1. The Color Change of the Indicator Bromthymol Blue
Fig. 19.5

2. Acid-Base Indicators
Usually dyes that are weak acids and display different
colours in protonated/deprotonated forms.
HIn(aq.) H+ (aq.) +In- (aq.)
Ka =
[H ][In ]
+ -
[ HIn]
In general we seek an indicator whose transition range
(±1pH unit from the indicator pKa) overlaps the steepest part
of the titration curve as closely as possible

3. pKa
H Cl Cl + H -7
O O
CH3 C CH3 C + H 4.74
O H O
H H
H C H H C + H ~50
H H
Strong acids give weak bases.
Weak acids give strong bases.

4. H Cl Cl + H
basicity
O O
CH3 C CH3 C + H
O H O
acidity
H H
H C H H C + H
H H
Strong acids give weak bases.
Weak acids give strong bases.

5. Colors and Approximate pH Range of
Some Common Acid-Base Indicators

6. Which molecule is the stronger acid, ethanol
or acetic acid? Ka
10-16
ethanol
10-4.74
more stable anion
acetic acid
because of
the stronger
O resonance and
acid
CH3 C inductive effects
O

7. Predict whether trifluoroacetic acid will be a
stronger or weaker acid than acetic acid.
Ka
H O H O
H C C H C C + H 10-4.74
H OH H O
acetic acid
F O F O
F C C F C C + H 10-0.23
F OH F O
Fluorine is more electro-
trifluoroacetic acid
negative than hydrogen. Anion
more acidic acid is more stable.

8. It is a general principle that the more stable
the anion the more acidic is the acid.
The principle is also successful across a row of
the periodic table.
increasing
electronegativity
pKa
H H
H C H H C + H 48
H H
H H
H N H H N + H 38
H H
O H O + H 15.7

9. The relative strengths of acids and bases is
given by the Ka of the acid. Ka
H Cl Cl + H 107
O O
CH3 C CH3 C + H 10-4.74
O H O
CH3 CH3
CH3 N H CH3 N + H 10-9.81
CH3 CH3

10. Identify the strongest acid.
Ka
H Cl Cl + H 107
O O
CH3 C CH3 C + H 10-4.74
O H O
CH3 CH3
CH3 N H CH3 N + H 10-9.81
CH3 CH3

11. Identify the weakest acid.
Ka
H Cl Cl + H 107
O O
CH3 C CH3 C + H 10-4.74
O H O
CH3 CH3
CH3 N H CH3 N + H 10-9.81
CH3 CH3

12. Identify the weakest base.
Ka
H Cl Cl + H 107
O O
CH3 C CH3 C + H 10-4.74
O H O
CH3 CH3
CH3 N H CH3 N + H 10-9.81
CH3 CH3

13. Identify the strongest base.
Ka
H Cl Cl + H 107
O O
CH3 C CH3 C + H 10-4.74
O H O
CH3 CH3
CH3 N H CH3 N + H 10-9.81
CH3 CH3

14. In chemistry, particularly biology, a large
number of compounds are acids and bases.
HO NH2 HO NH3
+ H
HO HO
dopamine
HO CO2H HO CO2
+ H
CO2H CO2H CO2H CO2H
citric acid
Biological fluids are often buffered (constant pH) an
it is useful to know the predominant species present
at a given pH.

15. Consider acetic acid with a Ka = 10-4.74
at pH = 4.74 [CH3CO2H] = [CH3CO2]
H
at lower pH, more acidic
than 4.74, acetic acid is the
major species present

16. Consider acetic acid with a Ka = 10-4.74
at pH = 4.74 [CH3CO2H] > [CH3CO2]
H
at lower pH, more acidic
than 4.74, acetic acid is the [CH3CO2H]
major species present

17. Consider acetic acid with a Ka = 10-4.74
at pH = 4.74 [CH3CO2H] < [CH3CO2]
OH
at lower pH, more acidic
than 4.74, acetic acid is the [CH3CO2H]
major species present
at higher pH, less acidic
than 4.74, acetate ion is
the major species present [CH3CO2]

18. If acetic acid is introduced into the blood
what will be the predominant species present?
Will it be acetate ion or acetic acid?
The pH of blood is maintained at ∼ 7.4
If the pH of blood was 4.74 then the
acetate ion would be equal to the acetic
acid ion concentration.
O O
[HOCCH3] = [ OCCH3] at pH = 4.74
If the pH is raised to 7.4 will the
concentration of acetate ion increase or
decrease?

