SOLUTION EQUILIBRIA
BUFFER – A solution of about equal amounts of a weak acid and its weak
conjugate base
A buffer is resistant to changes in pH because it can neutralize any strong
acid or base added to it
2A-1 (of 16)
A solution contains both HF and FWeak acid:
HF
Weak conjugate base: FIf a strong acid is added to the buffer:
1 HCl molecule
2A-2 (of 16)
A solution contains both HF and FWeak acid:
HF
Weak conjugate base: FIf a strong acid is added to the buffer:
H3O+ + F- → HF + H2O
The STRONG ACID (H3O+) is reacted away
completely by the WEAK CONJUGATE
BASE (F-)
 The original [H3O+] is virtually unchanged
2A-3 (of 16)
1 HCl molecule
A solution contains both HF and FWeak acid:
HF
Weak conjugate base: FIf a strong base is added to the buffer:
OH- + HF → H2O + FThe STRONG BASE (OH-) is reacted away
completely by the WEAK ACID (HF)
 The original [H3O+] is virtually unchanged
2A-4 (of 16)
1 OH- ion
The higher the concentrations of the weak
acid and weak conjugate base in the
buffer, the more strong acid or strong base
it can neutralize
This buffer can neutralize 2 H3O+’s and 2 OH-’s
2A-5 (of 16)
CALCULATING THE pH OF A BUFFER SOLUTION
Calculate the pH of a solution 0.25 M in hydrofluoric acid (Ka = 7.2 x 10-4)
and 0.50 M in sodium fluoride
A buffer: contains a weak acid (HF) and its weak conjugate base (F-)
HF (aq)
+
H2O (l)
⇆
Initial M’s
Change in M’s
Equilibrium M’s
0.25 - x
Ka = [H3O+][F-]
7.2 x 10-4 = x(0.50 + x)
_____________
0.25
-x
______________
(0.25 – x)
[HF]
3.6 x 10-4 = x =
2A-6 (of 16)
[H3O+]
3.44 =
pH
H3O+ (aq)
+
F- (aq)
~0
+x
0.50
+x
x
0.50 + x
7.2 x 10-4 = x(0.50)
_________
0.25
CALCULATING THE pH OF A BUFFER SOLUTION
Calculate the pH of a solution 0.25 M in hydrofluoric acid (Ka = 7.2 x 10-4)
and 0.50 M in sodium fluoride
[H3O+][F-]
_____________
= Ka
[HF]
[H3O+] x [F-]
______
log [H3O+] + log [F-]
______
= log Ka
[HF]
= Ka
[HF]
- log [H3O+] - log [F-]
______
= - log Ka
[HF]
log [H3O+] x [F-]
______
[HF]
= Ka
pH - log [F-]
______
= pKa
[HF]
pH = pKa + log [F-]
______
[HF]
2A-7 (of 16)
CALCULATING THE pH OF A BUFFER SOLUTION
Calculate the pH of a solution 0.25 M in hydrofluoric acid (Ka = 7.2 x 10-4)
and 0.50 M in sodium fluoride
pH
=
pKa +
log [A-]
_____
[HA]
HENDERSON-HASSELBALCH
EQUATION
2A-8 (of 16)
CALCULATING THE pH OF A BUFFER SOLUTION
Calculate the pH of a solution 0.25 M in hydrofluoric acid (Ka = 7.2 x 10-4)
and 0.50 M in sodium fluoride
pH
=
pKa +
log [A-]
_____
[HA]
DAVID HASSELHOFF
EQUATION
This can be used when a solution
contains both a weak acid and its
weak conjugate base
2A-9 (of 16)
CALCULATING THE pH OF A BUFFER SOLUTION
Calculate the pH of a solution 0.25 M in hydrofluoric acid (Ka = 7.2 x 10-4)
and 0.50 M in sodium fluoride
Determine the pKa of the weak acid
pKa = -log Ka
= -log (7.2 x 10-4) = 3.143
Use the David Hasselholf Equation
pH = pKa + log [F-]
= 3.143 + log (0.50 M)
[HF]
(0.25 M)
______
2A-10 (of 16)
___________
= 3.44
CALCULATING THE pH OF A BUFFER SOLUTION
pH
=
pKa + log [weak conj base]
______________________
[weak acid]
pH
=
pKa + log (nweak conj base) / vol
________________________
(nweak ccid) / vol
pH
=
pKa + log (nweak conj base)
_________________
(nweak acid)
Useful when buffers are produced from mixing different solutions together
2A-10 (of 16)
Find the pH of a buffer solution that is prepared by mixing 300. mL of 2.50 M
acetic acid (Ka = 1.8 x 10-5) and 200. mL of 2.50 M sodium acetate.
