

6. Water Almost Never Produces 10-7 M H+ and OH-
- pH=7 only true for pure water ([H+]=[OH-]=1.0×10-7 M)
- Any acid or base suppresses water ionization
: Follows Le Châtelier’s principal
Ex. In 10-4 M HBr
:In 10-4 M HBr solution, water dissociation produces only M OH- and H+

7. 8-2 Weak Acids and Bases Ka · Kb = Kw
Weak acid/base do not completely dissociate
- Dissociation constant (Ka ) for the weak acid HA: Ka
(8-3)
- Base Hydrolysis constant (Kb ) for the weak base B: Kb
(8-4)
- As Ka or Kb increase  pKa or pKb decrease
- Smaller pKa  stronger acid
Formic acid stronger acid than benzoic acid
(HA)
(A-)
(HA)
(A-)
Conjugate acid-base pair
Ka · Kb = Kw
(8-5)

8. Weak Is Conjugate to Weak
The conjugate base of a weak acid is a weak base.
The conjugate acid of a weak base is a weak acid.
The conjugate base of a strong acid is so weak acid(Not a base in water).
KaㆍKb = Kw
(for conjugate acid-base pair)
Ka=10-4 ⇒ Kb ?
Ka=10-5 ⇒ Kb ?
Using Appendix G
pKa = -logKa
pKa
Group Ka
1.4
POH
0.04
3.51 OH
3.1×10-4
6.04
9.1×10-7
8.25 NH
5.6×10-9

9. A Typical Weak-Acid Problem
8-3 Weak Acid Equilibria
A Typical Weak-Acid Problem
General Systematic Treatment of Equilibrium
- Unlike concentrated strong acid, need to account for water ionization
- Find pH for a solution of a general weak acid (HA)
Step 1: Pertinent reactions:
Step 2: Charge Balance:
(8-6)
Step 3: Mass Balance:
(8-7)
Step 4: Equilibrium constant expression (one for each reaction):
(8-8)
Step 5: Count equations and unknowns:
Four Equations:
(8-6),
(8-7),
(8-8), Kw
Four Unknowns:

10. Step 6: Solve (Not easy to solve  cubic equation results!):
- Again, need to make assumptions to simplify equations
- The goal is to determine [H+], so we can measure pH
Make Some Initial Assumptions:
- For a typical weak acid, [H+] from HA will be much greater than [H+] from H2O
- If dissociation of HA is much greater than H2O, [H+] ≫ [OH-] [OH-] neglect ⇒ for CB;
(8-9)
From (8-9):
Set [H+]=x  [A-]=x
Ex) F=0.050M, Ka=1.07x10-3, pH=?
For MB:
Use quadratic equation
For (8-8):
Use quadratic equation

11. Step 7: Verify Approximation:
Was the approximation, [H+] >>[OH-] justified ([H+] ≈ [A-])?
Setting F = M and Ka = 1.07x10-3 for o-hydroxybenzoic acid:
[H+] >> [OH-]
6.8x10-3M >> 1.5x10-12M
assumption is justified!
Determine [OH-] from water dissociation:

12. Summery for Weak-Acid Problem
HA A- + H+
Initial(M) F
F: Total concentration
Change(M) -x x +x
Approximation: [A-]≈[H+]=x
Equili.(M) (F-x) (x) (x)
Assump.
Weak acid: F ≫ [H+]  F - [H+] ≈ F
Use quadratic equation
Caution for assumption (F≫[H+]) : F/Ka > 102 → Can be assump. : Confirm when large Ka or low F : Should be justified

13. Ex) 0.1M CH3COOH(Ka=1.8×10-5), pH=?
Ex) M benzoic acid, pH=?
COOH
Benzoic acid
Ka = 6.28 ×10–5
Using approximation: F-x ≒ F

14. Fraction of Dissociation
(8-10)
Stronger acid
Ex:
What is the percent fraction dissociation for F = M o-hydroxybenzoic acid (Ka = 1.07x10-3 ) ?
① The α of the stronger acid is more higher than weaker acid. ② The α of weak acid(weak electrolyte) increases as diluted.
Weaker acid