19. If acetic acid is introduced into the blood
what will be the predominant species present?
Will it be acetate ion or acetic acid?
pH 4.74 ⇒ 7.4
O H
O O
[HOCCH3] = [ OCCH3]
If the pH is raised to 7.4 (more basic)
will the concentration of acetate ion
increase or decrease?

20. If acetic acid is introduced into the blood
what will be the predominant species present?
Will it be acetate ion or acetic acid?
Acetate ion is the major species present
if acetic acid is introduced into blood.
O O
[HOCCH3] = [ OCCH3] at pH = 4.74
If the pH is raised to 7.0 will the
concentration of acetate ion increase or
decrease?

21. If cocaine is introduced in the blood what will
be the major species present?
Ka
CH3 H CH3
10-8.81
N N
CO2CH3 CO2CH3
+ H
O O
O O
protonated form neutral form

22. If cocaine is introduced in the blood what will
be the major species present?
Ka
CH3
10-8.81
N
at pH = 8.81
=
CO2CH3
+ H
O
O
protonated form neutral form
pH 8.81 ⇒ 7.4 (more acidic)

23. The Henderson-Hasselbalch Equation
Take the equilibrium ionization of a weak acid:
HA(aq) + H2O(aq) = H3O+(aq) + A-(aq) [H3O+] [A-]
Ka =
[HA]
Solving for the hydronium ion concentration gives:
[HA]
[H3O+] = Ka x
[A-]
Taking the negative logarithm of both sides:
-log[H3O +] = -log Ka - log ( )
[HA]
[A-] ( )
pH = -log Ka - log [HA]
[A-]
Generalizing for any conjugate acid-base pair :
[base]
pH = -log Ka + log
( [acid] ) Henderson-Hasselbalch
equation

24. H-A H + A p H = -log [H⊕ ]
[A-]
pH = pKa + log
[HA]
a useful concept:
when [H-A] = [ A ]
Biological fluids are often buffered (constant pH) an
it is useful to know the predominant species present
at a given pH.

25. HENDERSON-HASSELBALCH EQUATION
For acids:  [A − ]  When [A-] = [HA],
pH = pK a + log 
 [HA] 

 
pH = pKa
[B] pKa applies
For bases: pH = pK a + log
[BH+ ] to this acid
Kb
→
B + H2O ← BH+ + OH-
Ka
base acid acid base
Derivation: HA  H+ + A-
[H+ ][A − ]  [H+ ][A − ]   [A − ]   [A − ] 
Ka = - log K a = − log   +
= − log [H ] − log   pK a = pH − log 
 [HA] 

[HA]  [HA]   [HA] 
     
 [A −] 
pH =pK a + log 
[HA] 

 

26. BUFFERS
Mixture of an acid and its conjugate base.
Buffer solution → resists change in pH when acids
or bases are added or when dilution occurs.
Mix:
A moles of weak acid + B moles of conjugate base
Find:
• moles of acid remains close to A, and
• moles of base remains close to B
⇒ Very little reaction
HA  H+ + A- Le Chatelier’s principle

28. Why does a buffer resist change in pH when small
? amounts of strong acid or bases is added?
The acid or base is consumed
by A- or HA respectively
A buffer has a maximum capacity to resist change to pH.
Buffer capacity, β:
→ Measure of how well solution resists change in
pH when strong acid/base is added.
dCb − dCa
β= =
dpH dpH
Larger β ⇒ more resistance to pH change

29. How a Buffer Works
Consider adding H3O+ or OH- to water and also to a buffer
For 0.01 mol H3O+ to 1 L water:
[H3O+] = 0.01 mol/1.0 L = 0.01 M pH = -log([H3O+]) = 2.0
So, change in pH from pure water: ∆pH = 7.00 – 2.00 = 5.0
For the H2CO3- / HCO3- system:
pH of buffer = 7.38
Addition of 0.01 mol H3O+ changes pH to 7.46
So change in pH from buffer: ∆pH = 7.46 – 7.38 = 0.08 !!!

