titration curves...a ph electrode in a solution of 0.100 m hcl (the analized compound) and...

TITRATION CURVES

To find a titration’s end point, we need to monitor some property of the reaction that has a well-defined value at the equivalence point. For example, the equivalence point for a titration of HCl with NaOH occurs at a pH of 7.0. A simple method for finding the equivalence point is to continuously monitor the titration mixture’s pH using a pH electrode, stopping the titration when we reach a pH of 7.0. Alternatively, we can add an indicator to the titrand’s solution that changes color at a pH of 7.0.

A titration curve provides us with a visual picture of how a property of the titration reaction changes as we add the titrant to the analyzed compound.

The titration curve in Figure, for example, was obtained by suspending a pH electrode in a solution of 0.100 M HCl (the analized compound) and monitoring the pH while adding 0.100 M NaOH (the titrant).

From an acid-base titration curve, we can deduce the quantities and values of acidic and basic substances in a mixture. In medicinal chemistry, the pKa and lipophilicity of a candidate drug predict how easily it will cross cell membranes. It is known that from pKa and pH, we can compute the charge of a polyprotic acid. Usually, the more highly charged a drug, the harder it is to cross a cell membrane. In this lecture, we will learn how to predict the shapes of titration curves and how to find end points with indicators.

….our goal is to construct a graph showing how the pH changes as titrant is added…..

ClHHCl

11.0lg)(lg HcpH

рН = 1

Let’s focus on the titration of 100.00 mL of 0.10 00 M HCl with 0.100 0 M NaOH

I. Titration of strong acid with strong base

At the start of the titration the solution is 0.100 M in HCl, which, because HCl is a strong acid, means that the pH is

The first step in each case is to write the chemical reaction between titrant and analyte. Then use that reaction to calculate the composition and pH after each addition of titrant.

OHNaClNaOHHCl 2

In the titration of any strong acid with any strong base, there are three regions of the titration curve that require different kinds of calculations: 1.Before the equivalence point, the pH is determined

by excess of acid in the solution.

2. At the equivalence point, is just sufficient to react with all to make. The pH is determined by dissociation of water. 3. After the equivalence point, pH is determined by excess of base in the solution.

We will do sample calculations for each region.

Before the equivalence point, HCl is present in excess and the pH is determined by the concentration of unreacted HCl.

1 Before the equivalence point

After adding X mL of NaOH the concentration of excess HCl is

volumetotal

addedNaOHmolesHClmolesinitialHcHClc )()(

)()(

)()()()(

)()(

NaOHVHClV

NaOHVNaOHcHClVHClc

volumetotal

addedNaOHmolesHClmolesinitialHcHClc

After adding 1 ml of 0,1 M NaOH:

lmolHc /098.01100

11.01001.0)( рН = 1,01

а

After adding 50 ml of 0,1 M NaOH: б

lmolHc /0338.050100

501.01001.0)( рН = 1,48

)()(

)()()()()(

NaOHVHClV

NaOHVNaOHcHClVHClcHc

lmolHc /103.590100

901.01001.0)( 3

After adding 90 ml of 0,1 M NaOH: b

рН = 2,3

After adding 99 ml of 0,1 M NaOH: c

After adding 99,9 ml of 0,1 M NaOH: d

lmolHc /10599100

991.01001.0)( 4

рН = 3,3

рН = 4,3 lmolHc /105

9.99100

9.991.01001.0)( 5

)()( NaOHnHCln

lmolOHcHc /10)()( 7

2 At the equivalence point

At the equivalence point the moles of HCl and the moles of NaOH are equal. Since neither the acid nor the base is in excess, the pH is determined by the dissociation of water.

рН = 7

The equivalence point occurs when the added titrant is exactly enough for stoichiometric reaction with the analyte. The equivalence point is the ideal result we seek in a titration. What we actually measure is the end point, which is marked by a sudden physical change, such as indicator color.

3

For volumes of NaOH greater than the equivalence point, the pH is determined by the concentration of excess OH–.

After the equivalence point

)()(

)()()()(

)()(

NaOHVHClV

HClVHClcNaOHVNaOHc

volumetotal

HClmolesinitialaddedNaOHmolesOHcNaOHc


lmolOHc /105100101

1001.01011.0)( 4

рН = 14 – рОН = 14 – 3,3 = 10,7


lmolOHc /02.0100150

1001.01501.0)( рН = 14 – рОН

= 14 – 2,7= 12,3

After adding 100,1 ml of 0,1 M NaOH: а

lmolOHc /1051001.100

1001.01.1001.0)( 5

рН = 14 – рОН = 14 – 4,3 = 9,7

Titration curve for the titration of 0.100 M HCl with 0.100 M NaOH.

