Analytical chemistry (2) - (1st 2 course) · effect on chemical equilibrium. An example is the...

{1} University of Anbar

Stage 1st// Analytical Chemistry 2 Education College for Pure Sciences

Dr. Bashar Abdulazeez Mahmood Department of Chemistry

Analytical chemistry (2) - (1st stage – 2

nd course)

مفردات المنهج

Chemical equilibrium, ionic equilibrium (the solubility of precipitates)

A-The application of solubility products constant

The solubility products constant

The common ion effect, effect of p on solubility

B- Acid-base equilibrium pH calculation

1-Solution of strong acid and strong base

2- Solution of weak acid and weak base

3- Solution of salts of weak acid and base

4- Solution of salts of weak base and acid

C-Buffer solutions

1-Properties of buffer solutions

2-Preparation of buffer solutions

3-Calculation of buffer capacity

4-Calculation of pH buffer solutions

5-Types of buffer solutions

An introduction of volumetric analysis

عتمدالكتاب الم

“Fundamental of Analytical Chemistry” by Doglas A. Skooge, Donald M. West and James

Holler, Eighth Edition, 1988.

كتة مساعدة

“Analytical Chemistry” by Gary D. Christian, Purnendu K. (Sandy) Dasgupta, Kevin A.

Schug, 7th

Edition, 2014.

“An Introduction to Analytical Chemistry” by Doglas A. Skooge & Donald M. West, 4th Edi-

tion, 1986.

“Analytical Chemistry” by Gary D. Christian, John Wiley and Sons, Inc. 6th Edition, 2004.

6891 ، هادي كاظم عوض واخرون،والحجمً الوزنًالكمً التحلٍل ،الاساسٍاث النظرٌت للكٍمٍاء التحلٍلٍت اللاعضوٌت.

التحلٍلٍتمفتاح الابداع للكٍمٍاء Analytical Chemistry، ،8009الاولى، الطبعتعمر جبر حلوة.




Chemical equilibrium

The reaction used in analytical chemistry never result in complete conversion of reactants

to products. Instead, they proceed to a state of chemical equilibrium in which the ratio of con-

centration of reactant and products is constant.

Chemical equilibrium refers to each reaction that reaches equilibrium state. For example if

CO2 in atmosphere is increased, some of it is dissolved in water of seas, rivers and lakes until

equilibrium attained:

CO2 + H2O H2CO3

But, if CO2 in atmosphere is decreased, carbonic acid in water is decomposed to liberate

CO2 to the air:

H2CO3 CO2 + H2O

Another example is the dissolving of NH3 in water:

NH3 + H2O NH4OH NH4+ + OH

–

These reactions are called reversible reactions. Therefore, the reversible reaction is the re-

action which functions in opposite directions one to the forward and the other to the opposite as

shown by the above equations. Really, it was found that the reactions continue even after equi-

librium achievement. The observed constant concentration relationship is thus the sequence of

an equality in the rates of the forward and reverse reactions ; that is chemical equilibrium is a

dynamic state. Both reactions are under the same conditions. Two arrows are used to separate

between reactants and products.

Another types of reactions are called Irreversible reactions which work in one direction

especially to the products. These reactions involve the formation of precipitate or liberation of

gas, or formation of slightly ionized species.

These reactions are called one-way reactions and one arrow is used to separate products

from reactants.




Homogenous reactions: They are reactions which involve reactants and products in the same

phase: gas, liquid or solid

H2 Gas + Cl2 Gas 2HCl Gas

2KOH Solution + H2C2O4 Solution K2C2O4 Solution + 2H2O Solution

Na2CO3 Solid + BaSO4 Solid Na2SO4 Solid + BaCO3 Solid

Heterogeneous reactions: They are reactions in which the reactants and products are not in the

same phase; at least one of the materials is in different phase:

CaCO3 Solid CaO Solid + CO2 Gas

FeS Solid + 2HCl Liquid FeCl2 Liquid + H2S Gas

MgCl2(l) + 2NaOH(l) Mg(OH)2(s) + 2NaCl(l)

Equilibrium - constant expressions

Chemical equilibrium and rate of reactions are expressed and explained by the law of mass

action. Let us consider the following reaction:

A + B C + D

Forward reaction Reversed reaction

Where A, B, C and D are different substances.

According to the law of mass action, the rate of forward reaction is directly proportional to

the molar concentrations of the reactants:

)1......(].........][[11 BAKV

K1 is proportionality constant of forward reaction

The rate of reversible reaction is directly proportional to the molar concentrations of the

products:

)2......(].........][[22 DCKV

K2 is proportionality constant of reversible reaction

At equilibrium V1= V2 and thus:

)3.....(..........]][[

]][[

]][[]][[

2

1

21

BA

DCK

K

K

DCKBAK

eq

Keq = equilibrium constant of the reaction

Therefore, the law of mass action states that:




At equilibrium, the product concentrations of reversible reaction divided by the product of

concentrations of forward reaction each raised to the power indicated by a number of ions or

molecules in the reversible reaction is equal to constant called equilibrium constant (Keq) at

constant temperature.

Therefore, if we have the following general reaction:

aA + bB cC + dD ……..(4)

The expression of equilibrium constant is:

)5.....(..........][][

][][baeq

BA

DCK

dc

If one or more of the species in equation 5 is a pure liquid, a pure solid, or the solvent pre-

sent in excess, no term for this species appears in the equilibrium constant expression. For ex-

ample, if D in equation 4 is the solvent H2O, the equilibrium constant expression simplifies to:

baeqBA

CK

c

][][

][

Where the square – bracketed terms have the following meanings:

1. Molar concentration if the species is a dissolved solute.

2. Partial pressure in atmospheres if the species is a gas.

The following figure shows the route of reactants and products for the reaction

A + B C + D

Route of reversible reaction until reaching its equilibrium dynamic case.

Rat

e o

f r

eact

ion

Time

Reaction of A+B

Formation of C+D

Equilibrium case




Factors affecting the chemical equilibrium

1- Concentration change: if we consider the following equilibrium :

CH3COOH + H2O H3O+ + CH3COO

–

An increase in CH3COO–

or H3O+ will shift the equilibrium to the left, but Keq stays or

remains constant. Also, an increase in CH3COOH concentration will shift the equilibrium to the

right, but Keq stays constant. Very dilute solution of CH3COOH in water will get complete dis-

sociation of the acid. This case is called infinite dilution of the electrolyte which becomes simi-

lar to strong electrolyte.

2- Temperature change: if the temperature of a system in equilibrium is raised, the equilibri-

um will be shifted in the direction which absorbs heat. For gaseous equilibrium an increase in

pressure results in a shift toward the decrease in the total number of molecules.

Any change in temperature will change the equilibrium constant values. An example is the

reaction :

H2 + I2 + heat 2HI Endothermic reaction

An increase in temperature will push the reaction to the right to get more HI. In opposite,

if the temperature is decreased, the reaction will shift to the left and little HI is obtained.

3- Pressure change: If there is a change in volume to smaller, there is a clear effect to get the

products when high pressure is used such as procedure of Haber in preparation or manufacture

ammonia industrially:

N2 + 3H2 2NH3

1v 3v 2v

On other hand, if there is no change in volumes between reactants and products, there is no

effect on chemical equilibrium. An example is the oxidation of N2 by O2 to get NO:

N2 + O2 2NO

1v 1v 2v There is no change in volume

2v

Catalyst: the role of catalyst is to enhance the chemical reaction and has no effect on the equi-

librium constant. When it hastens the forward reaction it hastens the reversible reaction as well.

For example the following reaction :

2H2 + O2 2H2O




Stays as it is unless a catalyst of palladium is available which pushes the reaction to occur

and gets drops of water :

2H2 + O2 2H2O

How to solve any equilibrium problem?