HC2H3O2 : Weak Acid
NaC2H3O2 : Salt w/ Weak Conj. Base
Solutions are mixed  calculate the moles of all species present
2.50 mol HC2H3O2 x 0.300 L solution
_______________________
= 0.750 mol HC2H3O2
L solution
2.50 M NaC2H3O2
 2.50 M Na+ and 2.50 M C2H3O2-
2.50 mol C2H3O2-
x 0.200 L solution
_______________________
L solution
2A-12 (of 16)
= 0.500 mol C2H3O2-
Find the pH of a buffer solution that is prepared by mixing 300. mL of 2.50 M
acetic acid (Ka = 1.8 x 10-5) and 200. mL of 2.50 M sodium acetate.
HC2H3O2 : Weak Acid
NaC2H3O2 : Salt w/ Weak Conj. Base
Determine the pKa of the weak acid
pKa = -log Ka
= -log (1.8 x 10-5) = 4.745
Use the David Hasselholf Equation
pH = pKa + log nC2H3O2-
= 4.745 + log (0.500 mol)
nHC2H3O2
(0.750 mol)
_________
2A-13 (of 16)
______________
= 4.57
Find the pH of a solution that is prepared by mixing 150. mL of 0.300 M
nitrous acid (Ka = 4.0 x 10-4) and 100. mL of 0.200 M sodium nitrite.
HNO2 : Weak Acid
NaNO2 : Salt w/ Weak Conj. Base
Solutions are mixed  calculate the moles of all species present
0.300 mol HNO2
_____________________
x 0.150 L solution
= 0.0450 mol HNO2
L solution
0.200 M NaNO2
 0.200 M Na+ and 0.200 M NO2-
0.200 mol NO2-
x 0.100 L solution
____________________
L solution
2A-14 (of 16)
= 0.0200 mol NO2-
Find the pH of a solution that is prepared by mixing 150. mL of 0.300 M
nitrous acid (Ka = 4.0 x 10-4) and 100. mL of 0.200 M sodium nitrite.
HNO2 : Weak Acid
NaNO2 : Salt w/ Weak Conj. Base
Determine the pKa of the weak acid
pKa = -log Ka
= -log (4.0 x 10-4) = 3.398
Use the David Hasselholf Equation
pH = pKa + log nNO2-
= 3.398 + log (0.0200 mol)
nHNO2
(0.0450 mol)
_______
2A-15 (of 16)
________________
= 3.05
What must be the [NO2-]/[HNO2] to make a buffer with a pH = 3.00?
HNO2 : Weak Acid
3.00 =
3.398 +
NO2- : Weak Conjugate Base
log [NO2-]
_________
[HNO2]
-0.398 =
log [NO2-]
_________
[HNO2]
0.40
=
[NO2-]
_________
[HNO2]
=
0.40
______
1.0
When the pKa of the weak acid is within 1 pH unit of the desired pH, then
the weak acid / weak conjugate base combination is a good choice for
preparing the buffer
2A-16 (of 16)
Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and
1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added.
HC2H3O2 : Weak Acid
NaOH
NaC2H3O2 : Salt w/ Weak Conj. Base
: Strong Base
Look up the Ka of the weak acid and determine its pKa
pKa = -log Ka
2B-1 (of 18)
= -log (1.8 x 10-5)
= 4.745
Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and
1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added.
HC2H3O2 : Weak Acid
NaOH
NaC2H3O2 : Salt w/ Weak Conj. Base
: Strong Base
Calculate the moles of all species present
1.50 mol HC2H3O2 x 0.500 L solution
_______________________
= 0.750 mol HC2H3O2
L solution
1.00 M NaC2H3O2
 1.00 M Na+ and 1.00 M C2H3O2-
1.00 mol C2H3O2-
x 0.500 L solution
_______________________
= 0.500 mol C2H3O2-
L solution
0.100 mol NaOH
2B-2 (of 18)
 0.100 mol Na+ and 0.100 mol OH-
Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and
1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added.
HC2H3O2 : Weak Acid
NaOH
NaC2H3O2 : Salt w/ Weak Conj. Base
: Strong Base
Strong bases react completely with acids
OH- (aq) + HC2H3O2 (aq) →
the weak acid is reacted
away by the strong base
2B-3 (of 18)
H2O (l) + C2H3O2- (aq)
the weak conjugate base
is produced
Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and
1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added.