№

1 0 1

2 1 1,01

3 50 1,48

4 90 2,3

5 99 3,3

6 99,9 4,3

7 100 7

8 100,1 9,7

9 101 10,7

10 150 12,3

)(NaOHV pH

40 80 120 160 180 0

2

4

6

8

10

12

рН

V(NaOH), ml

Equivalence point

Larg

e p

H c

hang

e

kno

wn

as

a b

reak

Indicator Colour pH1-pH2 pT

Phenolphthalein

Litmus

Methyl red

Methyl orange

uncoloured - pink

red-blue

red-yellow

red-yellow

8,2-10

5-8

4,2-6,2

3,1-4,4

9

7,0

5,0

4

CONCLUSIONS

•As the concentrations of acid and titrant decrease, the end point break decreases. So the selection of indicator becomes more critical.

•Near the equivalence point, the pH increases very rapidly. A strong acid – strong base titration curve has a large end point break.

• At the equivalence point only a neutral salt remains in solution so the pH is 7.00.

You can use this same approach to calculate the titration curve for the titration of a strong base with a strong acid, except the strong base is in excess before the equivalence point and the strong acid is in excess after the equivalence point.

II. Titration of weak acid with strong base

рН = 2,88

)()( 3 HcCOOHCHc

Let’s consider the titration of 100.00 mL of 0.1000 M CH3COOH with 0.1000 M NaOH .

lmolcKHc aa /1032.11.01075.1)( 35

Before adding the titrant, the pH is determined by the analyte, which in this case is a weak acid.

5

3 1075.1)( COOHCHKд

In the titration of any weak acid with any strong base, there are three regions of the titration curve that require different kinds of calculations: 1.Before the equivalence point, the pH is determined

by a buffer containing the weak acid and its salt.

2. At the equivalence point, the pH is determined by hydrolysis of a salt. 3. After the equivalence point, pH is determined by excess of base in the solution.


1

Before the equivalence point, the pH is determined by a buffer containing

the weak acid and its salt

Adding NaOH converts a portion of the acetic acid to its salt. Any solution containing comparable amounts of a weak acid, HA, and its salt, A–, is a buffer.

OHCOONaCHNaOHCOOHCH 233

Before the equivalence point

COONaCH

COOHCH

COOHCHc

cKHc

3

3

3)(

NaOHCOOHCH

NaOHNaOHCOONaCH

VV

Vc

volumetotal

addedNaOHmolesc

3

3

We can calculate the pH of a buffer using the Henderson–Hasselbalch equation.

Before the equivalence point the concentration of unreacted acetic acid is

)()(

)()()()(

)(

3

33

33

NaOHVCOOHCHV

NaOHVNaOHcCOOHCHVCOOHCHc

volumetotal

addedNaOHmolesCOOHCHmolesinitialCOOHCHc

And the concentration of acetate is

55

50 1075.11.050

501.01001.01075.1)(Hc

65

90 1094.11.090

901.01001.01075.1)(Hc

75

99 1076.11.099

991.01001.01075.1)(Hc

85

9.99 1075.11.09,99

9,991.01001.01075.1)(Hc

рН = 4,76

рН = 5,71

рН = 6,75

рН = 7,77

After adding 50 ml of 0,1 M NaOH: а



After adding 99,9 ml of 0,1 M NaOH: d

NaOHCOOHCHOHCOONaCH 323

2

COONaCHCOOHCH cpKpH33

lg2

1

2

17

lmolVV

Vcc

NaOHCOOHCH

NaOHNaOHCOONaCH /05.0

100100

1001.0

3

3

73,805,02

176,4

2

17pH

рН = 8,73

At the equivalence point the moles of acetic acid initially present and the moles of NaOH added are identical. Because their reaction effectively proceeds to completion, the pH will depend on hydrolysis of sodium acetate.

At the equivalence point

)()(

)()()()()(

3

33

NaOHVCOOHCHV

COOHCHVCOOHCHcNaOHVNaOHcOHc

3

lmolOHc /1051.200

1.01001.01.100)( 5

1.100

рОН = 4,3 рН = 9,7

After the equivalence point, the titrant is in excess and the titration mixture is a dilute solution of NaOH. We can calculate the pH using the same strategy as in the titration of a strong acid with a strong base.

