1. Buffers Titrations and the Henderson Hasselbach Equation

2. CH3COONa (s) Na+ (aq) + CH3COO- (aq)
The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.
The presence of a common ion suppresses the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
CH3COONa (s) Na+ (aq) + CH3COO- (aq)
common ion
CH3COOH (aq) H+ (aq) + CH3COO- (aq)

3. A buffer solution is a solution of:
A weak acid or a weak base and
The salt of the weak acid or weak base
Both must be present! (and should NOT neutralize EACH OTHER!!!!)
A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
Consider an equal molar mixture of CH3COOH and CH3COONa
CH3COOH (aq) H+ (aq) + CH3COO- (aq)
Adding more acid creates a shift left IF enough acetate ions are present

4. Which of the following are buffer systems? (a) KF/HF
(b) KCl/HCl, (c) Na2CO3/NaHCO3
(a) KF is a weak acid and F- is its conjugate base
buffer solution
(b) HCl is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it conjugate acid
buffer solution

5. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30
0.00
0.52 -x +x +x x x x
Ka for HCOOH = 1.8 x 10 -4
x = X 10 -4
[H+] [HCOO-] Ka = [HCOOH]
pH = 3.98

6. Henderson-Hasselbach equation
OR…… Use the Henderson-Hasselbach equation
Consider mixture of salt NaA and weak acid HA.
NaA (s) Na+ (aq) + A- (aq)
Ka =
[H+][A-]
[HA]
pKa = -log Ka
HA (aq) H+ (aq) + A- (aq)
[H+] =
Ka [HA]
[A-]
Henderson-Hasselbach equation
-log [H+] = -log Ka - log
[HA]
[A-]
pH = pKa + log
[conjugate base]
[acid]
-log [H+] = -log Ka + log
[A-]
[HA]
pH = pKa + log
[A-]
[HA]

7. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
HCOOH (aq) H+ (aq) + HCOO- (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30
0.00
0.52 -x +x +x x x x
pH = pKa + log
[HCOO-]
[HCOOH]
Common ion effect
0.30 – x  0.30
pH = log
[0.52]
[0.30]
= 3.98
x  0.52
HCOOH pKa = 3.77

8. HCl + CH3COO- CH3COOH + Cl-
HCl H+ + Cl-
HCl + CH3COO CH3COOH + Cl-

9. Calculate the pH of the 0. 30 M NH3/0. 36 M NH4Cl buffer system
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution?
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
[NH4+] [OH-]
[NH3]
Kb =
= 1.8 X 10-5
0.36
Initial
0.30
+ x
+ x
Change
- x
End x x x
(.36 + x)(x)
(.30 – x)
1.8 X 10-5 =
0.36x
0.30
1.8 X 10-5 
x = 1.5 X 10-5
pOH = 4.82
pH= 9.18

10. final volume = 80.0 mL + 20.0 mL = 100 mL
What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution?
final volume = 80.0 mL mL = 100 mL
NH M x L = mol / .1 L = 0.29 M
OH x L = mol / .1 L = 0.01M
NH M x = mol / .1 L = 0.24M
start (M)
0.29
0.01
0.24
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
end (M)
0.28
0.0
0.25
Equilibrium Calculation NH4+ (aq) H+ (aq) + NH3 (aq)
[H+] [NH3]
[NH4+]
[H+] 0.25
0.28
= 5.6 X 10-10
Ka= = 5.6 X 10-10
[H+] = 6.27 X
pH = 9.20

11. Calculate the pH of the 0. 30 M NH3/0. 36 M NH4Cl buffer system
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq) H+ (aq) + NH3 (aq)
pH = pKa + log
[NH3]
[NH4+]
pH = log
[0.30]
[0.36]
pKa = 9.25
= 9.17
final volume = 80.0 mL mL = 100 mL
start (M)
0.29
0.01
0.24
NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)
end (M)
0.28
0.0
0.25
pH = log
[0.25]
[0.28]
= 9.20

12. Equivalence point – the point at which the reaction is complete
Titrations
In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at the
endpoint (hopefully close to the equivalence point)
Slowly add base
to unknown acid
UNTIL
The indicator
changes color
(pink)

13. Titration Curve
A titration curve is a plot of pH vs. the amount of titrant added. Typically the titrant is a strong (completely) dissociated acid or base. Such curves are useful for determining endpoints and dissociation constants of weak acids or bases.