HC2H3O2 : Weak Acid
NaOH
NaC2H3O2 : Salt w/ Weak Conj. Base
: Strong Base
Strong bases react completely with acids
OH- (aq) + HC2H3O2 (aq) →
Initial moles
0.100
Reacting moles – 0.100
0
Final moles
2B-4 (of 18)
0.750
– 0.100
0.650
H2O (l) + C2H3O2- (aq)
0.500
+ 0.100
0.600
Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and
1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added.
HC2H3O2 : Weak Acid
NaOH
NaC2H3O2 : Salt w/ Weak Conj. Base
: Strong Base
David Hasselholf Equation
pH = pKa + log
nC2H3O2_________
nHC2H3O2
= 4.745 + log 0.600 mol
_____________
0.650 mol
2B-5 (of 18)
= 4.71
Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and
1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added.
HC2H3O2 : Weak Acid
NaOH
NaC2H3O2 : Salt w/ Weak Conj. Base
: Strong Base
Amend the David Hasselholf Equation
pH = pKa + log
nC2H3O2- + nOH_________
__________
nHC2H3O2 – nOH-
= 4.745 + log (0.500 mol + 0.100 mol)
______________________________
(0.750 mol – 0.100 mol)
2B-6 (of 18)
= 4.71
200. mL of a solution 0.40 M in ammonia and 0.30 M in ammonium chloride
are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH.
NH3 : Weak Base
NH4Cl : Salt w/ Weak Conj. Acid
HCl : Strong Acid
Look up the Ka of the weak acid and determine its pKa
pKa = -log Ka
2B-7 (of 18)
= -log (5.6 x 10-10)
= 9.252
200. mL of a solution 0.40 M in ammonia and 0.30 M in ammonium chloride
are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH.
NH3 : Weak Base
NH4Cl : Salt w/ Weak Conj. Acid
HCl : Strong Acid
Calculate the moles of all species present
0.40 mol NH3
_________________
x 0.200 L solution
= 0.080 mol NH3
L solution
0.30 mol NH4+ x 0.200 L solution
__________________
= 0.060 mol NH4+
L solution
0.45 mol H3O+ x 0.100 L solution
__________________
L solution
2B-8 (of 18)
= 0.045 mol H3O+
200. mL of a solution 0.40 M in ammonia and 0.30 M in ammonium chloride
are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH.
NH3 : Weak Base
NH4Cl : Salt w/ Weak Conj. Acid
HCl : Strong Acid
Strong acids react completely with bases
2B-9 (of 18)
200. mL of a solution 0.40 M in ammonia and 0.30 M in ammonium chloride
are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH.
NH3 : Weak Base
NH4Cl : Salt w/ Weak Conj. Acid
HCl : Strong Acid
Amend the David Hasselholf Equation
pH = pKa + log nNH3 – nH3O+
___________
______
nNH4+ + nH3O+
= 9.255 + log (0.080 mol – 0.045 mol)
_______________________________
(0.060 mol + 0.045 mol)
2B-10 (of 18)
= 8.78
A buffer can be made by mixing solutions of:
(1) a weak acid and a salt containing its weak conjugate base
(2) a weak acid and a lesser amount of a strong base
(3) a weak base and a lesser amount of a strong acid
2B-11 (of 18)
(1) If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium nitrite
are mixed, a buffer is produced. Find the pH.
HNO2 : Weak Acid
NaNO2 : Salt containing the Conjugate Base
 the solution is a buffer
Look up the Ka of the weak acid and determine its pKa
pKa = -log Ka
2B-12 (of 18)
= -log (4.0 x 10-4) = 3.398
(1) If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium nitrite
are mixed, a buffer is produced. Find the pH.
HNO2 : Weak Acid
NaNO2 : Salt containing the Conjugate Base
 the solution is a buffer
Calculate the moles of all species present
0.50 mol HNO2 x 0.600 L solution
____________________
= 0.30 mol HNO2
L solution
0.50 mol NO2-
__________________
L solution
2B-13 (of 18)
x 0.400 L solution
= 0.20 mol NO2-
(1) If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium nitrite
are mixed, a buffer is produced. Find the pH.
HNO2 : Weak Acid
NaNO2 : Salt containing the Conjugate Base
 the solution is a buffer
Use the David Hasselholf Equation
pH = pKa + log nNO2-
= 3.398 + log (0.20 mol)
nHNO2
(0.30 mol)
_______
2B-14 (of 18)
_____________
= 3.22
(2) If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium
hydroxide are mixed, a buffer is produced. Find the pH.
HNO2 : Weak Acid
NaOH : Strong Base
Calculate the moles of all species
0.50 mol HNO2 x 0.600 L solution
____________________
= 0.30 mol HNO2
L solution
0.50 mol OH-
_________________
L solution
2B-15 (of 18)
x 0.400 L solution
= 0.20 mol OH-
(2) If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium
hydroxide are mixed, a buffer is produced. Find the pH.
