Acid – Base Titrations EQUIVALENCE POINT The equivalence point, or stoichiometric point, of a chemical reaction is the point at which an added titrant

Acid – BaseTitrations

EQUIVALENCE POINT

• The equivalence point, or stoichiometric point, of a chemical reaction is the point at which an added titrant is stoichiometrically equal to the number of moles of substance (known as analyte) present in the sample: the smallest amount of titrant that is sufficient to fully neutralize or react with the analyte. In some cases there are multiple equivalence points, which are multiples of the first equivalence point, such as in the titration of a diprotic acid.

• Acid-Base Equivalence Point - the point at which chemically equivalent quantities of acid and base have been mixed, can be found by means of an indicator

Ca·Va·nº H = Cb·Vb·nº OH

http://en.wikipedia.org/wiki/Stoichiometry

http://en.wikipedia.org/wiki/Diprotic_acid



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.

Features of the Strong Acid-Strong Base Titration Curve

1. The pH starts out low, reflecting the high [H3O+] of the strong acid and increases gradually as acid is neutralized by the added base.

2. 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.

3. Beyond this steep portion, the pH increases slowly as more base is added.

Sample Calculation: Strong Acid-Strong Base Titration Curve

Problem 24-1. Consider the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH.

Region 1. Before the equivalence point, after adding 20.0 mL of 0.100 M NaOH. (Half way to the equivalence point.)

Initial moles of H3O+ =

- Moles of OH- added =

base added of volume acid of volumeoriginal

remaining OH of (mol)amount ]O[H 3

3

Sample Calculation: Strong Acid-Strong Base Titration Curve (Cont. I)

Region 2. At the equivalence point, after adding 40.0 mL of 0.100 M NaOH.

Initial moles of H3O+ = 0.0400 L x 0.100 M = 0.00400 M H3O+

- Moles of OH- added = 0.0400 L x 0.100 M =0.00400 mol OH-


remaining OH of (mol)amount ]O[H 3

3

Sample Calculation: Strong Acid-Strong Base Titration Curve (cont. II)

Region 3. After the equivalence point, after adding 50.0 mL of 0.100 M NaOH. (Now calculate excess OH-)

Total moles of OH- = 0.0500 L x 0.100 M = 0.00500 mol OH- -Moles of H3O+ consumed = 0.0400 L x 0.100 M =0.00400 mol


remaining OH of (mol)amount ][OH

HPr = Propionic Acid

The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak

Acid-Strong Base Titration Curve

1. The initial pH is higher.

2. A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point.

3. The pH at the equivalence point is greater than 7.00.

4. The steep rise interval is less pronounced.

Sample Calculation: Weak Acid-Strong Base Titration Curve

Problem 24-2. Consider the titration of 40.0 mL of 0.100 M HPr (Ka = 1.3 x 10-5) with 0.100 M NaOH.

Region 1. The solution of weak acid to be titrated, before any base is added.

Solution:

Ans:

Sample Calculation: Weak Acid-Strong Base Titration Curve (Cont.I)


Region 2. After 30. mL of base (total) has been added. This is clearly in the buffer region of the titration curve.

Solution: Refer to Lecture 23.

Can use the calculator program, ‘Buf’ developed in lecture 23. But first must calculate the nominal amounts of acid and base forms of the weak acid created by addition of the strong base. These are: [HA]0 =

[A-]0 =

Ans: From buffer program: pH =

Sample Calculation: Weak Acid-Strong Base Titration Curve (Cont.ll)


Region 3. After 40. mL of base (total) has been added. This is clearly at the equivalence point of the titration curve.

Solution: Refer to Lecture 23.