1. Write balanced chemical reactions

2. Write equilibrium constant expressions

3. Write all mass balance expressions

4. Write the charge balance expression

5. Equations >= Chemical species solution possible

6. Make assumptions where possible

7. Calculate answer

8. Check validity of assumptions

Types of equilibrium constants in analytical chemistry

Table show summarizes the types of chemical equilibrium and equilibrium constants that

are importance in analytical chemistry

Type of equilibrium

Name and

symbol of

equilibrium

Typical example Equilibrium constant

expression

1. Dissociation of water Ion-product

constant , Kw

2H2O

H3O+ + OH

-

Kw=[H3O+][OH

-]

2. Heterogeneous equilibrium be-

tween a slightly soluble substance

and its ion in as saturated solution

Solubility

product , Ksp

BaSO4(s) Ba+2

+

SO4-2 Ksp=[Ba

+2][SO4

-2]

3. Dissociation of a weak acid and

base

Dissociation

constant Ka

or Kb

CH3COOH+H2O

H3O++ CH3COO

-

CH3COO- + H2O

CH3COOH + OH-

Ka=[H3O+][CH3COO

-]

/[CH3COOH]

Kb=[CH3COOH][OH-]

/[CH3COO-]

4. Formation of a complex ion Formation con-

stant βn

Ni+2

+ 4CN-

Ni(CN-)-2

4

β4=[Ni(CN)-2

4] /

[Ni+2

][CN]4

5.Oxidation reduction equilibrium Kredox

MnO4+5Fe+2

+ 8H+

Mn

+2 + 5Fe

+3 + 4H2O

Kredox=[Mn+2

][Fe+2

]5

/[MnO4-][Fe

+2]5[H]

8

6. Distribution equilibrium for a so-

lute between immiscible solvents Kd I2(aq) I2(org) Kd=[I2]org / [I2]aq




Applying solubility – product constants

When an aqueous solution is saturated with slightly soluble salt, an equilibrium is estab-

lished between the dissolved part of the salt and its ions in the solution:

AgCl Ag+ + Cl

–

][

][][

AgCl

ClAgKeq

[AgCl] is considered to be constant and equal =1

∴ Keq [AgCl] = Ksp = [Ag+] [Cl

–]

Where Ksp is the solubility product constant of AgCl which equals 1.2×10–10

at 25 oC.

Ba(IO3)2(s) Ba+2

(aq) + 2IO3(aq)

K = [Ba+2

] [IO-3]

2 / [Ba(IO3)2] solid

K[Ba(IO3)2(s) = Ksp = [Ba+2

][IO-3]

2

Where anew constant is called the solubility- product constant

The solubility of a precipitate in pure water

Example: How many grams of Ba(IO3)2 (487 g/mol) can be dissolved in 500 mL of water at 25

Co ? Ksp Ba(IO3)2 = 1.57×10

-9

Ba(IO3)2 Ba+2

+ 2IO-3

Ksp = [Ba+2

] [IO-3]

2

1.57×10-9

= [Ba+2

] [IO-3]

2

Molar solubility of Ba(IO-3) = Ba

+2 since two moles of iodate are produced for each mole

of barium ion, the iodate concentration is twice the barium ion concentration.

[IO3-] = 2[Ba

+2]

[Ba+2

] (2[Ba+2

]) 2 = 4[Ba

+2]

3 = 1.57×10

-9

[Ba+2

]3 = (1.57×10

-9/4)

1/3 = 7.32×10

-4 M

1mol Ba+2

is produced for every mole of Ba(IO3)2

Solubility = 7.32×10-4

M

10-4

w = 7.32×10-4

× 487×500/1000 = 0.178 g




The effect of a common ion on the solubility of a precipitate

Example: Calculate the molar solubility of Ba(IO3)2 in a solution that is 0.02 M in Ba(NO3)2.

Molar solubility = ½ [IO3-] of Ba(IO3)2

Ba(IO3)2 Ba+2

+ 2IO-3

[Ba+2

] = 0.02 + ½ IO-3

[0.02 + ½ IO3-] neglected [IO3

-]

2 = 1.57×10

-9

Ba+2

= [0.02] = 0.02 M

[0.02][IO-3]

2 = 1.57×10

-9 → [IO

-3] = 2.8×10

-6 M

Example: Calculate the solubility of Ba(IO3)2 in a solution prepared by mixing 200 mL of 0.01

M Ba(NO3)2 with 100 mL of 0.1 M NaIO3 .

M

№ mmol Ba+2

= 200 mL × 0.01 mmol/mL = 2.0

№ mmol IO-3 = 100 mL × 0.1 mmol/mL = 10

If the formation of Ba(IO3)2 is complete

№ mmol excess NaIO3 = 10-2.0 = 8.0

Thus [IO-3] = 8.0 mmol / (200+100) mL = 0.02 M.

Molar solubility of Ba(IO3)2 = Ba+2

IO-3 = 0.02 + 2Ba

+2 → IO

-3 = 0.02

Solubility of Ba(IO3)2 = [Ba+2

] = Ksp / [IO3-]

2

= 1.57×10-9

/ (0.02)2 = 3.93×10

-6 mol/L

Conjugate acids and bases

A conjugate base is formed when an acid loses a proton. For example, acetate ion is the

conjugated base of acetic acid, and ammonium ion is the conjugated acid of the base ammonia.

acid1 base1 + proton

Acid1, and base1, are conjugate acid /base pair

CH3COOH CH3COO- + H

+

conjugate base

base2 + proton acid2

NH3 + H2O NH4+ + OH

-

conjugate acid




When these process are combined the result is an acid/ base or neutralization reaction

acid1 + base2 base1 + acid2

A conjugate acid is formed when base accepts a proton. A substance acts as an acid only in

the presence of a base and vice versa.

Example: NH3 + H2O NH+

4 + OH-

base1 acid2 (conjugate) (conjugate)

(acid1) (base2)

Example: H2O + HNO2 H3O+ + NO

-2

base1 nitrous (conjugate) conjugate base2

acid2 acid1 (hydronium ion) (nitrite ion)

Example: NO-2 + H2O HNO2 + OH

-

base1 acid2 (conjugate) (conjugate)

(acid1) (base2)

Amphiprotic species

Species that possess both acidic and basic properties are amphiprotic. An example is dihy-

drogen phosphate ion H2PO-4 which behaves as a base in the presence of an acid (proton donor

such as H3O+).

H2PO-4 + H3O

+ H3PO4 + H2O

base1 acid2 acid1 base2

In the presence of a proton accepter such as hydroxide ion, however H2PO-4 behaves as an

acid and donates a proton to form the conjugate base HPO-2

4

H2PO-4 + OH

- HPO

-24 + H2O

acid1 base2 base1 acid2

(conjugate base)

Here example for amphiprotic compound that contain both a weak acid and a weak base

functional group, when dissolve amino acid like glycine in water will produce a zwitter ion a

species that bears both a positive and a negative charge. Thus

NH2CH2COOH NH+

3CH2COO-

(glycine) (zwitter ion)

Amphiprotic solvents

Addition to the water there are other solvents acts as amphiprotic such as methanol, etha-

nol, and anhydrous acetic acid




NH3 + CH3OH NH+

4 + CH3O-

base1 acid2 conjugate conjugate

acid1 base2

CH3OH + HNO2 CH3OH+

2 + NO-2

base1 acid2 conjugate conjugate

acid1 base2

Autoprotolysis (also called auto ionization): involves the spontaneous reaction of molecules of

a substance to give a pair of ions.

Examples: base1 + acid2 acid1 + base2

H2O + H2O H3O+ + OH

-

CH3OH + CH3OH CH3OH+

2 + CH3O-

HCOOH + HCOOH HCOOH+

2 + HCOO-

NH3 + NH3 NH+

4 + NH-2

Strengths of acids and bases

The strong acids and bases reacting with solvent completely that no undissociated solute

molecules are left in aqueous solutions. Such as HClO4, HCl for acids and NaOH , KOH for ba-

ses . But the weaker acids and bases dissociated partially in water. Figure shows Dissociation

reactions and relative strengths of some common acids and their conjugate bases.

HClO4 + H2O H3O+ + ClO

-4

HCl + H2O H3O+ + Cl

-

H3PO4 + H2O H3O+ + H2PO

-4

Al(H2O)6 +3

+ H2O H3O+ + AlOH(H2O)5

+2

HC2H3O2 + H2O H3O+ + C2H3O

-2

H2PO-4 + H2O H3O

+ + HPO

2-4

NH+

4 + H2O H3O+ + NH3

Applying acid-base dissociation constants

When a weak acid or a weak base is dissolved in water, partial dissociation occurs – Thus

for nitrous acid, we can write

HNO2 + H2O H3O+ + NO

-2

Ka = [H3O+] [NO

-2] / [HNO2]

Where Ka is the acid dissociation constant for nitrous acid

Strongest

acid

Weakest

base

Weakest

acid

Strongest

base




The base dissociation constant for ammonia is

NH3 + H2O NH4+ + OH

-

Kb = [NH4+] [OH

-]/ [NH3]

The [H2O] does not appear in the equation because the concentration of water is so large

relative to the concentration of the weak acid or base.

Dissociation constants for conjugate acid/base pairs

Consider the base dissociation – constant expression for ammonia and the acid dissocia-

tion – constant expression for its conjugate acid, ammonium ion.