HNO2 : Weak Acid
NaOH : Strong Base
Use the Amended David Hasselholf Equation
pH = pKa + log nNO2- + nOH-
_________
__________
nHNO2 – nOH= 3.398 + log
(0 mol + 0.20 mol)
___________________________
(0.30 mol – 0.20 mol)
2B-16 (of 18)
= 3.70
(3) If 500. mL of 0.60 M sodium nitrite and 250. mL of 0.80 M hydrochloric
acid are mixed, a buffer is produced. Find the pH.
NaNO2 : Salt containing a Weak Base
HCl : Strong Acid
Calculate the moles of all species
0.60 mol NO2-
___________________
x 0.500 L solution
= 0.30 mol NO2-
x 0.250 L solution
= 0.20 mol H3O+
L solution
0.80 mol H3O+
_________________
L solution
2B-17 (of 18)
(3) If 500. mL of 0.60 M sodium nitrite and 250. mL of 0.80 M hydrochloric
acid are mixed, a buffer is produced. Find the pH.
NaNO2 : Salt containing a Weak Base
HCl : Strong Acid
Use the Amended David Hasselholf Equation
pH = pKa + log nNO2- – nH3O+
_________
___________
nHNO2 + nH3O+
= 3.398 + log
(0.30 mol – 0.20 mol)
_____________________________
(0 mol + 0.20 mol)
2B-18 (of 18)
= 3.10
ACID-BASE TITRATIONS
Acid-base reactions with either a strong base or strong acid run to
completion
Determinations of volumes or molarities are stoichiometric calculations
2C-1 (of 18)
If a 10.0 mL sample of an acetic acid solution is titrated with 24.5 mL of a
0.250 M lithium hydroxide solution, find the molarity of the acetic acid
This is a stoichiometric calculation
HC2H3O2
xM
10.0 mL
1 mol
+
→
LiOH
H(OH)
LiC2H3O2
0.250 M
24.5 mL
1 mol
0.250 mol LiOH x 0.0245 L
__________________
L
x 1 mol HC2H3O2
___________________
1 mol LiOH
= 0.613 M HC2H3O2
2C-2 (of 18)
+
x
1
____________
0.0100 L
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(a) Calculate the volume of NaOH solution needed to neutralize the HZ
solution.
NaOH
0.400 M
x mL
1 mol
+
HZ
x 1 mol NaOH
________________
L
= 0.0500 L NaOH solution
2C-3 (of 18)
NaZ
+
H(OH)
1.00 M
20.0 mL
1 mol
1.00 mol HZ x 0.0200 L
__________________
→
1 mol HZ
x
L
_____________________
0.400 mol NaOH
= 50.0 mL NaOH solution
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(b) Calculate the pH of the HZ solution before the titration if the Ka of HZ
is 1.0 x 10-5.
The pH of a weak acid is determined from an ICE table for the ionization
reaction of the weak acid
HZ (aq)
Initial M’s
Change in M’s
Equilibrium M’s
Ka = [H3O+][Z-]
____________
+
H3O+ (aq)
+
Z- (aq)
1.00
-x
0
+x
0
+x
1.00 - x
x
x
1.00 x 10-5 =
[HZ]
pH = -log (3.162 x 10-3) = 2.50
2C-4 (of 18)
⇆
H2O (l)
x2
____________
(1.00 – x)
x = 3.162 x 10-3 M
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(c) Calculate the pH of the solution after 10.0 mL of the NaOH was added.
This solution will contain both a weak acid and its conjugate base
Calculate moles of all species
1.00 mol HZ
________________
x 0.0200 L solution = 0.0200 mol HZ
L solution
0.400 mol OH- x 0.0100 L solution = 0.00400 mol OH__________________
L solution
2C-5 (of 18)
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(c) Calculate the pH of the solution after 10.0 mL of the NaOH was added.
Use the Amended David Hasselholf Equation
pH = pKa + log
nZ- + nOH_________
__________
nHZ – nOH-
= 5.00 + log
(0 mol + 0.00400 mol)
___________________________________
(0.0200 mol – 0.00400 mol)
2C-6 (of 18)
= 4.40
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(d) Calculate the pH of the solution after 25.0 mL of the NaOH was added.
Calculate moles of all species
1.00 mol HZ
________________
x 0.0200 L solution = 0.0200 mol HZ
L solution
0.400 mol OH- x 0.0250 L solution = 0.0100 mol NaOH
__________________
L solution
= 0.0100 mol OH-
2C-7 (of 18)
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(d) Calculate the pH of the solution after 25.0 mL of the NaOH was added.