After adding 100,1 ml of 0,1 M NaOH: а


№

1 0 2,8

2 50 4,76

3 90 5,71

4 99 6,75

5 99,9 7,77

6 100 8,73

7 100,1 9,7

)(NaOHV pH

Titration curve for the titration of 0.100 M CH3COOH with 0.100 M NaOH.

pH

V(ml)

Calculated curves showing the titration of HA of different concentrations with 0.100 M NaOH. As the acid becomes more dilute, the end point becomes less distinct.

Conclusions

• Low initial pH, means that analyt was acid. Break is small, that means analyte is weak acid. There is one break – acid is monoprotic.

•pH is basic (about 9) at equivalence point (middle of break).

•As HA becomes a weaker acid, or as the concentrations of analyte and titrant decrease, the inflection near the equivalence point decreases, until the equivalence point becomes too shallow to detect. It is not practical to titrate an acid when its strength is too weak (Ка <10

-7 ) or its concentration too dilute.

Calculated curves showing the titration of acids with NaOH. As the acid becomes weaker, the end point becomes less distinct.

5

3 1076.1)(NHKд

lmolcKOHc bb /1032.11.01076.1)( 35

)(lg OHcpOH

12,1114 pOHpH рН = 11,12

Let’s focus on the titration of 100.00 mL of 0.1000 M NH3 with 0.1000 M HCl

)()( 3 OHcNHc

III. Titration of weak base with strong acid

Before acid is added, the solution contains just the weak base, NH3, in water. The pH is determined by the pKb reaction.

In the titration of any weak base with any strong acid, there are three regions of the titration curve that require different kinds of calculations: 1.Before the equivalence point, the pH is determined

by a buffer containing the weak base and its salt.

2. At the equivalence point, the pH is determined by hydrolysis of a salt. 3. After the equivalence point, pH is determined by excess of acid in the solution.


Between the initial point and the equivalence point, there is a mixture of NH3 and its salt —Aha! A buffer!

1 Before the equivalence point

ClNHHClNH 43

ClNH

NH

NHc

cKOHc

4

3

3)(

HClNH

HClHClClNH

VV

Vc

volumetotal

addedHClmolesc

3

4

Before the equivalence point the concentration of unreacted ammonia solution is

)()(

)()()()(

)(

3

33

33

HClVNHV

HClVHClcNHVNHc

volumetotal

addedHClmolesNHmolesinitialNHc

And the concentration of ammonium chloride is

65

90 1094.11.090

901.01001.01076.1)(OHc

75

99 1076.11.099

991.01001.01076.1)(OHc

85

9.99 1075.11.09,99

9,991.01001.01076.1)(OHc

рН = 8,29

рН = 7,25

рН = 6,23

After adding 90 ml of 0,1 M HCl: a

After adding 99 ml of 0,1 M HCl: b

After adding 99,9 ml of 0,1 M HCl: c

2

HClNHOHClNH 324

ClNHNH cpKpH43

lg2

1

2

17

lmolVV

Vcc

HClNH

HClHClClNH /05.0

100100

1001.0

3

4

27.505,02

176,4

2

17pH

рН = 5,27

At the equivalence point, NH3 is converted completely into a weak salt NH4Cl. The pH is calculated by considering the salt hydrolysis reaction

At the equivalence point

After the equivalence point, the excess strong acid determines the pH.

3

)()(

)()()()()(

3

33

NHVHClV

NHVNHcHClVHClcHc

lmolHc /1051.200

1.01001.01.100)( 5

1.100

рН = 4,3

After adding 100,1 ml of 0,1 M HCl: а


№

1 0 11,12

2 90 8,29

3 99 7,25

4 99,9 6,23

5 100 5,27

6 100,1 4,30

)(HClV pHpH

V(ml)

Titration curve for the titration of 0.100 M NH3 with 0.100 M HCl.

CONCLUSIONS

•High initial pH, means that analyt was base. Break is small, that means analyte is weak base. There is one break – base is monobasic.

The pH at the equivalence point is below 7.

As B becomes a weaker base, or as the concentrations of analyte and titrant decrease, the inflection near the equivalence point decreases, until the equivalence point becomes too shallow to detect. It is not practical to titrate a base when its strength is too weak (Ка <10

-7 ) or its concentration too dilute.

Calculated curves showing the titration of bases with HCl. As the base becomes weaker, the end point becomes less distinct.

Titration of mixture of acids

а) titration of strong acids mixture

In this case no separation occurs. The sum of both acids can easily be determined.