NH3 + H2O NH4+ + OH

- Kb = [NH4

+] [OH

-]/ [NH3]

NH4+ + H2O NH3 + H3O

+ Ka = [NH3] [H3O

+]/ [HN4

+]

Multiplication of one constant expression by other gives.

KaKb = [NH3] [H3O+]/ [NH4

+] × [NH4

+] [OH

-]/ [NH3] = [H3O

+] [OH

-]

Kw = [H3O+] [OH

-]

Kw = KaKb = 10-14

Example: The dissociation constant for ammonium ion (acid) is 5.7×10-10

. Find the Kb for

NH3?

NH3 + H2O NH4+ + OH

-

Kb = Kw/Ka = 10-14

/ 5.7×10-10

= 1.75×10-5

Relative strengths of conjugate acid/base pairs

Example: what is Kb for the equilibrium?

CN- + H2O HCN + OH

-

Ka HCN = 6.2×10-10

, thus

Kb = Kw / Ka = 10-14

/ 6.2×10-10

= 1.61×10-5

Hydronium ion concentration of solutions of weak acids

When the weak acid HA is dissolved in water, two equilibrium are established that yield

hydronium ions.

HA + H2O H3O+ + A

- Ka = [H3O

+] [A

-]/[HA]

2H2O H3O+ + OH

- Kw = [H3O

+] [OH

-]

[A-]= [H3O

+]




The sum of the molar concentration of the weak acid and its conjugate base must equal the

analytical concentration of the acid HA because the solution contains no other source of A-

ions. Thus.

CHA = [A-] + [HA]

Substituting [H3O+] = [A

-]

CHA = [H3O+] + [HA]

Which rearranges to

[HA] = CHA – [H3O+]

When [A-] and [HA] are replaced by their equivalent terms the equilibrium – constant ex-

pression becomes

Ka = [H3O+]

2/ CHA – [H3O

+]

Which rearranges to:

CHA – [H3O+] = CHA

Ka = [H3O+]

2 / CHA

[H3O+] = (Ka × CHA)

1/2

Example: Calculate the hydronium ion concentration in 0.12 M nitrous acid (Ka = 7.1×10-4

).

The principal equilibrium is:

HNO2 + H2O H3O+ +NO

-2

Ka = [H3O+] [NO

-2] / [HNO2] = 7.1×10

-4

NO-2 = H3O

+

[HNO2] = 0.12 – [H3O+]

Ka = [H3O+]

2 / 0.12 – [H3O

+] = 7.1×10

-4

[H3O+] Neglected >> 0.12

= [H3O+]

2 / 0.12 = 7.1×10

-4

[H3O+] = (0.12 × 7.1×10

-4)

1/2 = 9.2 × 10

-3 M

Hydronium ion concentration of solutions of weak bases

NH3 + H2O NH+

4 + OH-

Kb = [NH4+] [OH

-] / [NH3]

Example: Calculate the hydroxide ion concentration of a 0.075 M NH3 solution (Ka=5.7×10-10

).

The predominant equilibrium is:




NH3 + H2O NH4+ + OH

-

Kb = [NH4+] [OH

-] / [NH3]

Kw = KaKb

Kb= Kw/Ka

= 10-14

/ 5.7×10-10

= 1.75×10-5

The chemical equation shows that [NH4+] = [OH

-]

Both NH4+ and NH3 come from the 0.075 M solution, thus.

[NH4+] + [NH3] CNH3 = 0.075

If we substitute [OH-] for [NH4

+] in the second of these equations and rearrange, we find:

[NH3] = 0.075 – [OH-]

Kb = [OH-]

2 / 7.5×10

-2 – [OH

-] = 1.75 × 10

-5

[OH-] >> 7.5×10

-2

Kb = [OH-]

2 / 7.5×10

-2 = 1.75×10

-5

[OH-] = (7.5×10

-2 × 1.75×10

-5)

1/2 = 1.15×10

-3 M

Example: calculate the hydroxide Ion concentration in a 0.01 M sodium hypochlorite solution

(where Ka = 3×10-8

).

The equilibrium between OCl- and water is:

OCl- + H2O HOCl + OH

-

Kb = [HOCl] [OH-] / [OCl

-],

Kb = Kw/Ka = 10-14

/ 3×10-8

= 3.33×10-7

[OH-] = [HOCl]

[OCl-] + [HOCl] = 0.01, OCl

- = 0.01 – [OH

-] = 0.0l

[OH-] >> 0.01

[OH-]

2/0.01 = 3.33×10

-7, [OH

-] = 5.8×10

-5 M

Ionization of weak acids and weak bases

CH3COOH + H2O H3O+ + CH3COO

–

]][[

]O][HCOOCH [

32

3

-

3

COOHCHOHKeq

[H2O] is constant




][

]O][HCOOCH [][

3

3

-

3

2COOHCH

KOHK aeq

Where Ka the ionization constant of acetic acid which =1.75×10–5

at 25 oC.

NH3 + H2O NH4 OH NH4+ + OH

–

]][[

]][NHOH [

32

4

-

NHOHKeq

But [H2O] is constant.

][

]OH ][[NH][

3

-

42

NHKbOHKeq

Where Kb is ionization constant of ammonia which =1.75×10-5

at 25 oC.

Note: if the concentration of an acid or base is much less than 10-7

M, then its contribution to

the acidity or basicity will be negligible compared with the contribution from water. The pH of

a 10-8

M sodium hydroxide solution would therefore not differ significantly from 7. If the con-

centration of the acid or base is around 10-7

M, then its contribution is not negligible and neither

is that from water, hence the sum of the two contributions must be taken.

Example: Calculate the pH and pOH of a 1.010-3

M solution acetic acid, Ka= 1.75 10-5

?

[ ][ ]

[ ]

If CHA > 100 Ka, x can be neglected compared to CHA

[ ]

[ ]




[ ]

[ ] √

Example: The basicity constant kb for ammonia is 1.75 10-5

at 25 C, (H is only coincidental

that this is equal to Ka for acetic acid). Calculate the pH and pOH for a 1.0 10-3

M solution of

ammonia?

[

][ ]

[ ]

√ [ ]

[ ]

General law to calculate [OH-] for weak bases:

[ ] √

Example: Calculate the pH of a 0.1 M solution of CH3COONa, Ka=1.75 10-5

?

[ ][ ]

[ ]

kb

[ ]

[ ]

For salt of weak acids:[ ] √

√

Example: Calculate the pH of a 0.25 M solution of ammonium chloride NH4Cl?




[ ][ ]

[ ]

For salt of weak bases:[ ] √

Buffer solution

It is a solution consisting of a mixture of a weak acid and its salt or a weak base and its salt

that resists change in pH when a small amount of an acid or base is added or when the solution

is diluted.

A buffer solution that is widely used in the biological or clinical laboratory and in bio-

chemical studies and in the physiological pH range is that prepared from tris (hydroxymethyl)

aminomethane (HOCH2)3CNH2, and its conjugate acid (the amino group is protonated).

Many biological reactions of interest occur in the pH range (6-8). A number, particularly

specific enzyme reactions that might be used for analyses, may occur in the pH range (4-10).

One example of a buffer solution found in nature is blood. The normal pH of human

blood is 7.4.

One useful series of buffers are phosphate buffers (H3PO4/H2PO4-) or (H2PO4

-/HPO4

-2) or

(HPO4-2

/PO4-3

). Many reactions occur inside the living cells at certain pH values. Also, many chemical re-

actions such as precipitation, oxidation-reduction reactions and complex formations reactions

are not occurred unless the pH is stabilized at certain values. These pH values are determined

experimentally. Many household and cosmetic products need to control their pH values such as

shampoo to counteract the alkalinity of the soap and prevent irritation, baby lotion to maintain a

pH of about 6 to prevent bacteria multiplying, washing powder, eye drops, fizzy lemonade ..etc

Common ion: It is the ion which is liberated from weak and strong electrolytes and its role is

to reduce the ionization of the weak electrolyte.

Example:




Rearrangement of equation (3), we get:

[CH3COOH] is approximately equal the concentration of original acid because the ionized

species is very little because of the effect of common ion (CH3COO -).

Also, [CH3COO-] is equal to the concentration of the salt (CH3COONa).

Therefore, [CH3COOH] = Ca ≈ the concentration of acid and [CH3COO-] = Cs ≈ the concentra-

tion of salt.

Cs

CaKH a loglog]log[

Cs

CapKpH a log

or

Ca

CspKpH a log

or

][

][log

acid

saltpkapH

Equation (5) is used to calculate pH solutions of weak acid and its salt.

Another example: Derivation the pH solution of weak base and its salt.




