# Acid Value Calculator Weak Acid Titration Calculations Paul

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```					Weak Acid Titration Calculations

Paul Gilletti Ph.D.
Mesa Community College
Mesa, AZ

1
Strong and Weak acids

Equilibrium and pH

Strong Acids:       Weak Acids:
HClO4 H2SO4         “The Rest”
HNO3      HI
HBr       HCl
HClO3
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
14
12
10
8
pH

Weak
6
4                    Strong
2
0
0           20         40     60
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+ + Cl-
Note the “one way arrow”.

Weak Acids:
Only a small % (dissociated) in water.

HC2H3O2 + H2O  H3O+ + C2H3O2-
Note the “2-way” arrow.

Why are they different?
7
Strong Acids:

HCl HCl        (H2O)
HCl
HCl
HCl

8
Strong Acids:
Cl-
H 3O +
(H2O)                       Cl-
H 3O +
H3O+ Cl-       Cl-
H 3O +
H 3O +
Cl-

Note: No HCl molecules remain in
solution, all have been ionized in water.
9
Weak Acid Ionization:

HC2H3O2
HC2H3O2
(H2O)

HC2H3O2
HC2H3O2                  
HC2H3O2

Add water to MOLECULES of WEAK Acid

10
Weak Acid Ionization:

HC2H3O2
HC2H3O2
(H2O)
HC2H3O2
         H30+     C2H3O2-

HC2H3O2        HC2H3O2

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+ + A-

Ka = _ [H3O+][A-]     These are equilibrium
[HA]         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
[OH  ][HA ]
[H  ][ A  ]   Kb 
1   Ka                          [A  ]
[HA ]
[H+]=[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  H+ + OAC-
[HOAc]        I 0.10       0   0
C -x        +x  +x
Ka = [x][x]
E 0.10-x     x   x
0.10-x
Small,drop, WHY?
Ka = x 2         x = 0.00134 = [H+]
0.10
pH = -log 0.00134 =2.87
17
25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5

This is also the “buffer region”
4
3

pKa = pH at this point (halfway point)
1.

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 + A-
3. 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 + A-

Equilibrium XS HA + H2O  H3O+ + A-             2 sources

always
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-:
0.01400LNaOH 0.080molNaOH 1molOH- 0.00112 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
0.02500L           HOAc
0.10mol _____ = 0.0025 mol HOAc
1L
2. We found mol of OH-: which =„s initial moles of OAC- anion.
0.01400LNaOH 0.080molNaOH 1molOH-
_____________________________________= 0.00112 mol OH-
1L          1molNaOH
3. Subtraction gives what?

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
0.02500L            HOAc
0.10mol _____ = 0.0025 mol HOAc
1L
2. find moles of OH-: which =„s initial moles of OAC- anion.
0.01400LNaOH 0.080molNaOH 1molOH-
_____________________________________= 0.00112 mol OH-
1L          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**

[H  ][ A  ]
[HOAc]  [H+] + [OAc-]
Ka                          I** 0.0354    0 0.0287
[HA ]
C      -x     +x     +x
[x][ 0.0287  x]
1.8 x 10 
-5                           E 0.0354-x x 0.0287+x
[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 + OH-
Since 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                 = 0.0025 mol HOac
1000mL

2. Find mol of OH-

31.25mL 0.080mol NaOH 1molOH- = 0.0025 mol OH-
__
1000mL       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?
1. hydrolysis of the acetate anion:
OAc- + H2O  HOAc + OH-

[OH ][ HOAc ]              [OAc-] [HOAc] [OH-]
Kb 
[OAc  ]            I  0.0444       0       0
[x 2 ]             C   -x         +x      +x
Kb 
[0.0444  x]           E 0.0444-x       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  ]                   I   0.0444      0               0
[x 2 ]                  C     -x       +x              +x
Kb 
[0.0444  x]                E 0.0444-x       x               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  ]                     I  0.0444   0                 0
C   -x     +x                +x
[x 2 ]
5.56 x 10   -10
                      E 0.0444-x   x                 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 OH-

What 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            = 0.0025 mol HOAc
1000mL
2. Find moles of OH-:

40.00mL    0.080molOH-            = 0.0032 mol OH-
1000mL
3. Subtract to find XS:
0.0032 - 0.0025 = 0.0007 mole XS OH-
4. 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-]
4
2          STUDY FOR QUIZ     pOH=-Log[XS OH-]
3
[OH  ][ HA ]
[H  ][ A  ]   Kb 
1   Ka                          [A  ]
[HA ]
[H+]=[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 ] 
Region     4:            totalvolum e
Region 2:
acid + baseXS acid + salt                           pOH=-Log[OH-]
HA+NaOH HA(XS)+NaA
 moleOH 
[H  ]
s
 totalLiter         Region 3: mol acid=mol base
Ka 
 molHA  molOH 

     totalLiter s     

hydrolysis of the salt anion
A- + H2OHA + OH
-
[OH ][HA ]            x2
Kb             
[ A ]  [HA ]  initialmoleHA 
 totalLiter 
           s 

[H  ][A  ]      x2        x2
Region 1: only acid (HA)    Ka                            
[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  ]
s
 totalLiter 
Ka 
 molHA  molOH 
                                                        Region 3: mol acid=mol base
     totalLiter s     
hydrolysis of the salt anion
A- + H2OHA + OH-

[OH ][HA ]              x2
Kb                   
[ A  ]  [HA ]  initialmoleHA 
 totalLiter 
           s 

Region 1: only acid (HA)
[H  ][A  ]      x2        x2
Ka                            
[HA ]        [HA ]0  x [HA ]0

41

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