15. Essence of a Weak-Acid Problem
F: Total concentration
(8-11)
Approximation: [A-]≈[H+]=x
Example
A Weak-Acid Problem
Find the pH of 0.050M trimethylammonium chloride.
Solution
From App. G: Ka=10-pKa = = 1.59x10-10
(CH3)3NH+Cl- → (CH3)3NH+ + Cl- Ka
(CH3)3NH+ = (CH3)3N + H+
(8-12)
Using approximation: F-x ≒ F

16. 8-4 Weak Base Equilibria
Treatment of Weak Base (B) is Very Similar to Weak Acid
- Assume all OH- comes from base and not dissociation of water
B + H2O BH+ + OH-
Initial(M) F
Change(M) -x x x
Equili.(M) (F-x) (x) (x)
Approx.
Weak base: F ≫ [OH-]  F - [OH-] ≈ F
(8-13)
Use quadratic equation
Caution for approx. (F≫[OH-]) : F/Kb > 102 → Can be approx. : Confirm when large Kb or low F : Should be justified

17. • Fraction of Dissociation(α)
Ex) What is the pH of cocaine dissolved in water? (F = M and Kb=2.6x10-6 for cocaine)
Initial(M) Change(M) -x x x
Eqlili.(M) ( x) (x) (x)
• Fraction of Dissociation(α)
(8-14)

18. 8-5 Buffers
A buffered solution resists changes in pH when acids or bases are added
- Buffer is a mixture of a weak acid(HA) and its conjugate base(A-)
: Must be comparable amounts of acid & base (>0.1M)
Composition of buffer solution
① Weak acid(HA) + Conjugate base (A-) ② Weak base(B) + Conjugate acid (BH+)
pH dependence of the rate of cleavage of an amide bond by the enzyme chymotrypsin.

19. Mixing a Weak Acid and Its Conjugated Base
① HA = H++ A-
② A-+ H2O = HA + OH-
HA + A-
1) Weak acid(HA) reacts “incompletely (very little)“ with a weak base(A-).
2) Very little change of moles (or concentrations) for the mixing HA and A-
Consider FHA=0.10M HA (pKa=4.00) α
Low α little dissociation of HA and A-
Consider FA-=0.10M A- (pKb=10.00)
Mixing A-  Reaction ① goes to the left
Le Chatelier’s principle
Mixing HA  Reaction ② goes to the left
⇒ From the approximation 1) and 2), the equilibrium concentration of HA and A- remain unchanged.
[HA] ≒ FHA
[A-] ≒ FA-

20. Henderson-Hasselbalch Equation
pH of buffer solution(HA + A-)
Take log of both sides:
Rearrange: pH
pKa
(8-16)
pH of buffer solution(B + BH+)
(8-17)
When activities are includes,
(8-18)

21. Properties of the Henderson-Hasselbalch Equation
Determines pH of buffered solution
⇒ Need to know ratio of conjugate [acid, HA] and [base, A-]
If [A-] = [HA]  pH = pKa
All equilibria must be satisfied simultaneously in any solution at equilibrium
Only one concentration of H+ in a solution
[A-]/[HA] pH
100:1
pKa + 2
10:1
pKa + 1
1:1
pKa
1:10
pKa - 1
1:100
pKa - 2

23. Buffer in Action Ex. Tris buffer BH+ B (This form is “tris”)
Tris (hydroxymethyl)aminomethane
BH+
(pKa=8.072)
B (This form is “tris”)
- Dissociation of Tris hydrochloride(BH+Cl– )
BH+Cl–  BH+ + Cl– BH B + H+

24. - The pH of a buffer is nearly independent of volume
Example
A Buffer Solution
Find the pH of 1.00-L aqueous solution containing 12.43g of tris(FM , B) plus 4.67 g of tris hydrochloride(FM , BH+) (pKa = 8.07).
Solution
[B]=(12.43g/ g/mol)/1.00L = M
[BH+]=(4.67g/ g/mol)/1.00L = M
- The pH of a buffer is nearly independent of volume
Example
Effect of Adding Acid to a Buffer
If we add 12.0 mL of 1.00 M HCl to the solution in previous solution, what will be the new pH?
pH does not change very much (ΔpH=-0.20)
Solution
B H+ → BH+
Initial(mol)
Final(mol)
( )
( )

