Weak Acid Titration Calculations
Paul Gilletti Ph.D.
Mesa Community College
Mesa, AZ
1
Strong and Weak acids
Equilibrium and pH
Strong Acids:
HClO4 H2SO4
HNO3
HI
HBr
HCl
HClO3
Weak Acids:
“The Rest”
2
25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5
Starting pH
Equivalence point
0.10 M NaOH
3
Acetic acid is a weak acid. What would be the starting pH
of the solution if it were a strong acid? (25.0 mL of 0.10M
Acetic acid)
Do a quick calculation as if acetic acid were a strong acid and
compare this value to that on the graph.
How do the two values compare?
Weak Acid: Graph starting pH = 2.87
Calculated as Strong acid: pH = -log(0.10) = 1.0
How and Why are they different?
“Let’s investigate!”
4
pH
14
12
10
8
6
4
2
0
Weak
Strong
0
20
40
60
mL of Base Added
5
How do the two values compare?
Graph starting pH = 2.87
Strong acid calculation: pH = -log(0.10) = 1.0
What is the significance of this difference?
Since the two values differ by ~2 pH units and pH is a log
scale, the concentration of H3O+ in the strong acid
calculation is 1 x 102 or 100 times greater than that
observed on the graph.
Let’s see WHY they are different?
6
Strong Acids:
100% ionized (completely dissociated) in water.
HCl + H2O  H3O+ + ClNote the “one way arrow”.
Weak Acids:
Only a small % (dissociated) in water.
HC2H3O2 + H2O  H3O+ + C2H3O2Note the “2-way” arrow.
Why are they different?
7
Strong Acids:
HCl HCl
HCl
HCl
HCl
(H2O)
ADD WATER to MOLECULAR ACID
8
Strong Acids:
Cl-
(H2O)
H 3O +
H 3O +
H 3O +
Cl-
H3O+ ClH 3O +
ClCl-
Note: No HCl molecules remain in
solution, all have been ionized in water.
9
Weak Acid Ionization:
HC2H3O2
HC2H3O2
HC2H3O2
HC2H3O2
HC2H3O2
(H2O)

Add water to MOLECULES of WEAK Acid
10
Weak Acid Ionization:
HC2H3O2
HC2H3O2
HC2H3O2
HC2H3O2
HC2H3O2
(H2O)

H30+
C2H3O2-
Note: At any given time only a small portion of the acid
molecules are ionized and since reactions are running in
BOTH directions the mixture composition stays the same.
This gives rise to an Equilbrium expression, Ka
11
pH calculations:
To do pH calculations one must consider
both the nature and condition (amount of
ionization) of the species in the solution and
then calculate the concentration of the
hydronium ion (H+ or H3O+).
12
Strong acid: [H3O+] = concentration of acid
so: pH = -log [H3O+] = -log[acid]
Weak Acid: one must calculate the [H3O+]
from an equilibrium ionization expression.
HA + H2O  H3O+ + AKa = _ [H3O+][A-]
[HA]
These are equilibrium
concentrations.
13
25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5
2
4
3
1.Starting pH, only acid and water.
0.10 M NaOH
14
Approximations:
[H  ][ A  ]
Ka 
[HA ]
[H+][A-]
4
2
pOH=-Log[XS OH-]
3
1
[H  ][ A  ]
Ka 
[HA ]
[H+]=[A-]
[OH  ][HA ]
Kb 
[A  ]
[OH-]=[HA]
15
Problem: Calculate the pH of 25.0 mL of 0.10M
acetic acid (HOAc).
The Ka of HOAc = 1.8 x 10-5
Which region of a titration curve would this be?
16
Problem: Calculate the pH of 25.0 mL of 0.10M
acetic acid (HOAc).
The Ka of HOAc = 1.8 x 10-5
Ka = [H+][OAc-]
[HOAc]
Ka = [x][x]
0.10-x
Ka = x 2
0.10
HOAc  H+ + OACI 0.10
0
0
+x
+x
C -x
x
x
E 0.10-x
Small,drop, WHY?
x = 0.00134 = [H+]
pH = -log 0.00134 =2.87
17
25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5
2.Some base has been added
This is also the “buffer region”
4
3
1.
pKa = pH at this point (halfway point)
0.10 M NaOH
18
Region 2 Theory and Calculations:
Base has been added and some of the acid has been
neutralized.
What has changed and WHY?
1. The moles of acid is decreased.
2. The moles of acid anion (A-) is increased.
HA + OH-  H2O + A3. The total volume has increased.
HA + OH-  H2O + A- + XS HA
note: the 1:1 ratio
19
Region 2 Theory and Calculations:
Base has been added and some of the acid has been
neutralized. intial RXN
HA + OH-  H2O + AEquilibrium XS HA + H2O  H3O+ + A-
2 sources
always
Leads to:
(calculate from equilibrium)
Ka= [H+][A-]
[HA]
and
[H+][A-]
Why?
20
Region 2
Problem: Calculate the pH of a solution that is
prepared by combining 25.00mL of 0.10M acetic
acid (HOAc) with 14.00mL of 0.080M NaOH.
(Ka of HOAc = 1.8 x 10-5).
Solution: HOAc + OH-  H2O + OAc- 1:1 ratio
1. Find initial moles of Acid:
0.02500L 0.10mol HOAc = 0.0025 mol HOAc
1L
2. find moles of OH-:
0.01400L
NaOH 0.080molNaOH 1molOH0.00112 mol OH_____________________________=
1L
1molNaOH
21
Problem: Calculate the pH of a solution that is prepared by
combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of
0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).
Solution:
1. We found initial mol of Acid: HOAc + OH-  H2O + OAc- 1:1 ratio
0.10mol _____
HOAc = 0.0025 mol HOAc
1L
2. We found mol of OH-: which =‘s initial moles of OAC- anion.
0.02500L
0.01400LNaOH 0.080molNaOH 1molOH_____________________________________=
0.00112 mol OH1molNaOH
1L
3. Subtraction gives what?
No, not the numerical answer!
moles of XS acid (initially present for OUR Eq. calculation).
22
Problem: Calculate the pH of a solution that is prepared by
combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of
0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).
Solution:
1. Find initial moles of Acid: HOAc + OH-  H2O + OAc- 1:1 ratio
0.10mol _____
HOAc = 0.0025 mol HOAc
1L
2. find moles of OH-: which =‘s initial moles of OAC- anion.
0.02500L
0.01400LNaOH 0.080molNaOH 1molOH_____________________________________=
0.00112 mol OH1L
1molNaOH
3. Subtraction gives moles of XS acid (initial for I.C.E.).
0.0025 - 0.00112 = 0.00138 mole acid INITIAL
4. and..we have 0.00112 mole OAc- INITIAL
for I.C.E.
23
Problem: Calculate the pH of a solution that is prepared by
combining 25.00mL of 0.10M acetic acid (HOAc) with 14.00mL of
0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).
Solution: FROM PREVIOUS SLIDE.
1. initial M of Acid: HOAc + OH-  H2O + OAc- 1:1 ratio
0.00138 mole/0.039L = 0.0354 M acid after RXN =INITIAL
2. M of OH-: which =‘s initial M of OAC- anion.
0.00112 mole/0.039L = 0.0287 M OAc- after RXN =INITIAL**
[HOAc]  [H+] + [OAc-]