Use the Amended David Hasselholf Equation
pH = pKa + log nZ- + nOH-
_________
__________
nHZ – nOH= 5.00 + log
(0 mol + 0.0100 mol)
_________________________________
= 5.00
(0.0200 mol – 0.0100 mol)
At the half equivalence point, pH = pKa
2C-8 (of 18)
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(e) Calculate the pH of the solution at the equivalence point.
(remembering 50.0 mL of base needed) Calculate moles of all species
1.00 mol HZ
________________
x 0.0200 L solution = 0.0200 mol HZ
L solution
0.400 mol OH- x 0.0500 L solution = 0.0200 mol NaOH
__________________
L solution
= 0.0200 mol OH-
2C-9 (of 18)
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(e) Calculate the pH of the solution at the equivalence point.
Use the Amended David Hasselholf Equation
pH = pKa + log nZ- + nOH-
_______
__________
nHZ – nOH= 5.00 + log
(0 mol + 0.0200 mol)
_________________________________
= ???
(0.0200 mol – 0.0200 mol)
At the equivalence point, the solution contains only the conjugate base, Z-
2C-10 (of 18)
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(e) Calculate the pH of the solution at the equivalence point.
The pH of a weak base is determined from an ICE table for the ionization
reaction of the weak base
Z- (aq)
+
H2O (l)
⇆
HZ (aq) +
OH- (aq)
Calculate the molarity of the Z- in the solution at the equivalence point
0.0200 mol Z-
__________________
= 0.2857 M mol Z-
0.0700 L
Calculate the Kb of the weak base ZKb = Kw = 1.00 x 10-14
____
______________
Ka
1.0 x 10-5
2C-11 (of 18)
= 1.0 x 10-9
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(e) Calculate the pH of the solution at the equivalence point.
Z- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
Kb = [HZ][OH-]
_____________
2C-12 (of 18)
H2O (l)
⇆
0.2857
-x
0.2857 - x
1.0 x 10-9 =
HZ (aq) +
OH- (aq)
0
+x
x
0
+x
x
x2
_______________
(0.2857 – x)
[Z-]
1.69 x 10-5 =
+
x =
[OH-]
4.77 =
pOH
1.0 x 10-9 =
x2
__________
(0.2857)
9.23 =
pH
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(f) Calculate the pH of the solution after 60.0 mL of the NaOH was added.
Calculate moles of all species
1.00 mol HZ
________________
x 0.0200 L solution = 0.0200 mol HZ
L solution
0.400 mol OH- x 0.0600 L solution = 0.0240 mol NaOH
__________________
L solution
= 0.0240 mol OH-
2C-13 (of 18)
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(f) Calculate the pH of the solution after 60.0 mL of the NaOH was added.
Use the Amended David Hasselholf Equation
pH = pKa + log nZ- + nOH______
__________
nHZ – nOH= 5.00 + log
(0 mol + 0.0200 mol)
_________________________________
= ???
(0.0200 mol – 0.0240 mol)
Impossible – the strong base did not all react away
There are 0.0040 moles of excess OH- in the solution
2C-14 (of 18)
20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution.
(f) Calculate the pH of the solution after 60.0 mL of the NaOH was added.
Calculate the molarity of the OH- in the solution
0.0040 mol OH-
____________________
0.0800 L
pOH =
pH
1.30
= 12.70
2C-15 (of 18)
= 0.050 M mol OH-
Weak Acid-Strong Base Titration (HZ and NaOH)
At the start of the titration : pure HZ
At the half-neutralization point, half HZ, half Z-  pH = pKa
At the equivalence point : pure ZA weak acid-strong base titration produces an equivalence point pH > 7
because at the equivalence point the solution contains a weak base (Z-)
2C-16 (of 18)
INDICATORS
Phenolphthalein is a weak organic acid (pKa = 9.75)
HInd (aq) + H2O (l) ⇆ H3O+ (aq) + Ind- (aq)
colorless
pink
colorless if more HInd, pink if more IndKa =
[H3O+][Ind-]
_______________
[Hind]
Color change occurs when [Ind-] = [Hind] , so when
Ka = [H3O+]
or
pKa
=
pH
Phenolphthalein will change color at pH = 9.75
 good for titrations in which the equivalence point pH is slightly basic
2C-17 (of 18)
Strong Acid-Strong Base Titration (HA and NaOH)
A strong acid-strong base titration produces an equivalence point pH = 7
because at the equivalence point the solution contains a non base (A-)
An indicator with a pKa = 7 should be used for this titration
2C-18 (of 18)