Rearrangement of equation (3), we get:

[NH4OH] the concentration of original base = Cb.

[NH4+] the concentration of the salt = Cs.

s

b

bC

CKOH loglog]log[

s

bb

C

CpKpOH log or

b

sb

C

CpKpOH log

[ ]

[ ]

[base]

[salt]logpK14pH b

Equations (5) and (6) are used to calculate the pH of buffer solutions formed from weak

base and its salt.

Note: The high concentrated solutions of acids and bases are considered buffer solutions.

Therefore, 1 M NaOH is buffer solution because it resists the pH change when little amounts of

strong acid or strong base are added. But their pH is changed by dilution by addition of water

which changes pH value. Table contains some standard buffer solutions.

Table show a group of standard buffer solutions.

№ Buffer solution pH value

1 0.01 M potassium tetra oxalate 1.48

2 0.01 M potassium dihydrogen citrate 3.72

3 0.01 M CH3COOH + 0.01 M CH3COONa 4.64

4 0.01 M K2HPO4 + 0.01 M Na2HPO4 6.85

5 0.05 M Na2B4O7.10H2O 9.18

6 0.1 M NH4OH + 0.1 M NH4Cl 9.26

7 0.025 M NaHCO3 + 0.025 M Na2CO3 10.00

8 0.01 M Na3PO4 11.72




Calculation of pH buffer solutions

Example: Calculate the pH of a buffer prepared by adding 10 mL of 0.1 M acetic acid to 20 mL

of 0.1 M sodium acetate?

Henderson-Hasselbalch equation

[ ]

[ ]

[ ]

[ ]

[ ]

[ ]

[ ]

[ ]

We can use millimoles of acids and salt in place of molarity.

Example: Calculate the pH of a solution prepared by adding 25 mL of 0.1 M sodium hydroxide

to 30 mL of 0.2 M acetic acid (this would actually be a step in a typical titration?

these react as follows:-

]

[ ]

[ ]

The amount of acid or base that can be added without causing a large change in pH is gov-

erned by the buffering capacity (B), the buffering capacity increases with the concentration of

the buffering species. The buffering capacity is maximum at pH= pKa.

Example: A buffer solution is 0.2 M of acetic acid and sodium acetate. Calculate the change in

pH upon adding 1.0 mL of 0.1 M hydrochloric acid to 10 mL of this solution?

[ ] [ ]

[ ] [ ]




[ ] [ ]

[ ] [ ]

[ ]

[ ]

The change in pH is -0.05. This is rather small especially if we consider that had the HCl

been added to unbuffered neutral solution the final concentration would have been approxi-

mately 10-2

M, and the pH would be 2.0.

Example: Calculate the volume of concentrated ammonia and the weight of ammonium chlo-

ride you would have taken to prepare 100 mL of a buffer at pH 10.0 if the final concentration of

salt is to be 0.2 M?

[ ]

[ ]

[ ]

[ ]

[ ]

[ ]

[ ]

[ ]




Example: How many grams ammonium chloride and how many milliliters 3.0 M sodium hy-

droxide should be added to 200 mL water and diluted to 500 mL to prepare a buffer of pH 9.50

with salt concentrated of 0.10 M?

[ ]

[ ]

[

]

[ ]

[

]

[ ]

[

]

[ ]

[ ]⇒ [ ]

The NH3 formed from an equal number of millimoles of NH4Cl. Therefore, 50 + 90 = 140

mmol NH4Cl must be taken.

The millimoles of NaOH needed are equal to the millimoles of NH3:

Example: Calculate pH of buffer solution formed by adding 10 mL of 0.1M CH3COOH

to 10 mL of 0.1 M CH3COONa. Also, calculate the change in pH when 1 mL of 0.1 M HCl and

1 mL of 0.1 M NaOH is added separately to this mixture of buffer solution.

Concentrations of the acid and its salt are reduced to 0.05 M as a result of dilution effect.

∴ [CH3COOH] = 0.05 M and [CH3COONa] = 0.05 M.

[ ]

[ ]

74.4074.405.0

05.0log74.4 pH

When 1mL of 0.1 M HCl is added to this buffer:




∴ M HCl = 0.005 M

Therefore, the added acid will increase the concentration of unionized CH3COOH accord-

ing to Lechatelie principle, but decrease the concentration of the salt as a result of combination

of [H+] with [CH3COO

-] to form CH3COOH.

M CH3COOH

[CH3COOH] = 0.05 + 0.005 = 0.055 M

[CH3COONa] = 0.05 - 0.005 = 0.045 M

65.409.074.4055.0

045.0log74.4 pH

Therefore, the pH value is reduced by the unit of 0.09 by addition of 1mL of 0.1 M HCl

which is very little change. Now, when 1mL of 0.1 M NaOH is added.

M NaOH = 0.005 M

OH – is combined with H

+ to form water.

Therefore, the equilibrium (1) will be replaced to the right with increasing the ionization

of CH3COOH. Thus, a decrease of [CH3COOH] occurs and increase in [CH3COONa] as a re-

sult of addition more acetate to the solution.

[CH3COOH] = 0.05 - 0.005 = 0.045 M

[CH3COONa] = 0.05 + 0.005 = 0.055 M

83.409.074.4045.0

055.0log74.4 pH

Therefore, a little increase in pH value is occurs which equals 0.09 units.




Example: A buffer solution of 0.1 M NH4OH and 0.08 M of NH4Cl its volume = 100 mL

Calculate the pH of this solution. Also, calculate the change in pH when 1mL of 1 M HCl and

1mL of 1 M NaOH is added separately to this solution

.

Solution:

[base]

[salt]logpKpOH b

64.41.074.41.0

08.0log74.4 pOH

∴ pH = 14 – 4.64 = 9.36

Now, when 1mL of 1M HCl is added:

M HCl = 10-2

M

H+ will combine with OH

– to form water and the equilibrium (1) is replaced to right with

ionization of new amount of NH4OH. Therefore, there will be a decrease in [NH4OH] which

equals = 0.1 - 0.01 = 0.09 M.

On other hand, there will be an increase in salt of [NH4Cl]:

0.08 + 0.01 = 0.09 M

74.409.0

09.0log74.4 pOH

∴ pH = 14 – 4.74 = 9.26

Therefore, a little decrease in pH occurs when 1mL of 1 M HCl is added.

When 1mL of 1M NaOH is added:

M NaOH = 10-2

M

Therefore, OH–

combines with NH4+ to form NH4OH accompanied with increase in

[NH4OH] and decrease in [NH4Cl] .

[NH4OH] = 0.1 + 0.01 = 0.11 M




[NH4Cl] = 0.08 – 0.01 = 0.07 M

54.42.074.411.0

07.0log74.4 pOH

pH = 14 – 4.54 = 9.46

Here, a little increase in pH value occurs which confirms approximately the constant value

of pH of buffer solution.

Calculation of buffer capacity

Buffer Capacity

It is the number of moles or mmoles of a strong base (y), which when added to one liter of

a buffer raises his own pH by one unit , or it is the number of moles or mmoles of a strong

acid (x) which, when added to one liter of this buffer reduced its pH by one unit. The higher the

capacity the higher the amounts of a strong acid or a strong base which can be added to the

buffer without significantly changing his own pH. The capacity of a buffer solution can be in-

creased by increasing both Cs , Ca and Cb and it will be at maximum when Cs = Ca or Cs = Cb

and in this case :

pH = pKa + 0 Or pOH = pKb + 0

Example: Calculate the buffer capacity of a buffer solution containing 0.2 M NH3 and 0.1 M

NH4Cl [ pKb (NH3) = 4.76 ] ?

Solution: First: we calculate the pH of the buffer :

[ ]

[ ]

5.95.414 pH

Second : we calculate the buffer capacity by one of two ways:

(1) according to the above buffer capacity definition , suppose that x moles of strong acid such

as HCl (buffer capacity) have been added to one liter of the buffer solution. HCl will covert the

base NH3 to salt (NH4+), so, the base will decrease and the salt will increase by the same num-

ber of HCl moles (all reactions are 1:1) and the pH of the buffer will decrease by one to become

8.5 instead of 9.5 (or the pOH will increase by one to become 5.5 instead of 4.5) thus:




(2) Suppose that Y moles of strong base such as NaOH (buffer capacity) have been added to

one liter of the buffer solution . NaOH will covert the salt (NH4+) into base NH3, so, the base

will increase and the salt will decrease by the same number of NaOH moles (all reactions are

1:1) and the pH of the buffer will increase by one to be-

come 10.5 instead of 9.5 (or the pOH will decrease by one

to become 3.5 instead of 4.5) thus:

Types of buffer solutions

1- Acidic buffer solutions

An acidic buffer solution is simply one which has a pH less than 7. Acidic buffer solutions

are commonly made from a weak acid and one of its salts - often a sodium salt.