25. Calculating How to Prepare a Buffer Solution
Example
Calculating How to Prepare a Buffer Solution
How many milliters of M NaOH should be added to 10.0 g of trisH+(BH+) to give a pH of 7.60 in a final volume of 250 mL?
Solution
Adding OH-  reduce [BH+] and increase [B]
Moles of trisH+(BH+) = (10.0 g)/ g/mol) = mol
BH OH- → B H2O
Initial(mol) x
Final(mol) x x
Mole of OH- = MV ⇒ mol=(0.500M)(a mL)
∴ a=mL of NaOH = 32.0 mL ⇒ dilute to mL

26. Preparing a Buffer in Real Life!
0.1 M tris containing Tris buffer (pH = 7.60, 1.0 L)
Prepare 0.1 mol trisH+ using beaker (~800mL)
- Monitor pH using pH meter
- Add NaOH solution until the pH is exactly 7.60
Transfer to volumetric flask of 1.0 L
Dilute to the mark (1.0L) with water and mix.
The pH should be adjusted by pH meter!!!
(Do not simply add the calculated quantity of NaOH)

27. Buffer Capacity Why Does a Buffer Resist Changes in pH?
- Strong acid or base is consumed by B or BH+
- Buffer Capacity (β): measure of a solutions resistance to pH change(“+” value)
(8-19)
:where Ca and Cb are the number of moles of strong acid and strong base per liter needed to produce a unit change in pH
- Greater β  more resistant to pH change
- Maximum capacity to resist pH change occurs at pH=pKa
Choosing a Buffer
- Choose a buffer with pKa as close as possible to desired pH
- Useful buffer range is pKa ± 1 pH units
- Buffer pH depends on temperature
and ionic strength  activity coefficients

28. [HA]=[A-]→[A-]/[HA]=1 ⇒ pH = pKa : max β
Buffer capacity depends on:
- Absolute concentration of HA and A-
- Relative concentration(molar ratio) of HA and A-
[HA]=[A-]→[A-]/[HA]=1 ⇒ pH = pKa : max β
[A-]/[HA] pH
100:1
pKa + 2
10:1
pKa + 1
1:1
pKa
1:10
pKa - 1
1:100
pKa - 2
Maximum buffer capacity : pH = pKa
Effective buffer range : pKa ± 1

29. Buffer pH Depends on Ionic Strength and Temperature
- Changing ionic strength (μ) changes pH.
(∵ Buffer solution: high concentration (ionic strength ≠ 0, <1)  reduce effective concentration  use activity (A))
- Changing temperature changes pH.
(∵ pKa depends on temperature.)
Tris buffer: pH 8.07 (25 0C) pH 8.7 (4 0C) pH 7.7 (37 0C)
When What You Mix Is Not What You Got
: In dilute solution or extremes of pH, don’t use Henderson-Hasselbalch Eq.
In dilute solution (∵increase α values of HA and A-)
In extremes of pH, [H+]≫[OH-] or [H+]≪[OH-]
FHA ≠ [HA]
FA- ≠ [A-]

30. A Dilute Buffer Prepared from a Moderately Strong Acid
Example
A Dilute Buffer Prepared from a Moderately Strong Acid
Mixing of mol HA (pKa=2.00) and mol of A- in 1.00 L  pH = ?
Solution
Initial(M)
Change(M) x x x
Eqlili.(M) (0.010-x) (x) (0.010+x)
(8-22)
Use quadratic equation
(correct)
Applying approximation ([HA]≒FHA and [A-]≒FA-) : Henderson-Hasselbalch Eq.
(incorrect)
HA is too strong and the concentrations are too low for HA and A-
⇒ [HA]≠FHA and [A-] ≠FA-

31. Effect of dilution of a buffer solution
# Very diluted buffer solution changes pH.
Dilution pH constant
Constant
Dilution pH increase
dilution
- Dilution will not alter the pH of a buffer solution. (It will, however, decrease the buffer capacity, since the concentration has been reduced.)