Can use the calculator program developed in lecture 23. But first must calculate the nominal amounts of acid and base forms of the weak acid created by addition of the strong base. These are:

[HA]0 =

[A-]0 =

Ans: From buffer program: pH =

Sample Calculation: Strong Acid-Strong Base Titration Curve (cont. IIl)

Region 4. After the equivalence point, after adding 50.0 mL of 0.100 M NaOH. (Now calculate excess OH-)

Total moles of OH- = -Moles of weak acid consumed =

Moles of OH- remaining =


tinneutralizaafter remaining OH of (mol)amount ][OH

The four Major Differences Between a Weak Acid-Strong Base Titration Curve and a Weak

Base-Strong Acid Titration Curve

1. The initial pH is above 7.00.

2. A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point.

3. The pH at the equivalence point is less than 7.00.

4. Thereafter, the pH decreases slowly as excess strong acid is added.

Features of the Titration of a Polyprotic Acid with a Strong Base

1. 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).

2. The pH at the midpoint of the buffer region is equal to the pKa of that acid species.

3. The same volume of added base is required to remove each mole of H+.

Acid-Base Indicators and the Measurement of pH

• Definition: A weak organic acid, HIn that has a different color than its conjugate base, In-, with the color change occurring over a specific and relatively narrow pH range.

• Typically, one or both forms are intensely colored, so only a tiny amount of indicator is needed, far too little to perturb the pH of the solution.

• Since the indicator molecule is a weak acid, the ratio of the two forms governed by the [H3O+] of the test solution:

a

3-

3a32

K

OH

In

HIn :Therefore

HIn

InOHHIn of K )(In)(OH)O(H)HIn(

aqaqlaq

Let us consider quantitatively, the case of titrating a weak acid with a strong base. If the weak acid has one dissociable proton, then the overall reaction is:

HA + OH- = A- + H20

We will assume that the strong base NaOH and the weak acid anion NaA are completely dissociated in solution.

Furthermore, we will not neglect the contribution of the dissociation of water.

Our Titration System is Governed by Four Equations

balance material

balance charge

hydrolysiser wat

mequilibriu base-acid

• Such a system has 8 experimentally measurable variables: Ka, Kw, [HA]0 and [NaOH]0

• If we assume that the first four (Ka, Kw, [HA]0 and [NaOH]0) are known, then we are left with 4 equations in 4 unknowns.

• Of the 4 unknowns, the only one we can conveniently measure is [H+]. This suggests that we solve the four equations for [H+] by successive elimination.

The Exact Solution to the Titration ProblemWe now proceed to solve the four equations for [H+] by successive elimination of variables. The result for [H+] is

Finding [H+] requires that we find the three roots of the above cubic equation and then selecting the one root that is consistent with physical reality, i.e. leads to all positive concentrations. Fortunately, in the above case only one root is positive and the other two are negative.

This positive root can be found by the ‘Solver’ function of your calculator. You can make a wild guess that is positive and the calculator will converge to the correct answer.

Answers

1. Region 1, pH = 1.477, Region 2, pH = 7.000, Region 3, pH = 12.046

2. Region 1, pH = 2.95 , Region 2, pH = 5.36 , Region 3,

pH = 8.79 , Region 4 = 12.05

26

Weak Acid Titration Calculations

27

Strong and Weak acids

Equilibrium and pH

Strong Acids:HClO4 H2SO4

HNO3 HIHBr HClHClO3

Weak Acids:“The Rest”

28

Titration Curve of a Weak Acid with a Strong Base

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50

Volume of Strong Base Added

equivalence pointbuffer region note volume ratiobetween these points25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5

0.10 M NaOH

Starting pH

Equivalence point

29

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?

How and Why are they different?

Weak Acid: Graph starting pH = 2.87

Calculated as Strong acid: pH = -log(0.10) = 1.0

“Let’s investigate!”

30

02468

101214

0 20 40 60

mL of Base Added

pH

Strong

Weak

31

How do the two values compare?

What is the significance of this difference?

Graph starting pH = 2.87

Strong acid calculation: pH = -log(0.10) = 1.0

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?

32

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?

33

Strong Acids:

HCl HCl HClHCl HCl

ADD WATER to MOLECULAR ACID

(H2O)

34

Strong Acids:

(H2O)H3O+

H3O+

H3O+

H3O+

H3O+

Cl-

Cl-

Cl-

Cl-

Cl-

Note: No HCl molecules remain in solution, all have been ionized in water.