A common example would be a mixture of acetic acid and sodium acetate in solution. You

can change the pH of the buffer solution by changing the ratio of acid to salt, or by choosing a

different acid and one of its salts.

2- Alkaline (basic) buffer solutions

An alkaline buffer solution has a pH greater than 7. Alkaline buffer solutions are common-

ly made from a weak base and one of its salts. A frequently used example is a mixture of am-

monia solution and ammonium chloride solution. A buffer solution has to contain things which

will remove any hydrogen ions or hydroxide ions that you might add to it - otherwise the pH

will change. Acidic and alkaline buffer solutions achieve this in different ways.

Polyprotic acids and their salts

[ ][ ]

[ ]

[ ][

]

[ ]

[ ][

]

[ ]




[ ] [ ]

[ ]

The stepwise Ka values of polyprotic acids get progressively smaller as the increased nega-

tive charge makes dissociation of the next proton more difficult. We can titrate the first two

protons of H3PO4 separately. The third is too weak to titrate.

Example: The pH of blood is 7.40. What is the ratio of [HPO42-

]/ [H2PO4-] in the blood (assume

25 C) ?

[ ]

[ ]

[ ]

[ ]

[

]

[ ]

[

]

[ ]

[

]

[ ]

[

]

[ ]

Salts of polyprotic (acids or bases)

1. Amphoteric salts: -

H2PO4- possesses both acidic and basic properties. That is, it is amphoteric. It ionizes as a

weak acid and it is a Bronsted base that hydrolyzes:-

[ ][

]

[ ]

[ ][ ]

[ ]

2. Unprotonated salt in a fairly strong Bronsted base in solution and ionizes as follows:-

Example: The total carbon dioxide content (HCO3- + CO2) in a blood sample is determined by

acidifying the sample and measuring the volume of CO2 evolved with a Van Slyke manometry

apparatus. The total concentration was determined to be 28.5 mmol/L. The blood pH at 37 C

was determined to be 7.48. What are the concentration of HCO3- and CO2 in the blood?




[

]

[ ]

[

]

[ ]⇒

[ ]

[ ]

[

]

[ ]⇒

[ ]

[ ]

[ ] [ ]

[ ] [ ]

[ ] [ ]

[ ] ⇒ [ ]

[ ]

H. W.:

1- Calculate the pH of 0.1 M KHCO3.

2- Calculate the pH of the following buffer solutions:

a) 0.0405 M NH4OH + 0.02 M (NH4)2SO4 .

b) 0.0176 M phenol + 0.0254 M sodium phenolate.

c) 1 M Cl3CCOOH + 0.5 M Cl3CCOONa .

d) 0.164 M ethylamine + 0.272 M ethylamine hydrochloride.

3- Calculate the ratio of salt/acid or salt/base for the following buffer solutions which give

pH = 9.8 :

a) NH3, NH4Cl.

b) CH3NH2, CH3NH3Cl.

c) HCN, NaCN.

d) HOCl, NaOCl.

4- A liter of solution contains 0.25 M HCOOH and 0.3 M HCOONa, calculate the pH of

this solution. Calculate the change of pH of this solution when 10 mL of 0.1 M HCl and

10 mL of 0.1 M NaOH are added separately to this solution.

5- A mixture of 0.123 M NH4Cl and 0.246 M NH4OH has volume of 20 mL, calculate its

pH value. 1 mL of 1 M HCl and 1mL of 1 M NaOH are added separately to this solu-

tion. Calculate the change in pH value.




Volumetric analysis (titration analysis)

Are the most useful and accurate analytical techniques, especially for millimole amounts

of analyte. They are rapid and can be automated, and they can be applied to smaller amounts of

analyte when combined with a sensitive instrumental technique for detecting the completion of

the titration reaction, for example, pH measurement.

In a titration the test substance (analyte) in a flash react with a reagent added from a bu-

rette as a solution of known concentration.

This is referred to as a standard solution and is called the titrant. The volume of titrant re-

quired to just completely react with the analyte is measured. Since we know the concentration

as well as the reaction between the analyte and the reagent, we can calculate the amount of ana-

lyte.

Titration : It is gradual addition of titrant which is always standard solution to the titrand which

is always the analyte (the solution under test or unknown), in the presence of certain indicator.

The titration is finished by the sudden change of indicator color.

Titrant : It is the solution which is contained in the burette and it is always the standard solu-

tion, but in some cases it refers to the analyte or unknown.

Titrand : It is the solution which is always contained in a conical flask and it is always the ana-

lyte, but sometimes it refers to standard solution. The burette and pipette and volumetric flasks

are used to measure and transfer the exact volumes of solutions. Beakers, conical flasks and

cylinders are used to measure or transfer the approximate volumes of solutions.




Standard and standardized solutions

Standard solution: A solution is prepared by dissolving an accurately weighed quantity of a

highly pure material called a primary standard by direct weighing.and diluting to an accurately

known volume in a volumetric flask.

Standardized solution is a solution of approximate concentration which can be known ex-

actly by standardizing it with standard solution. Such as preparation of approximately 0.1 M or

0.1 N HCl and standardizing it with standard solution of Na2CO3 or Borax

When the standardized solution is prepared and standardized, its properties become identi-

cal to the properties of standard solution.

Characteristics of standard solution

1- Its concentration remains constant for months or years, or at least within the period of ti-

tration.

2- It rapidly reacts with the analyte and the reaction is complete within the period of the ex-

periment.

3- Its reaction with the analyte can be expressed as balanced equation in order to get the exact

weight of the analyte.

4- A sudden change of the reaction should occur in order to identify the equivalence point of

the reaction by suitable chemical indicator.

Examples: solutions of oxalic acid (0.1 N), sodium carbonate (0.1 N), sodium chloride (0.1 N)

and borax (0.1 N).

Primary standard materials

It is a material or chemical of high purity and characterized by the following requirements:

1- Its purity should not be less than 99.5%, otherwise purification methods should be avail-

able to confirm its purity.

2- It should be stable and not be hydrated or efflorescent.

3- It can be easily obtained and not expensive.

4- It is preferred to have high equivalent weight. For example, if we compare the equivalent

weights of Na2CO3 (53) and borax (191), we find that the equivalent w of borax is four times




larger than sodium carbonate. If we want to prepare 0.1 N of both solutions we should use:

1.325 g Na2CO3 and 4.775 g borax.

If an error of 0.02 g is occurs in weighing, therefore the percentages of error equal:

respectively.

Therefore, the percentage of error with Na2CO3 is four times higher than borax. As the

weight is increased the percentage of error is decreased.

A high formula weight means a larger weight must be taken for a given number of moles.

This reduces the error in weighing.

5- The primary standard material is easily soluble in water or the applicable solvent.

A solution standardized by titrating a primary standard is itself a secondary standard. It

will be less accurate than a primary standard solution due to the errors of titrations.

The requirements of a titration

(1) The reaction must be stoichiometric: That is, there must be a well defined and known reac-

tion between the analyte and the titrant.

(2) The reaction should be rapid. Most ionic reactions.

(3) There should be no side reaction, and the reaction should be specific.

(4) There should be a marked change in some property of the solution when the reaction is

complete. This may be a change in color of the solution or in some electrical or other

physical property of the solution (by used indicator or pH meter).

(5) The point at which an equivalent or stoichiometric amount of titrant is added is called the

equivalence point. The point at which the reaction is observed to be complete is called the

end point, that is, when a change in some property of the solution.

(6) The reaction should be quantitative. That is, the equilibrium of the reaction should be far to

the right so that a sufficiently sharp change will occur at the end point to obtain the desired

accuracy. The equivalence point is the theoretical end of the titration where the number of

moles of titrant = number of moles of analyte. The end point is the observed end of the ti-

tration.




Chemical calculations of volumetric titrations

Chemical calculations which associate with titration process in volumetric analysis are

very significant since they illustrate the idea about the accuracy of titration data which refer to

the precision of measurements that lead to accurate results.

The reaction in volumetric analysis should be rapid and complete. The reactions are ex-

pressed in balanced equations and from these reactions one can know the ratio of reactants

since they involve equivalents of the reactants at equivalence point.

Normal concentration gives simple and rapid calculations because they involve equivalent

amounts of the reactants.

Molar concentrations require some attention and recognition.