б) titration of a mixture of one strong and one weak acids

The strongest acid is titrated first, resulting in the first endpoint.

в) titration of weak acids mixture

4243 POHHPOH

2

442 HPOHPOH

3

4

2

4 POHHPO

Example:

Mixtures of acids can be determined in one titration if there is a significant difference in pKa value. The strongest acid is titrated first, resulting in the first endpoint, the second endpoint can represent the weaker acid. If the difference in pK is too small no separate endpoints will be found, only the sum of the acids can be determined.

It can be viewed as a mixture of three acids

;105

;102,6

;107

13

3

8

2

3

1

K

K

K

Phosphoric acid reacts with water in steps and during the titration H3PO4 reacts with hydroxide in three steps

42243 POHOHOHPOH

2

4242 HPOOHOHPOH

3

42

2

4 POOHOHHPO

A titration could theoretically result in three equivalent points. The last one however cannot

be detected because of the low Ka number

phenolphthalein

Methyl orange

NaH2PO4

Na2HPO4

Na3PO4

H3PO4

Titration curve

1 Before adding the titrant, the pH is determined by the weak acid.

4343)( POHPOH cKHc

Adding NaOH converts a portion of the phosphoric acid to its salt NaH2PO4. Before the equivalence point, the pH is determined by a buffer containing the weak acid and its salt :

2

42

43

43)(

POH

POH

POHc

cKHc

3

2

21 pKpKpH

;36.12

;21.7

;12.2

3

2

1

pK

pK

pK 66.42

21.712.2pH

H3PO4 :

NaH2PO4

4 The second equivalent point will be determine be the hydrolysis of

The first equivalent point can be calculated as following:

Na2HPO4

79.92

36.1221.7

2

32 pKpKpH

This equivalent point can be detected using methyl orange

This equivalent point can be detected using phenolphthalein

5

)(lg2

1

2

1

2

14332

PONacpKpKpH OH

6.121.0lg2

136.12

2

17pH

At the third equivalence point, H3PO4 is converted completely into a weak salt Na3PO4. The pH is calculated by considering the salt hydrolysis reaction

.

This equivalent point however cannot be detected because of the low Ka number (Ka ≈ 10

-13)

•Low initial pH, means that analyt was acid. Breaks are small, that means analyte is weak acid. There are two breaks – acid is polyprotic.

•Phosphoric acid can be titrated as monoprotic acid in the presence of methyl orange and its equivalent is one n(H3PO4), and as diprotic in the presence of phenolphtaleine and in this case its equivalent is n( 1/2 H3PO4).

•Phosphoric acid cannot be titrated as triprotic acid, because its К3 is too low 10-13, at this step the acid is too weak.

Conclusions:

ANALYSIS OF SALTS

Acid –base titrimetry can be used for the quantitative analysis of many inorganic salts, formed by weak acid and strong base or by weak base and strong acid. A standard solution of NaOH can be used to determine the concentration of inorganic salts, such as NH4Cl, and inorganic salt, such as Na2CO3 can be analyzed using a standard solution of HCl.

EXAMPLES OF ACID-BASE DETERMINATIONS

A salt, formed by strong acid and weak base (with рКB), is titrated as a weak acid which pKa can be calculated using

the expression рКа = 14 - рКв.

A general rule is that only salts with pK < 7 ( or К ≥10-7) can be determined accurately in aqueous solution.

A salt, formed by strong base and weak acid (with рКа), is titrated as a weak base which pKB can be calculated using the

expression рКв = 14 - рКа.

In both cases the titration curves are similar.

In both cases the titration curves are similar.

Example:

NaCN can be titrated as a weak base with рКв = 14 - рКа = 14 – 9,2 = 4,8;

КB ≈ 10-5

but NH4Cl cannot be titrated as a weak acid with рКа = 14 - рКв = 14 – 4,76 = 9,24,

Ка ≈ 10-9.

TEST

Indicate which salt can be directly titrated in aqueous solution by 0,1 M HCl, (HCOOH (pKa = 3,75), CH3COOH (pKa=4,76), C6H5COOH (pKa = 4,20), HCN (pKa = 9,30))?

A. HCOONa;

B. KCN;

C. CH3COOK;

D. C6H5COONa;

E. CH3COONa.

Indicate which salt can be directly titrated in aqueous solution by 0,1 M HCl, (HCOOH (pKa = 3,75), CH3COOH (pKa=4,76), C6H5COOH (pKa = 4,20), HCN (pKa = 9,30))?