35

HC2H3O2

HC2H3O2

HC2H3O2

HC

2H3O

2

HC2H3O2

(H2O)

Weak Acid Ionization:

Add water to MOLECULES of WEAK Acid

36

HC2H3O2

HC2H3O2

HC2H3O2

HC

2H3O

2

HC2H3O2

H30

+ C2H3O2-

(H2O)

Weak Acid Ionization:

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

37

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+).

38

Strong acid: [H3O+] = concentration of acidso: pH = -log [H3O+] = -log[acid]

Weak Acid: one must calculate the [H3O+] from an equilibrium ionization expression.

HA + H2O H3O+ + A-

Ka = _ [H3O+][A-] [HA]

These are equilibrium concentrations.

39


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50



0.10 M NaOH

1. Starting pH, only acid and water.

2

34

40


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50


equivalence pointbuffer region note volume ratiobetween these points

1

2

3

4

[H+]=[A-]

[H+][A-]

pOH=-Log[XS OH-]

[OH-]=[HA]

]HA[

]A][H[K a

]HA[

]A][H[K a

]A[

]HA][OH[Kb

Approximations:

41

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?

42

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]

HOAc H+ + OAC-ICE

0.10 0 0-x +x +x0.10-x x xKa = [x][x]

0.10-x Small,drop, WHY?

Ka = x2 0.10

x = 0.00134 = [H+]

pH = -log 0.00134 =2.87

43


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50



0.10 M NaOH

1.

2.

34

Some base has been addedThis is also the “buffer region”

pKa = pH at this point (halfway point)

44

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 HAnote: the 1:1 ratio

45

Region 2 Theory and Calculations:

Base has been added and some of the acid has been neutralized. intial RXN

Equilibrium XS HA + H2O H3O+ + A-

Ka= [H+][A-] [HA]

Leads to:

and [H+][A-] Why?

always(calculate from equilibrium)

HA + OH- H2O + A-

2 sources

46

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:

= mol HOAc0.02500L1L

0.10mol HOAc 0.0025

2. find moles of OH-:

_____________________________= mol OH-0.01400LNaOH

1L0.080molNaOH

1molNaOH

1molOH- 0.00112

Region 2

47


Solution:1. We found initial mol of Acid: HOAc + OH- H2O + OAc- 1:1 ratio

_____ = mol HOAc0.02500L1L

0.10mol HOAc 0.0025

2. We found mol of OH-: which =‘s initial moles of OAC- anion._____________________________________= mol OH-0.01400LNaOH

1L

0.080molNaOH

1molNaOH1molOH- 0.00112

3. Subtraction gives what?No, not the numerical answer!

moles of XS acid (initially present for OUR Eq. calculation).

48


Solution:1. Find initial moles of Acid: HOAc + OH- H2O + OAc- 1:1 ratio

_____ = mol HOAc0.02500L1L

0.10mol HOAc 0.0025

2. find moles of OH-: which =‘s initial moles of OAC- anion.

_____________________________________= mol OH-0.01400LNaOH

1L

0.080molNaOH

1molNaOH

1molOH- 0.00112

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.

49


Solution: FROM PREVIOUS SLIDE. 1. initial M of Acid: HOAc + OH- H2O + OAc- 1:1 ratio

2. M of OH-: which =‘s initial M of OAC- anion.

0.00138 mole/0.039L = 0.0354 M acid after RXN =INITIAL

0.00112 mole/0.039L = 0.0287 M OAc- after RXN =INITIAL**

[HOAc] [H+] + [OAc-]I**

CE

0.0354 0 0.0287 -x +x +x0.0354-x x 0.0287+x

]HA[

]A][H[Ka

]x0354.0[

]x0287.0][x[10x 8.1 5 -

Can we drop these?

x = 0.0000222 =[H+] pH = - Log [0.0000222] = 4.65

50

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.