For example, if NaOH solution is titrated with HCl solution, there is no difference in using

normality or molarity because the reaction is performed in 1:1 ratio between NaOH and HCl

NaOH + HCl NaCl + H2O

But the case is different in titration of Na2CO3 solution with HCl solution.

Na2CO3 + 2HCl 2NaCl + CO2 + H2O

Where two moles of HCl react with one mole of Na2CO3. This means that the strength of

Na2CO3 is twice of HCl. If the molar concentration is used, a precaution should be given to the

factor 2 in HCl.

This equality is incorrect if molar concentration is used. Therefore, a factor of 2 should be

used at the left side of the above reaction in order to express correctly the reaction between

Na2CO3 and HCl.

But if the normal concentration is employed for both solutions, the following reaction is

correct because the reaction is performed on the basis of equivalents or milliequivalents.

The same treatment is considered in precipitation, oxidation-reduction and complex for-

mation titrations.

Calculations of acid-base or neutralization titrations (one of the products is water).




The solid material may be primary standard material, therefore, the prepared solution is

standard. If the solid material is not primary standard, the prepared solution is not standard (has

an approximate concentration).

Example: Show by calculation how could you prepare 250 mL of 0.1 N Na2CO3 from the sol-

id primary standard of Na2CO3.

Number of equivalents of solid Na2CO3 = Number of equivalents of Na2CO3 in solution.

№ of milliequivalents of solid Na2CO3 = № of milliequivalents of Na2CO3 in solution.

№ of meq. of solid Na2CO3 = № of meq. of Na2CO3 in solution.

Therefore, 1.325 g of Na2CO3 is exactly weighed by sensitive balance and dissolved in

250 mL of solution in 250 mL size volumetric flask to get 0.1 N of Na2CO3 solution. This solu-

tion is a standard solution which is prepared from high purity of solid Na2CO3.

Example: Show by calculation how could you prepare 500 mL of 0.1 N H2SO4 from its con-

centrated solution has density of 1.84 g/mL and percentage of acid equals 98% (w/w).

Solution:

w of H2SO4 in liter of solution = 1000× 1.84× 0.98

=36.8 eq/L Or meq/mL

N1V1 = N2V2 36.8 × V1 = 500 × 0.1




Example: Calculate the weight of H2SO4 in 5 liters if 25 mL of this solution requires 22.5

mL of 0.095 N KOH.

Solution:

№ of meq. of KOH = № of meq. of H2SO4

= 0.0855 × 49 × 5 = 20.9475 g

Example: A solution of Na2CO3 contains 795 mg per liter of solution. Calculate the nor-

mality of this solution. What is the volume of H2SO4 of 0.1 N that equivalent to 10 mL of

Na2CO3 solution.

Solution:

Example: 30 g of KHC2O4.H2C2O4.2H2O is dissolved in distilled water and completed to li-

ter. 40 mL of this solution is titrated with KOH which required 20 mL Calculate the normali-

ty of KOH.

Solution:

meq. of this acidic solution = meq. of KOH

Example:10 mL of vinegar has density of 1.055 g/mL and requires 39.82 mL of 0.225 N of a




base to reach equivalence point. Calculate the percentage of acetic acid in vinegar (w/w).

(

)

(

)

Example: A 0.471 g sample containing sodium bicarbonate was dissolved and titrated with

standard 0.1067 M hydrochloric acid solution, requiring 40.72 mL The reaction is:

Calculate the percent sodium bicarbonate in the sample.

Solution:

Example: A 0.2638 g soda ash sample is analyzed by titrating the sodium carbonate with the

standard 0.1288 M hydrochloric solution, requiring 38.27 mL The reaction is:

Calculate the percent sodium carbonate in the sample.

Solution:

Example: How many milliliters of 0.25 M solution of H2SO4 will react with 10 mL of a 0.25

M solution of NaOH.

One half as many millimoles of H2SO4 will react one millimole NaOH




(

)

Example: An approximate 0.1 M hydrochloric acid solution is prepared by 120-fold dilution

of concentration hydrochloric acid. It is standardized by titrating 0.1876 g of dried primary

standard sodium carbonate:

The titration required 35.86 mL acid. Calculate the molar concentration of the hydro-

chloric acid.

Solution: millimoles HCl equal to twice the millimoles of Na2CO3

Example: The iron (II) in an acidified solution is titrated with a 0.0206 M solution of

KMnO4?

If the titration required 40.2 mL, how many milligrams iron are in the solution.

Example: Oxalic acid, H2C2O4, is a reducing agent that reacts with KMnO4 as follows:

Its two protons are also treatable with a base. How many milliliters of 0.100 M NaOH

0.100 M KMnO4 will react with 500 mg H2C2O4?

Solution:




(

)

Example: A 0.4671 g sample containing sodium bicarbonate (a monoacidic base) is dis-

solved and titrated with a standard solution of hydrochloric acid, requiring 40.72 mL The hy-

drochloric acid was standardized by titrating 0.1876 g sodium carbonate, which required

37.86 mL acid. Calculate the percent sodium bicarbonate in the sample.

№ meq of HCl = № meq of .

Back titration

Sometimes a reaction is slow to go to completion, and a sharp end point cannot be ob-

tained. A back titration will often yield useful results. In this technique, a measured amount of

the reagent, which would normally be the titrant, is added to the sample so that there is a

slight excess. After the reaction with the analyte is allowed to go to completion, the amount of

excess (unreacted) reagent is determined by titration with another standard solution.In back-

titration, a known number of millimoles of reaction it is taken, in excess of the analyte. The

unreacted portion is titrated.

mmol reagent reacted = mmol taken – mmol back titrted

mg analyte = mmol reagent reacted x factor (mmol analyte/mmol reagent)




(

) (

)

Example: A 0.50 g sample containing Na2CO3 plus inert matter in analyzed by adding 50.0

mL of 0.1 M HCl, a slight excess, boiling to remove CO2, and then back-titrating the excess

acid with 0.1 M NaOH. If 5.6 mL of NaOH is required for the back-titration, what is the per-

cent Na2CO3 in the sample?

Solution: (

)

Example: Chromium (III) is slow to react with EDTA (H4Y) and is therefore determined by

back-titration. A pharmaceutical preparation containing chromium (III) is analyzed by treat-

ing a 2.63 g sample with 5.00 mL of 0.0103 M EDTA. Following reaction, the unreacted

EDTA is back-titrated with 1.32 mL of 0.0122 M zinc solution. What is the percent chromi-

um chloride in the pharmaceutical preparation?

Solution:

Example: 0.3542 g of Na2CO3 was dissolved in water and titrated with HCl which consumed

30.32 mL of the acid. Calculate the normality of HCl solution.

Solution: Na2CO3 + 2HCl 2NaCl + CO2 + H2O




1 mol of Na2CO3 requires two moles of HCl.

∴ 2×№ of mmol of Na2CO3 = № of mmol of HCl.

H.W.: 1- Show by calculations how could you prepare the following solutions:

a) 525 mL of 0.4 F BaCl2 from solid BaCl2.2H2O.

b) 2.30 liters of 0.2 M K+ from solid K2SO4.

c) 100 mL of 0.1 F AgNO3 from a solution that was 0.441 F in the salt.

d) 20 liters of a solution that is 0.202 M in Na+ from a 2.42 F solution of Na2SO4.

2- A solution was prepared by dissolving 1.68 g of K4Fe(CN)6 in water and diluting exactly

to 500 mL Calculate:

a) the formal concentration of K4Fe(CN)6 .

b) The molar concentration of K+ assuming complete dissociation.

c) the weight-volume percent of K4Fe(CN)6 .

d) the weight-weight percent of K4Fe(CN)6 if the density of solution = 1.008 g/mL .

e) The number of moles Fe(CN)64-

in 16 mL of the solution.

3- An impure 1 g sample of arsenious acid (H3AsO3) is oxidized to H3AsO4 by titration with

45 mL of 0.08 N iodine. Calculate the percentage of H3AsO3 (F. w =125.9) and percentage

of As.

4- 20 mL of 0.05 M EDTA is added to 700 mg of iron(II). The excess of EDTA is back ti-

trated with 5.08 mL of 0.042 M Cu2+

solution. Calculate the percentage of Fe2+

as Fe2O3 in

the sample.

5- Tin(II) is titrated with dichromate according to the equation:

Cr2O72-

+ 3Sn2+

+ 14H+ 3Sn

4+ + 2Cr

3+ + 7H2O .

Calculate the weight of tin(II) in a sample that requires 20 mL of 0.1 M Cr2O72-

.

Classification of volumetric or titrimetric methods

(1) Neutralization (acid-base) titrations.

(2) Precipitation titrations.

(3) Complexation titrations.