[H ][ A ]
Ka 
I** 0.0354
0 0.0287
[HA ]
-x
+x
+x
C
[x][0.0287  x]
-5
E 0.0354-x x 0.0287+x
1.8 x 10 
[0.0354  x]
Can we drop these?
x = 0.0000222 =[H+]
pH = - Log [0.0000222] = 4.65
24
Region 2 summary:
1. write balanced equation for initial weak/strong RXN.
2. find moles of acid
3. find moles of base
4. subtract to determine XS (and limiting reactant which equals
initial moles of salt formed)
5. Change XS to M and limiting Reactant to M
6. Write balanced Chemical Equilibrium Equation:
+
HA  H + A
7. Use values from #5 a initial values in I.C.E. Chart
8. Work as Equilibrium problem.
25
25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5
2.
4
3. Equivalence point
1.
0.10 M NaOH
26
Region 3. Theory and Calculations.
Mole of acid = moles of OH- added. Equivalence pt.
HA + OH-  H2O + A-
no XS of either
The titration curve indicates the pH to be above 7.
What causes this?
Hydrolysis of the A- anion: A- + H2O  HA + OHSince this equilibrium involves OH- being formed,
it is a Kb problem.
[OH  ][ HA ]
Kb 
[A  ]
º
27
Region 3 problem: Calculate the pH at the equivalence
point of a solution formed by the titration of 25.00mL of
0.10M acetic acid with 31.25mL of 0.080M NaOH
Ka(HOAc)=1.8 x 10-5.
1. Find moles of Acid (which =‘s moles of base at Eq pt.)
25.00mL 0.10mol HOAc
1000mL
= 0.0025 mol HOac
2. Find mol of OH31.25mL 0.080mol NaOH 1molOH
__ - = 0.0025 mol OH1000mL
1mol NaOH
3. Since the # of moles are the same, this is at the Eq. Pt. and
therefore the initial moles of OAc- ion = 0.0025 mol
28
Region 3 problem: Calculate the pH at the equivalence
point of a solution formed by the titration of 25.00mL of
0.10M acetic acid with 31.25mL of 0.080M NaOH
Ka(HOAc)=1.8 x 10-5.
moles of Acid = moles of base = 0.0025 = Eq. Pt.
New Volume = 0.05625L..and [OAc-] = 0.0025/0.05625=0.0444M