51


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50



0.10 M NaOH

1.

2.

3.

4

Equivalence point

52

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-

º

Since this equilibrium involves OH- being formed, it is a Kb problem.

]A[

]HA][OH[Kb

53

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 = mol HOac1000mL0.10mol HOAc 0.0025

2. Find mol of OH-

31.25mL __ = mol OH-0.080mol NaOH1000mL

1molOH-

1mol NaOH

0.0025

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

54


moles of Acid = moles of base = 0.0025 = Eq. Pt.

New Volume = 0.05625L..and [OAc-] = 0.0025/0.05625=0.0444MQuestion: What equilibrium(ia) gives rise to the pH?

1. hydrolysis of the acetate anion:

OAc- + H2O HOAc + OH-

[OAc-] [HOAc] [OH-]ICE

0.0444 0 0 -x +x +x 0.0444-x x x

]OAc[

]HOAc][OH[Kb

]x0444.0[

]x[K

2

b

55


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?



OAc- HOAc OH-

ICE

0.0444 0 0 -x +x +x0.0444-x x x

]OAc[

]HOAc][OH[Kb

]x0444.0[

]x[K

2

b

Can we drop this x?

Also: KaKb=Kw=1.0 x 10-14 @25oC so Kb= 5.56 x 10-10

How do we find Kb?

56


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?



OAc- HOAc OH-

ICE

0.0444 0 0 -x +x +x 0.0444-x x x

]OAc[

]HOAc][OH[Kb

]x0444.0[

]x[10x 56.5

2-10

x = 4.97 x 10-6 = [OH-]pOH = 5.30pH =14 - 5.30 = 8.70

57


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50



0.10 M NaOH

1.

2.

3.

4 XS base

58

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.

59

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.

25.00mL = mol HOAc

1. Find moles of acid:

1000mL

0.10mol HOAc

2. Find moles of OH-:

40.00mL = mol OH-1000mL0.080molOH-

0.0025

0.0032

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

60


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50



1

2

3

4

[H+]=[A-]

[H+][A-]

pOH=-Log[XS OH-]

[OH-]=[HA]

]HA[

]A][H[K a

]HA[

]A][H[K a

]A[

]HA][OH[Kb

Approximations:

STUDY FOR QUIZ

61

Titration Curves for Three Different Weak Acids

pH

mL of Strong Base Added

Ka = 1.4 x 10-5

Ka = 1.4 x 10-6

Ka = 1.4 x 10-7

62

63

CO32- + H+ HCO3

-

HCO3- + H+ HCO3

64

65


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50



0.10 M NaOH

stotalLitermolOHmolHA

stotalLitermoleOH

]H[

K a

Region 2:acid + baseXS acid + saltHA+NaOH HA(XS)+NaA

0

2

0

2

a ]HA[

x

x]HA[

x

]HA[

]A][H[K

Region 1: only acid (HA)

stotalLitereHAinitialmol

x

]HA[]A[

]HA][OH[K

2

b

Region 3: mol acid=mol basehydrolysis of the salt anion

A- + H2OHA + OH-

Region 4: etotalvolum

molHAmolesOH]OH[

pOH=-Log[OH-]

66


0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35 40 45 50



stotalLitermolOHmolHA

stotalLitermoleOH

]H[

K a

Region 2:acid + baseXSacid + saltHA+NaOH HA(XS)+NaA

0

2

0

2

a ]HA[

x

x]HA[

x

]HA[

]A][H[K

Region 1: only acid (HA)

stotalLitereHAinitialmol

x

]HA[]A[

]HA][OH[K

2

b

Region 3: mol acid=mol base hydrolysis of the salt anionA- + H2OHA + OH-

Region 4: XS base

etotalvolum

molHAmolesOH]OH[

pOH=-Log[OH-]

Documents

Acid – Base Titrations EQUIVALENCE POINT The equivalence point, or stoichiometric point, of a chemical reaction is the point at which an added titrant