A. HCOONa; рКв = 14 - рКа = 14 – 3,75 = 10,25 Ка ≈ 10-10

B. CH3COOK; рКв = 14 - рКа = 14 – 4,76 = 9,24 Ка ≈ 10-9

C. KCN; рКв = 14 - рКа = 14 – 9,30 = 4,7 Ка ≈ 10-5

D. C6H5COONa; рКв = 14 - рКа = 14 – 4,20 = 9,8 Ка ≈ 10-10

E. CH3COONa. рКв = 14 - рКа = 14 – 4,76 = 9,24 Ка ≈ 10-9

Indicate which compounds can be directly titrated in aqueous solution:

A. CH3COOH (pKa = 4,76);

B. NH3 (pKb = 4,75);

C. CH3COONa;

D. amidopyrine (pKb ≈ 9);

E. NH4Cl.

Titration of Na2CO3

2232 22 COOHNaClHClCONa

рКв1 = 14 – рКа1 = 14 – 6,35 = 7,65

The sodium carbonate Na2CO3 can be titrated in two steps. The carbonate and bicarbonate are stable bases and can be titrated with acid. Using the expression pKa + pKB = pKw =14 the pKB can be calculated. So the pKB of carbonate is 3.68 and the pKB of bicarbonate is 7.65

рКв2 = 14 – рКа2 = 14 – 10,32 = 3,68

NaClNaHCOHClCONa 332

323 COHNaClHClNaHCO

A titration could result in two equivalent points and the titration curve has two breaks.

Before adding the titrant, the pH is determined by the salt hydrolysis.

60,111,0lg2

1

2

10,327)COlgc(Na

2

1pK

2

17pH 32

COH

a32

2

1

Adding HCl converts a portion of the sodium carbonate to sodium bicarbonate NaHCO3. Before the equivalence point, the pH is determined by a buffer containing two salts :

2

)(3)(32 formedrestNaHCOCONa

3

34,82

32,1035,6

2

pKpKpH

32

2

32

1

COH

a

COH

a

After the first equivalence point, the pH is determined by the following buffer :

4

)(32)(3 formedrest COHNaHCO

5

The first equivalent point can be determined by the sodium bicarbonate NaHCO3 hydrolysis :

The second equivalent point can be determined by the formed carbonic acid H2CO3. Once HCO3

- gets protonated, we will be left with solution of carbonic acid, with pH around 4.0. This is close to the pH at which methyl orange starts to change color.

Titration curve of Na2СO3

phenolphthalein

Methyl orange

ANALYSIS OF BASES

OHCONaCONaOH 23222

Strong bases (alkalies) like NaOH or KOH - both solid and dissolved - easily reacts with atmospheric carbon dioxide. That means they are usually contaminated with carbonates. It is not a problem to determine sum of hydroxide and carbonates concentration by titration with a strong acid (although presence of dioxide means end point detection can be a little bit tricky). However, quite often we can be interested in the amount of pure base in a sample or even carbon dioxide absorbed.

The determination of the bases can be performed in two ways. 1. One of the suggested procedures calls for addition

of barium chloride, and subsequent titration of base in the presence of BaCO3 precipitate.

2. Determination of mixtures of base and carbonate by double indicator method.

1. Method based on the gravimetric precipitation of carbonates by BaCl2

32322 BaCONaClBaClCONa

OHNaClHClNaOH2

This method is considered to be more precise

2. Double indicator method

OHNaClHClNaOH 2


323 COHNaClHClNaHCO

рН ≈ 13-14

рН ≈ 11-12 рН ≈ 8-9

рН ≈ 8-9 рН ≈ 4-5

рТph = 9

рТm/о = 4,0 V

1ph(H

Cl)

V2m/о(H

Cl)

Results in this method not as precise as in the case of gravimetric method, nonetheless method is very fast and simple and often accurate enough.

OHNaClHClNaOH 2


323 COHNaClHClNaHCOV

1ph(H

Cl) V

2м/о(H

Cl)

3221

)(1

/

2 )()(CONa

HCl

phоm VHClVHClV1.

323221

)()(2CONa

HCl

CONa

HCl VV2.

NaOH

HCl

CONa

HCl

оm VVHClV )()(

/

232)(3.

Calculations

4. )()( HClnNaOHn

)()( 3221 HClnCONan …….

titration curves...a ph electrode in a solution of 0.100 m hcl (the analized compound) and...

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