Question: What equilibrium(ia) gives rise to the pH?
[OH  ][ HOAc ]
Kb 
[OAc  ]
[x 2 ]
Kb 
[0.0444  x]
1. hydrolysis of the acetate anion:OAc- + H2O  HOAc + OH
[OAc-]
[HOAc] [OH-]
I 0.0444
C -x
E 0.0444-x
0
+x
x
0
+x
x
29
Region 3 problem: Calculate the pH at the equivalence point of a solution
formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M
NaOH
Ka(HOAc)=1.8 x 10-5.
moles of Acid = moles of base = 0.0025 = Eq. Pt.
New Volume = 0.05625L...... and [OAc-] =0.0444M
Question: What equilibrium(ia) gives rise to the pH?
1. hydrolysis of the acetate anion:
OAc- + H2O  HOAc + OH
[OH  ][ HOAc ]
OAc
HOAc
OH
Kb 
[OAc  ]
0
0
I 0.0444
-x
+x
+x
C
[x 2 ]
Kb 
x
x
E 0.0444-x
[0.0444  x]
Can we drop this x?
How do we find Kb?
Also: KaKb=Kw=1.0 x 10-14 @25oC so Kb= 5.56 x 10-10
30
Region 3 problem: Calculate the pH at the equivalence point of a solution
formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M
NaOH
Ka(HOAc)=1.8 x 10-5.
moles of Acid = moles of base = 0.0025 = Eq. Pt.
New Volume = 0.05625L...... and [OAc-] =0.0444M
Question: What equilibrium(ia) gives rise to the pH?
1. hydrolysis of the acetate anion:
OAc- + H2O  HOAc + OH[OH  ][ HOAc ]
OAc
HOAc
OH
Kb 
[OAc  ]
0
0
I 0.0444
+x
+x
C -x
2
[
x
]
5.56 x 10-10 
x
x
E 0.0444-x
[0.0444  x]
x = 4.97 x 10-6 = [OH-]
pOH = 5.30
pH =14 - 5.30 = 8.70
31
25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5
2.
4
XS base
3.
1.
0.10 M NaOH
32
Region 4 theory and Calculations:
All the acid has been neutralized and XS base has been added
HA + OH-  H2O + A- + XS OHWhat is the dominating factor that controls the pH?
The XS strong base (OH-)
Calculations:
Find moles of XS OH- and then use the new
volume to find [OH-] and then the pOH and pH.
33
Region 4 problem: Calculate the pH of a solution formed
by the titration of 25.00mL of 0.10M acetic acid with
40.00mL of 0.080M NaOH. Ka(HOAc)=1.8 x 10-5.
1. Find moles of acid:
25.00mL 0.10mol HOAc
1000mL
2. Find moles of OH-:
0.080molOH
40.00mL
1000mL
3. Subtract to find XS:
= 0.0025 mol HOAc
= 0.0032 mol OH-
0.0032 - 0.0025 = 0.0007 mole XS OH4. Find OH- concentration, pOH, and pH:
pOH = -Log[0.0007/0.06500] = 1.97 and pH = 12.03
34
Approximations:
[H  ][ A  ]
Ka 
[HA ]
[H+][A-]
2
4
pOH=-Log[XS OH-]
STUDY FOR QUIZ
3
1
[H  ][ A  ]
Ka 
[HA ]
[H+]=[A-]
[OH  ][HA ]
Kb 
[A  ]
[OH-]=[HA]
35
Titration Curves for Three Different Weak Acids
Ka = 1.4 x 10-7
pH
Ka = 1.4 x 10-6
Ka = 1.4 x 10-5
mL of Strong Base Added
36
37
CO32- + H+  HCO3-
HCO3- + H+  HCO3
38
39
molesOH   molHA
[OH ] 
4:
totalvolum e

Region 2:
acid + baseXS acid + salt
HA+NaOH HA(XS)+NaA
 moleOH  
[H  ]

 totalLiter s 
Ka 
 molHA  molOH  


totalLiter s


Region
pOH=-Log[OH-]
Region 3: mol acid=mol base
hydrolysis of the salt anion
A- + H2OHA + OH
-
[OH  ][ HA ]
x2
Kb  

[ A ]  [HA ]  initialmol eHA 
 totalLiter s 


Region 1: only acid (HA)
Ka 
[H  ][ A  ]
x2
x2


[HA ]
[HA ]0  x [HA ]0
0.10 M NaOH
40
Region 4: XS base
molesOH   molHA
[OH ] 
totalvolum e

pOH=-Log[OH-]
Region 2:
acid + baseXSacid + salt
HA+NaOH HA(XS)+NaA
 moleOH  
[H  ]

 totalLiter s 
Ka 
 molHA  molOH  


totalLiter s


Region 3: mol acid=mol base
hydrolysis of the salt anion
A- + H2OHA + OH-
Kb 
Region 1: only acid (HA)
Ka 
[OH  ][ HA ]
x2

[ A  ]  [HA ]  initialmol eHA 
 totalLiter s 


[H  ][ A  ]
x2
x2


[HA ]
[HA ]0  x [HA ]0
41