30. How a Buffer Works
Consider a buffer made from acetic acid and sodium acetate:
CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
[CH3COO-] [H3O+]
Ka = or
[CH3COOH]
+
[CH3COOH]
[H3O ] = Ka x
[CH3COO-]

31. How a Buffer Works
Let’s consider a buffer made by placing 0.25 mol of acetic acid and
0.25 mol of sodium acetate per liter of solution.
What is the pH of the buffer?
And what will be the pH of 100.00 mL of the buffer before and
after 1.00 mL of concentrated HCl (12.0 M) is added to the buffer?
What will be the pH of 300.00 mL of pure water if the same acid
is added?
+
[CH3COOH] -5
(0.25)
[H3O ] = Ka x = 1.8 x 10 x = 1.8 x 10-5
[CH3COO-] (0.25)
pH = -log[H3O+] = -log(1.8 x 10-5) = pH = 4.74 Before acid added!

32. How a Buffer Works
What is pH if added to pure water?
1.00 mL conc. HCl 1.00 mL x 12.0 mol/L = 0.012 mol H3O+
Added to 300.00 mL of water :
0.012 mol H3O+
= 0.0399 M H3O+ pH = -log(0.0399 M)
301.00 mL soln.
pH = 1.40 Without buffer!

33. How a Buffer Works
After acid is added to buffer:
Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+
Initial 0.250 ---- 0.250 0
Change +0.012 ---- -0.012 0.012
Equilibrium 0.262 ---- 0.238 0.012
Solving for the quantity ionized:
Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+
Initial 0.262 ---- 0.238 0
Change -x ---- +x +x
Equilibrium 0.262 - x ---- 0.238 + x x
Assuming: 0.262 - x = 0.262 & 0.238 + x = 0.238
[CH3COOH]
+
[H3O ] = Ka x =1.8 x 10-5 x (0.262) = 1.982 x 10-5
[CH3COO-] (0.238)
pH = -log(1.982 x 10-5) = 5.000 - 0.297 = 4.70 After the acid is added!

34. How a Buffer Works
Suppose we add 1.0 mL of a concentrated base instead of an acid. Add
1.0 mL of 12.0 M NaOH to pure water and our buffer, and let’s see what
the impact is: 1.00 mL x 12.0 mol OH-/1000mL = 0.012 mol OH-
This will reduce the quantity of acid present and force the equilibrium
to produce more hydronium ion to replace that neutralized by the
addition of the base!
Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+
Initial 0.250 ---- 0.250 0
Change - 0.012 ---- +0.012 +0.012
Equilibrium 0.238 ---- 0.262 +0.012
Assuming: Again, using x as the quantity of acid dissociated we get:
our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238
0.238 = 1.635 x 10-5
[H3O+] = 1.8 x 10-5 x
0.262
pH = -log(1.635 x 10-5) = 5.000 - 0.214 = 4.79 After base is added!

35. How a Buffer Works
By adding the 1.00mL base to 300.00 mL of pure water we would get a
hydroxide ion concentration of:
- 0.012 mol OH- = 3.99 x 10-5 M OH-
[OH ] =
301.00 mL
The hydrogen ion concentration is:
+
Kw 1 x 10-14
[H3O ] = -
= = 2.506 x 10-10
[OH ] 3.99 x 10-5 M
This calculates out to give a pH of:
pH = -log(2.5 6 x 10-10) = 10.000 - 0.408 = 9.59 With 1.0 mL of the
base in pure water!
In summary:
Buffer alone pH = 4.74
Buffer plus 1.0 mL base pH = 4.79 Base alone, pH = 9.59
Buffer plus 1.0 mL acid pH = 4.70 Acid alone, pH = 1.40

36. Problem:
Calculate the pH of a solution containing 0.200 M NH3
and 0.300 M NH4Cl given that the acid dissociation
constant for NH4+ is 5.7x10-10.
NH3 + H2O  NH4+ + OH- pKa = 9.244
Ka
base acid
[B] pKa applies
pH = pK a + log
[BH+ ] to this acid
(0.200)
pH = 9.244 + log
(0.300)
pH = 9.07