(4) Reduction-Oxidation titrations.




Calculations of acid-base titrations

An Acid-Base titration involves a neutralization reaction in which an acid is reacted with

an equivalent amount of base at equivalence point or endpoint.

It is also called neutralization titration curves since H+ that comes from acids or acidic so-

lutions is neutralized with OH – which comes from bases or basic solutions to form very slightly

ionized water:

H+ + OH

– H2O

In acid-base titrations, there is a change in [H+] concentration or pH of the solution by in-

creasing or decreasing the pH of the solution until equivalence point where a sudden change in

pH is indicated by a sudden change in a certain acid-base indicator color.

The titrations are followed by calculation the pH during the process before addition of titrant

and after addition of titrant till exceeding of equivalence point.

The data obtained are used to set up the titration curves and identifying the pH at equiva-

lence point.

A) Titration of strong acid versus strong base

The equivalence point is where the reaction is theoretically complete while the endpoint

where the colour of indicator were changed.

Detection of the endpoint and equivalence point

Example: Calculate the pH at 0, 10, 90, 100, 110% titration of 50.0 mL of 0.10 M HCl with

0.10 M NaOH?

(1) At 0% titration: before addition of 0.1 M NaOH

[ ]

(2) At 10% titration: before equivalence point




(3) At 90% titration: before equivalence point

(4) At 100% titration: equivalence point

[ ][ ]

√ [ ] [ ]

(5) At 110% titration: after equivalence point




Construction (plot) titration curve of strong acid versus strong base

The relationship between pH calculated for HCl remaining or unreacted (excess) or NaOH

on Y axis and the volume of titrant (0.1M NaOH) added on X axis, this curve called titration

curve. This curve used for estimation the equivalence point (theoretically) and selection of the

indicator for detecting the endpoint reaction by the colour change of the indicator.

Note: The selection of the indicator become more critical as the solution become more dilute

and the sharpness endpoint decrease as the concentration.

The point at which the reaction is observed to be complete at the indicator colour where

changed is called endpoint.

Acid-base indicators (pH indicators)

Are substances which change colour with change pH. They are usually weak acids or ba-

ses, which when dissolved in water dissociate slightly and form ions.

Phenolphthalein indicator is a

colourless, weak acid which disso-

ciates in water forming pink anions.

Under acidic conditions, the equi-

librium is to the left, and the con-

centration of the anions is too low

for the pink colour to be observed.

1=(0.1M)versus (0.1M)

2=(0.1M)versus (0.1M)

3=(0.1M)versus (0.1M)




However, under alkaline conditions, the equilibrium is to the right, and the concentration of the

anion becomes sufficient for the pink colour to be observed.

Acid–base titration is performed with a phenolphthalein indicator, when it is a strong acid

– strong base titration, a bromthymol blue indicator in weak acid – strong base reactions, and a

methyl orange indicator for strong acid – weak base reactions. If the base is off the scale, i.e. a

pH of > 13.5, and the acid has a pH > 5.5, then an Alizarine yellow indicator may be used. On

the other hand, if the acid is off the scale, i.e. a pH of < 0.5, and the base has a pH < 8.5, then a

Thymol Blue indicator may be used.

Common acid – base indicators

Indicator Approximate pH range for color change Color change

Methyl orange 3.2-4.4 Red to yellow

Bromthymol blue 6.0-7.6 Yellow to blue

Phenolphthalein 8.2-10 Colorless to pink

Litmus 5.5-8.2 Red to blue

Bromcresol green 3.8-5.4 Yellow to blue

Thymol blue 8.0-9.6 Yellow to blue

Assume the indicator is a weak acid designated HIn, and assume that the nonionized form

is red while the ionized form is blue and designated In-.

[ ][ ]

[ ]

Acidic (nonionized form), red color Basic (ionized form), blue color

We can write a Henderson–Hasselbalch equation for this just like other weak acids:

[ ]

[ ]

Your eyes can generally discern only one color if it is 10 times as intense as the other.

When only the color of the nonionized form is seen:

When only the color of the ionized form is observed:

So, the pH is going from one color to the other has changed from pKa-1 to pKa+1. This

is a pH change of 2 units, and most indicators require a transition range of about two pH units.

Choose an indicator with a pKa near the equivalence point.




Two drops (0.1 mL) of 0.01 M indicator (0.1% solution with F. w =100) is equal to 0.01

mL of 0.1 M titrant.

B) Titration of weak acid versus strong base

Acetic acid with sodium hydroxide

Example: Calculate the pH at 0, 10, 25, 50, and 60 mL titrant in the titration of 50 mL of 0.1 M

acetic acid (Ka=1.75×10-5

) with 0.10 M NaOH?

(1) At 0 mL titrant (0.1 M NaOH): HOAc solution only

[ ] √ √

[ ]

(2) At 10 mL titrant (0.1 M NaOH): before equivalence point (buffer formation region)

[ ]

[ ]

(3) At 25 mL titrant (0.1 M NaOH): before equivalence point (buffer formation region)

[ ]

[ ]

(4) At 50 mL titrant (0.1 M NaOH): equivalence point

All NaOH reacted with all HOAc and converted it to its salt sodium acetate.

[ ] √ √

√




[ ]

(5) At 60 mL titrant (0.1 M NaOH): after equivalence point (NaOH solution alone)

Construction (plotting) titration curve of weak acid versus strong base

The sharpness endpoint decreases as the concentration decreases.

Titration curves for 50 mL 0.1 M weak acids of different Ka value versus 0.1 M NaOH

Titration curves of three multiprotic acids with standard 0.1M NaOH(A) 0.1M H3PO4 (B) 0.1M H2C2O4 (C) 0.1M H2SO4

& The sharpness of the endpoint decreases as Ka decreases.




C) Titration of weak base versus strong acid

Example: Calculate the pH at 0, 10, 25, 50, and 60 mL of titrant of 50 mL of 0.1 M NH3

(Kb=1.75×10-5

) with 0.1 M HCl?

(1) At 0 mL titrant (0.1 M HCl): NH3 solution alone

[ ] √ √

[ ]

(2) At 10 mL titrant (0.1 M HCl): before equivalence point (buffer formation region)

[ ]

[ ]

(3) At 25 mL titrant (0.1 M HCl): before equivalence point (buffer formation region)

[ ]

[ ]

(4) At 50 mL titrant (0.1 M HCl): equivalence point region

The all HCl added converted all NH3 to its salt NH4Cl

[ ] √ √

√

(5) At 60 mL titrant (0.1 M HCl): after equivalence point




It is worthy to note here that equivalence point occurs at pH = 5.27 which is acidic medi-

um as a result of hydrolysis of NH4Cl

NH4Cl + H2O NH4OH + HCl

Which gives weak base and strong acid. A plot of pH values against mL of HCl shows a

titration curve of equivalence point at pH = 5.28 (acidic medium).

Construction (plotting) titration curve

Titration curve for 50 mL 0.1 M weak base of different Kb values versus 0.1 M HCl. The sharpness of the endpoint decreases as

Kb decreases of weak bases.

Titration of sodium hydroxide versus mixture of acids

Mixtures of acids (or bases) can be titrated stepwise if its difference in Ka values of at least

104, unless perhaps a pH meter is used to construct the titration curve.

Note: one acid should be at least 104

weaker than other to titrate separately.

Example: A mixture of HCl and H3PO4

is titrated with 0.1 M NaOH. The first

endpoint (methyl red) occurs at 35.0 mL,

and the second endpoint (bromothymol

blue) occurs at total of 50.0 mL (15.0 mL




after the first point). Calculate the millimoles HCl and H3PO4 present in the solution?

The second endpoint corresponds to that the titration of one proton of H3PO4

(H2PO4-→ HPO4

2-).

The HCl and the first proton of H3PO4 titrate. A 15 mL portion of base was used to titrate

the first proton of H3PO4 (same as for the second proton), leaving 20 mL used to titrate the HCl.

The difference in Kb values must be at least 104 to titrate separately.

Example: A 0.527 g sample of mixture containing Na2CO3 and NaHCO3 and inert impurities is

titrated with 0.109 M HCl, requiring 15.7 mL to reach the phenolphthalein endpoint and a total

of 43.8 mL to reach the modified methyl orange endpoint, what is the percent each of Na2CO3

and NaHCO3 in mixture?

The phenolphthalein endpoint volume HCl equivalent for half amount of Na2CO3

(

)

(

)

The volume HCl at second endpoint (modified methyl orange) equal for second half

amount of Na2CO3 plus the amount of NaHCO3.

