Acid-Base Titration Things to learn : - strong acid – strong base titration - weak acid – strong...

Acid-Base Acid-Base TitrationTitration

Things to learn :

- strong acid – strong base titration

- weak acid – strong base titration

- strong acid – weak base titration

- prediction of titration curve

- acid-base indicator

Strong Acid - Strong Acid - Strong Base Strong Base

TitrationTitrationIn strong acid – strong base titration, there are three regions of the titration curve that represent different kinds of calculations :

- before equivalence point

- at equivalence point

- after equivalence pointExample :

Consider the titration of 50.00 ml of 0.020 M KOH with 0.10 M HBr

KOH + HBr H2O + KBr

Before titration :

Moles of HO- present = (0.050 l)(0.020 mol/l)

= 0.001 moles

What happens when 3.00 ml of HBr is added ?No. of moles in 3.00 ml HBr = (0.003)(0.10)

= 0.0003

Moles of HO- unconsumed = 0.001 – 0.0003

= 0.007

Since the volume in the flask in now 53 ml,

[HO-] in the flask = (0.007 mol)/(0.053 l)

= 0.0132 M

pH of solution =-log [H+] = -log

= 12.12

Kw

[HO-]

What happens at the equivalence point ?

At the equivalence point, the H+ is just sufficient to react with all the HO- to form water. The pH is determined by the dissociation of water.

Since there is 0.001 mol HO- in the flask, 0.001 mol H+ must be added to reach equivalence point

Volume of H+ added = =10 ml

Let : x = [H+] = [HO-]

Kw = [H+] [HO-] = x2

x = 1.00 x 10-7

pH = 7.00

0.001 mol0.10 mol/l

pH at the equivalence point in any strong acid – strong base titration will be 7.00 at 25oCWhat happens after the equivalence point ?

When 10.50 ml of HBr is added to the solution :

moles of excess H+ = (0.0005 l)(0.10 mol/l)

= 0.00005 mol

Concentration of excess H+ =

=8.26 x 10-4 M

pH = -log (8.26 x 10-4 )

= 3.08

0.00005 mol0.0605 l

Weak Acid - Weak Acid - Strong Base Strong Base

TitrationTitrationStrong + weak complete reaction

In weak acid – strong base titration, the titration curve consists of four regions :

- before any base is added

HA H+ + A-

- before the equivalence point : solution consists of a mixture of unreacted HA and A-

- at the equivalence point : all the HA has been converted to A-

A- + H2O HA + HO-

- beyond the equivalence point : excess strong base is added and the pH of the solution is determined by the amount of strong base

Ka

Kb

Example :

Consider a 50.0 ml solution of 0.020 M HA with pKa = 6.15 which is treated with

0.10 M NaOH

Before the addition of NaOH :

HA H+ + A- Ka = 10-6.15

Let x = [H+] =[A-]

Ka = = =10-6.15

x = 1.19 x 10-4

pH = -log (1.19 x 10-4)

= 3.93

[H+][A-]

[HA]

x2

0.020 - x2

What happens when 3.0 ml of 0.10 M NaOH is added ?

When NaOH is added, a mixture of HA and A- is created a buffer

Moles of NaOH added = (0.003 l)(0.10 M)

= 0.0003

Concentration of A- = = 5.66 x 10-3 M

Moles of HA left = (0.050 l)(0.020 M) – 0.0003

=0.0007

Concentration of HA = =0.0132 M

Using the Henderson-Hasselbalch equation

pH = pKa + log = 6.15 + log

= 5.78

0.00030.053

0.00070.053

[A-][HA]

5.66 x 10-3

0.0132

What happens at the equivalence point ?

At the equivalence point, sufficient amount of NaOH has been added to react with all the HA

Moles of HA present = (0.050 l)(0.020 M)

= 0.0010

Volume of NaOH added = (0.0010)/(0.10 M)

=0.01 l = 10 ml

Concentration of A- = = 0.0167 M

HA + HO- H2O + A-

HA + HO-

Since Kw = KaKb Kb = =1.43 x 10-8

Let x = [HA] = [HO-]

0.0010 mol0.060 l

Kb

Ka

Kw

Ka

Kb = = =1.43 x 10-8

x = 1.54 x 10-5 M

Using Kw = [H+][HO-] = 1x 10-14

[H+] =

pH = -log[H+] = 9.18

The pH at the equivalence point in not 7.00.

In weak acid – strong base titration, the

pH at the equivalence point is always

higher than 7

[HO-][HA][A-]

x2

0.0167 - x

1x 10-14

1.54 x 10-5

What happens after the equivalence point ?

Now there is excess NaOH in the solution

Since NaOH is a strong base, we can say that the pH is determined by the concentration of the excess NaOH

When 10.10 ml of NaOH is added there is an excess of 0.10 ml of NaOH.

[HO-] = = 1.66 x 10-4 M

pH = -log[H+]

= -log

= 10.22

Kw

[HO-]

(0.10 M)(0.0001 l)

0.06010 l

Weak Base - Weak Base - Strong Acid Strong Acid

TitrationTitrationB + H+ BH+

Since it is a strong acid, therefore reaction goes essentially to completion

In weak base – strong acid titration, the titration curve consists of four regions :

- before any acid is added:

B + H2O BH+ + HO-

- before the equivalence point – solution is a buffer

B + HA BH+ + A-

BH+ + H2O B + H3O+

pH = pKa (for BH+) + log

Kb

Ka

[B][BH+]

- at the equivalence point

B + HA BH+ + A-

BH+ + H2O B + H3O+

pH is obtained by considering the acid dissociation

[BH+] original [B] because of dilution

Since the solution at equivalence point contains BH+ , thus the pH should be below 7- after the equivalence point

There is excess HA in the solution. Since HA is a strong acid, it determines the pH (contribution from the hydrolysis of BH+ is comparatively small and can be neglected)

Titration in Titration in Diprotic SystemsDiprotic Systems

Treatment is an extension from the monoprotic system

Example :

Consider the titration of 10 ml of 0.10 M base, B, with 0.10 M HCl. The base is

dibasic with pKb1 = 4.00 and pKb2 = 9.00.