15. Features of the Strong Acid-Strong Base Titration Curve
The pH starts out low, reflecting the high [H3O+] of the strong acid and increases gradually as acid is neutralized by the added base.
Suddenly the pH rises steeply. This occurs in the immediate vicinity of the equivalence point. For this type of titration the pH is 7.0 at the equivalence point.
Beyond this steep portion, the pH increases slowly as more base is added.

16. Sample Calculation: Strong Acid-Strong Base Titration Curve
Consider the titration of 40.0 mL of M HCl with M NaOH.
Region 1. Before the equivalence point, after adding 20.0 mL of M NaOH. (Half way to the equivalence point.)
Initial moles of H3O+ = L x M = mol H3O+
- Moles of OH- added = L x M = mol OH-
pH=1.477

17. Sample Calculation: Strong Acid-Strong Base Titration Curve (Cont. I)
Region 2. At the equivalence point, after adding 40.0 mL of M NaOH.
Initial moles of H3O+ = L x M = mol H3O+
- Moles of OH- added = L x M = mol OH-
An equal number of moles of NaOH and HCl have reacted, leaving only a solution of their salt (NaCl). The pH=7.00 because the cation of the strong base and the anion of the strong acid have no effect on pH.

18. Sample Calculation: Strong Acid-Strong Base Titration Curve (cont. II)
Region 3. After the equivalence point, after adding 50.0 mL of M NaOH. (Now calculate excess OH-)
Total moles of OH- = L x M = mol OH Moles of H3O+ consumed = L x M = mol
pOH=1.954
pH=12.046

19. Weak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)
CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l)
At equivalence point (pH > 7):
CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq)

20. The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak Acid-Strong Base Titration Curve
The initial pH is higher.
A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point.
The pH at the equivalence point is greater than 7.00.
The steep rise interval is less pronounced.

21. Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq) NH4Cl (aq)
H+ (aq) + NH3 (aq) NH4Cl (aq)
At equivalence point (pH < 7):
NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)

22. The four Major Differences Between a Weak Acid-Strong Base Titration Curve and a Weak Base-Strong Acid Titration Curve
The initial pH is above 7.00.
A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point.
The pH at the equivalence point is less than 7.00.
Thereafter, the pH decreases slowly as excess strong acid is added.

24. Features of the Titration of a Polyprotic Acid with a Strong Base
The loss of each mole of H+ shows up as separate equivalence point (but only if the two pKas are separated by more than 3 pK units).
The pH at the midpoint of the buffer region is equal to the pKa of that acid species.
The same volume of added base is required to remove each mole of H+.

25. Acid-Base Indicators

27. The titration curve of a strong acid with a strong base.
16.5

28. Which indicator(s) would you use for a titration of HNO2 with KOH ?
Weak acid titrated with strong base.
At equivalence point, will have conjugate base of weak acid.
At equivalence point, pH > 7
Use cresol red or phenolphthalein

29. Finding the Equivalence Point (calculation method)
Strong Acid vs. Strong Base
100 % ionized! pH = 7 No equilibrium!
Weak Acid vs. Strong Base
Acid is neutralized; Need Kb for conjugate base equilibrium
Strong Acid vs. Weak Base
Base is neutralized; Need Ka for conjugate acid equilibrium
Weak Acid vs. Weak Base
Depends on the strength of both; could be conjugate acid, conjugate base, or pH 7

30. HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l) end (moles) 0.0 0.0 0.01
Exactly 100 mL of 0.10 M HNO2 are titrated with 100 mL of a 0.10 M NaOH solution. What is the pH at the equivalence point ?
start (moles)
0.01
0.01
HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)
end (moles)
0.0
0.0
0.01
[NO2-] =
0.01
0.200
= 0.05 M
Final volume = 200 mL
Equilibrium Calculation NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.05
0.00
0.00 -x +x +x x x x
Kb =
[OH-][HNO2]
[NO2-] = x2
0.05-x
pOH = 5.98
= 2.2 x 10-11
pH = 14 – pOH = 8.02
x  1.05 x 10-6
= [OH-]