(

)

(

)




D) Titration of weak acid with weak base

Example: 25 mL of 0.1 M CH3COOH solution is titrated with 0.1 M NH4OH solution. Calcu-

late the pH of this titration at the following stages:

(a) before addition of ammonia solution.

(b) after addition of 15 mL of 0.1 M ammonia solution.

(c) at equivalence point.

(d) after addition of 30 mL of 0.1 M ammonia solution.

Solution: (a) before addition of ammonia solution.

pH value before addition of 0.1 M NH4OH:

aa CpKpH log2

1

2

1

1.0log2

174.4

2

1

= 2.37 + 0.5 = 2.87

(b1) after addition of 15 mL of 0.1 M NH4OH.

VCH3COOH = 15 mL reacting volume of CH3COOH which is changed into CH3COONH4 .

unreacted volume of CH3COOH = 25 – 15 = 10 mL

Therefore, we have buffer solution of CH3COOH and its salt CH3COONH4.

Total volume = 15 + 25 = 40 mL




][

][log

acid

saltpKpH a

025.0

0375.0log74.4

= 4.74 + 0.18 = 4.92

(b2) after addition of 24 mL of 0.1 M NH4OH.

VCH3COOH = 24 mL reacting volume of CH3COOH which is altered into CH3COONH4 .

unreacted volume of CH3COOH = 25 – 24 = 1 mL


][

][log

acid

saltpKpH a

002.0

049.0log74.4

= 4.74 + 1.39 = 6.13

(c) pH at equivalence point (after addition of 25 mL of 0.1 M NH4OH):

Therefore, all CH3COOH is changed into its salt CH3COONH4.





CH3COONH4 is salt derived from weak acid and weak base, The following relation is used

to calculate its pH.

baw pKpKpKpH2

1

2

1

2

1

74.4

2

174.4

2

114

2

1

= 7

(d) After addition of 30 mL of 0.1 M NH4OH:

The reacting volume of NH4OH = 25 mL which reacts with 25 mL of CH3COOH to form

CH3COONH4 .

Unreacted volume of NH4OH = 30 – 25 = 5 mL

Thus we have another buffer solution which is NH4OH and its salt.


][

][log

base

saltpKpOH b

44.57.074.4009.0

045.0log74.4

∴ pH = 14 – 5.44 = 8.56

Therefore, we get the following titration data:

mL NH4OH 0 15 24 25 30

pH 2.87 4.92 6.13 7.0 8.56

A plot of pH values versus mL of NH4OH, will give a titration curve showing equivalence

point at pH = 7.0

A titration curve of weak acid with weak base

It is clear from the above titration data and titra-

tion curve that:

1- pH at equivalence point equals 7.0. This is a spe-

cial case where Ka=Kb. But if Ka>Kb, the pH of solu-

tion is less than 7.0 . When Ka<Kb, the pH of solution

is more than 7.0 .




2- The change in pH around equivalence point is not sharp and is not clear. For example, the

change in pH is less than one unit, when the volume increases from 24 to 25 mL Therefore, it is

difficult to find chemical indicator to identify the equivalence point. As a conclusion, another

methods are used for this titration such as potentiometric titration.

H.W.:

1) Calculate the pH of a solution contains 2.72 g of KH2PO4 and 3.48 g of K2HPO4 in 100

mL. pK2 for phosphoric acid = 7.21 .

2) Calculate the weight of NH4Cl that should be added to 100 mL of 1 M NH4OH to get pH =

9.3 . Consider no change in volume.

a. plot the titration data to get the titration curve and locate the equivalence point.

3) 30 mL of ammonia (0.1 M) is titrated with 0.1 M HCl. Calculate the pH at the following

stages:

a. before addition of HCl.

b. after addition of 20 mL of the acid.

c. after addition of 30 mL of the acid.

d. after addition of 40 mL of the acid.

e. plot the titration curve and locate the equivalence point.

4) Calculate the pH of titration 25 mL of 0.1 M acid with 0.1 M base after addition the fol-

lowing volumes of the corresponding base. 5 mL, 12.5 mL, 17.5 mL and 30 mL:

a. HCl with NaOH .

b. Benzoic acid with NaOH .

c. HCl with NH4OH.

5) 20 mL of acetic acid (0.1 M) is titrated with 0.1 M NaOH. Calculate the pH data at the fol-

lowing stages:

a. before addition of base.

b. after addition of 10 mL of NaOH.

c. after addition of 20 mL of base.

d. after addition of 25 mL of the base.

e. plot the titration data to get the titration curve and locate the equivalence point.

f. Mention the indicator which is convenient in this titration.




Calculations of precipitation titrations

In precipitation titrations, one of the products is slightly soluble salt called precipitate.

Example: Calculate the percentage of silver in silver alloy if a solution prepared by dissolving

0.3 g of the alloy requires 23.80 mL of 0.1N NH4SCN. (Ag=108)

Solution: № of meq. of Ag

+ = № of meq. of SCN

-

The same scheme of calculation is followed by any addition of volume of AgNO3 solution

after equivalence point. the following data are obtained for both titrations with AgNO3 solution.

pCl 1 1.6 4.96 7.88

pAg 0 8.32 4.96 2.04

pI 1 1.6 8 13.96

pAg 0 14.4 8 2.04

A plot of pCl and pI against volume of AgNO3 solution gives the following titration curves.

0

2

4

6

8

1 0

1 2

1 4

0 5 1 0 1 5 2 0 2 5 3 0 3 5

m l o f A g N O 3

p C l

0

2

4

6

8

1 0

1 2

1 4

0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0

m l o f A g N O 3

p I

(a) (b)




Equivalence point at pH =5.28. Therefore,

an indicator is chosen which its color is

changed in acidic medium which is used

such as methyl red or methyl orange.

Example: Find the weight of BaCl2 in 250 mL of solution where 40 mL in excess of 0.102 N

AgNO3 was added to 25 mL of BaCl2. The excess of AgNO3 was titrated with 0.098 N SCN- .

Solution: BaCl2 + 2AgNO3 Ba(NO3)2 + 2AgCl

AgNO3 + SCN- AgSCN + NO3

-

№ of meq. of AgNO3 = № of meq. of SCN - + № of meq. of BaCl2.

1.47 + 25 N = 4.08

Calculation of oxidation-reduction titrations

These titrations involve the reactions between oxidant and reductant which subject trans-

ferring of electrons from reductant to oxidant.

Example: A sample of Na2C2O4 weighs 0.2734 g is dissolved in water acidified with dilute

H2SO4 and titrated at 70 oC with KMnO4 solution, which requires 42.68 mL The titration exceeds

the equivalence point and the excess of KMnO4 is back titrated with 0.1024 N oxalic acid which

requires 1.46 mL Calculate the normality of KMnO4 solution.

0

2

4

6

8

10

12

14

0 5 10 15 20 25 30 35

ml HCl

pH




Solution:

№ of meq. of KMnO4 = № of meq. of Na2C2O4 + № of meq. of H2C2O4

Example: A sample of Iron ore weighs 0.6038 g was dissolved in an acid and Fe3+

is reduced to

Fe2+

and titrated with 0.1073 N K2Cr2O7 which requires 38.42 mL Calculate the percentage of

Fe in the sample and express the percentage as FeO, Fe2O3 and Fe3O4. Atomic w of Fe =56.

Solution:

№ of meq. of Fe2+

= № of meq. of K2Cr2O7




Calculations of complex formation titrations

In these titrations a stable complex is formed which is exploited in determination of most

metal ions with EDTA as a ligand.

Example: 25 mL of Ni2+

solution requires 0.521 g of KCN according to the following reaction:

Ni2+

+ 4CN -

Ni(CN)42-

complex ion

Calculate the molarity of Ni2+

solution?

Solution: 1 mol of Ni2+

requires 4 mol of KCN in order to get the above reaction. The strength

of Ni2+

is four times greater than KCN.

∴ 4×№ of mmol of Ni2+ = № of mmol of KCN.

Example: 3 g of EDTA (disodium) is dissolved in one liter. Calculate the volume of 0.05 M

Mg2+

solution required to react with 25 mL of EDTA solution.

Solution: The ratio of reaction is 1 Mg2+

: 1 EDTA

EDTA is ethylene diamine tetra acetic acid abbreviated as Na2H2Y.2H2O which is disodi-

um salt. Also, it is abbreviated as EDTA and has the following formula.

M. w of EDTA =2×23 +8×16 +14×2 +10×12 +2×18 +14×1 = 372 g/mol

Mg2+

reacts with EDTA in the ratio 1:1 ~~ Mg2+

+ EDTA Mg-EDTA

№ of mmol of EDTA = № of mmol of Mg2+

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