Calculate the pH at each point along the titration curveBefore an acid is added :

B + H2O BH+ + HO-

Let x = [BH+] = [HO-] . Thus

Kb1 =

= = 1.00 x 10-4

x = 3.11 x 10-3

[H+] = pH=11.49

Kb1

[BH+] [HO-] [B]

x2 0.10 - x

Kw

[HO-]

When acid is added and before the first equivalence point, we have a buffer containing B and BH+

B + H+ BH+

BH+ + H+ BH22+

BH22+ BH+ + H+

BH+ B + H+

If 1.5 ml of HCl has been added :

Moles of HCl added = (0.0015 l)(0.10 M)

= 1.5 x 10-4

= moles of BH+ formed

[ BH+] = = 1.304 x 10-2 M

1.5 x 10-4

0.0115

Kb1

Kb2

Ka1

Ka2

[B] [BH+]

Moles of B left =(0.010)(0.10) – (1.5 x 10-

4)

=0.00085

[B] = = 7.39 x 10-2 M

Using the Henderson-Hasselbalch equation:

pH = pKa2 + log

= p( )+ log

= 10.00 + log (5.667)

= 10.75

0.00085

0.0115

Kw

Kb1

7.39 x 10-2

1.304 x 10-2

At the first equivalence point, all the B has been converted to BH+ which is both an acid and a base

Moles of B present = (0.010)(0.10) = 0.001

Volume of acid used = = 10 ml

[BH+] = =0.05 M

Using : [H+] =

where K1 = Ka1 and K2 = Ka2

[H+] =3.16 x10-8

pH = 7.50

K1K2[BH+] + K1Kw

K1 +[BH+]

0.001 mol0.10 M

0.001 mol0.020 l

At regions between the first and second equivalence points, a buffer containing BH+ and BH2

+ is formed :

BH+ + H+ BH2+

BH2+ BH+ + H+

When 15 ml of HCl is added :

Moles of BH22+ = (0.005 l)(0.10 M)

= 0.0005

[BH22+] = = 0.02 M

Moles of BH+ left = 0.001 – 0.0005

= 0.0005

[BH+] = = 0.02 M

Ka1

0.0005 mol0.025 l

0.0005 mol0.025 l

Using the Henderson-Hasselbalch equation:

pH = pKa1 + log

= 5.00 + log 1 = 5.00

[BH+][BH2

2+]

At the second equivalence point, all the BH+ has been converted to BH2

+

BH+ + H+ BH22+

BH22+ BH+ + H+

pH of the solution is determined by the acid dissociation

Moles of BH+ present at first equivalence point = 0.001

Volume of acid added between first and second equivalence point = (0.001 mol)/((0.10 M)

= 10 ml

Ka1

Moles of BH22+ = moles of BH+ = 0.001

[BH22+] = = 0.033 M

Ka1 =

=

x = 5.72 x 10-4

pH = -log (5.72 x 10-4) = 3.24

0.001 mol0.030 l

[BH+][H+][BH2

2+]

Kw

Kb2

x2

0.033 - x

Beyond the second equivalence point, the pH is determined by the excess HCl

If the total volume of HCl added is 25.0 ml,

Moles of excess HCl = (0.005 l)(0.10 M)

= 0.0005

[H+] = = 1.43 x 10-2 M

pH = 1.85

0.00050.035

(a) Titration of 10.0 ml of 0.100 M base (pKb1 = 4.00, pKb2 = 9.00) with 0.100 M HCl

(b) Titration of 10.0 ml of 0.100 M nicotine (pKb1 = 6.15, pKb2 = 10.85) with 0.100 M HCl

Finding the End Finding the End PointPoint

Titrations are commonly performed to determine :

- how much analyte is present

- the equilibrium constants of the analyte

How would one determine the end point ?(i) Autotitrator

- pH is measured by electrodes immersed in the analyte solution

pH/ V and (pH/ V)/ V are computed

- when pH/ V is maximum and (pH/ V)/ V = 0 end point

(ii) Indicator

An acid-base indicator is itself an acid or base whose various protonated species have different colours : eg phenolphthalein

Hln In- + H+

pH = pK1 + log

The indicator changes color over a pH range

Generally only one color is observed if the ratio of the concentrations of the two forms is 10:1

[In-][Hln]

acid color base color

So the pH in going from one color to the other has changed from (pK1 - 1) to (pK1

+ 1) most indicators require a transition range of about two pH units

The pH range over which the color changes is called the transition range

0.7< pH< 2.7 8.0< pH< 9.6

When only the color of the unionized form is seen :

=

pH = pK1 + log = pK1

- 1When only the color of the ionized form is seen :

= pH = pK1 + 1

110

110

[In-][Hln]

101

[In-][Hln]

Choosing an Choosing an IndicatorIndicator

An indicator with a color change near pH 5.54 would be useful in determining the end point of the titration.

The closer the point of color change is to pH 5.54 the more accurate will be the end point

The difference between the observed end point (when there is a color change) and the true equivalence point is called the indicator error

Use only a few drops of dilute indicator solution for each titration