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Chemistry/Ionic Equilibria 1

IONIC EQUILIBRIA

1. Concepts of Acid/Base

1.1 Arrhenius Theory

1.1.1 Definition 1. An acid is a substance which dissolves in water to give hydrogen ions, H

+(aq).

2. A base is a substance which dissolves in water to give hydroxide ions, OH-(aq).

3. An acid-base reaction (i.e. neutralization) is the combination of H+(aq) and OH

-(aq) to form H2O(l).

Neutralization : H+(aq) + OH

-(aq) → H2O(l)

Summary : H+(aq) and OH

-(aq) are responsible for acidic and basic properties respectively.

1.1.2 Drawbacks of Arrhenius Theory

1. It confines to aqueous solution only. However, many acids are capable of giving hydrogen ions in solvents

other than water. e.g. nitric acid in ethanol ionizes to give hydrogen ions

2. Some compounds are not classified as acids because they do not contain the element hydrogen explicitly,

although they are capable of releasing hydrogen ions after reacting with water.

e.g. SO2(g) + H2O(l) 2H+(aq) + SO3

2-(aq)

3. Some substances are not classified as bases because they do not contain hydroxide explicitly, although

they are capable of releasing hydroxide ions by reacting with water.

e.g. NH3(g) + H2O(l) NH4+(aq) + OH

-(aq)

1.2 Bronsted-Lowry Theory

1.2.1 Introduction

This theory emphasizes on the role of hydrogen ion, H+ (i.e. proton).

Definition :

1. An acid is a molecule or ion that can donate a proton, H+ (i.e. a proton donor).

2. A base is a molecule or ion that can accept a proton, H+ (i.e. a proton acceptor).

3. An acid-base reaction is the transfer of proton from an acid to a base (i.e. a proton-transfer reaction).

Neutralization : H+(aq) + OH

-(aq) → H2O(l)

NH3(aq) + HCl(aq) → NH4Cl(aq) or NH4+(aq) + Cl

-(aq)

Implications :

1. An acid must possess the element hydrogen for donation.

2. A base can either be a negative ion or a neutral molecule with a lone pair of electrons to form dative

covalent bond with proton. Reason :

Since proton has no electrons, bonding between the base and proton must be dative covalent in nature, with

the base providing both electrons

⇒ a base must have a lone pair of electrons available for donation

N

H H

H + H+ N

H

H H

H

+

3. Bronsted-Lowry definition includes all substances that fit the Arrhenius definition. In addition, it includes

a. as acids, ions such as HSO4- and NH4

+ that can donate proton in solution, even though they cannot exist

alone; and

b. as bases, all substances that can accept proton such as CO32- or CN

-, even though they do not contain

hydroxide explicitly.


1.2.2 Conjugate Acid-Base Pair Consider the ionization of ethanoic acid in water :

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O

+(aq)

After CH3COOH(aq) has lost its proton, it becomes CH3COO

-(aq).

From reverse reaction, CH3COO-(aq) is capable of accepting a proton to form back the acid

⇒ CH3COO-(aq) is itself a

⇒ CH3COO-(aq) is called the of CH3COOH(aq)

⇒ CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O

+(aq)

acid conjugate base Similarly, consider the ionization of ammonia in water :

NH3(aq) + H2O(l) NH4+(aq) + OH

-(aq)

After NH3(aq) has accepted a proton, it becomes NH4

+(aq).

From reverse reaction, NH4+(aq) is capable of donating its proton to form back the base

⇒ NH4+(aq) is itself an

⇒ NH4+(aq) is called the of NH3(aq)

⇒ NH3(aq) + H2O(l) NH4+(aq) + OH

-(aq)

base conjugate acid Conclusion :

1. After an acid loses a proton, it becomes a base and is called the conjugate base of the given acid.

2. After a base accepts a proton, it becomes an acid and is called the conjugate acid of the given base.

These pairs of species are known as conjugate acid-base pairs;

they convert into each other by the transfer of , i.e. gain/loss of a . Remark :

1. An acid-base reaction must involve both acid and base.

Q Both donor and acceptor of proton must be present for transfer of proton to take place.

2. An acid only shows its properties in the presence of a base. Similarly, a base only shows its properties in the

presence of an acid.

Q For an acid to function, a base must be present to H+.

For a base to function, an acid must be present to H+.

3. In an acid-base reaction, since the given acid (acid1) will be converted into its conjugate base (base1) and the

given base (base2) will be converted into its conjugate acid (acid2), the process can be represented as :

As there are acids and bases on both sides of the equation, the

process can go either way and is represented by an . The forward reaction is the transfer of a proton from acid1 to base2,

and the reverse reaction is the transfer of the proton from acid2 to base1.

Essentially, the process can be taken as base1 and base2

competing to accept proton

⇒ an acid-base reaction is in fact the competition to gain/lose proton between bases

⇒ alternatively, an acid-base reaction can also be taken as the competition to gain/lose proton between acids

acid1 + base2 base1 + acid2

conjugateacid-base pair2

conjugateacid-base pair1


Example : HCl(aq) + H2O(l) Cl-(aq) + H3O

+(aq)

acid base conjugate base conjugate acid of HCl of H2O Exercise : 1. Identify the conjugate acid-base pairs in the following neutralization reactions.

a. CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO

-(aq)

b. NH4+(aq) + H2O(l) H3O

+(aq) + NH3(aq)

c. H2O(l) + CO32-(aq) HCO3

-(aq) + OH

-(aq)

2. Write an equation to show that HPO4

2-(aq) can act as a Bronsted base in water. (94, I)

Solution : 3. Outline the Bronsted-Lowry theory of acids and bases with respect to the following reaction :

H2O(l) + HCl(aq) → H3O+(aq) + Cl

-(aq) (89, II)

Solution : H2O(l) + HCl(aq) → H3O+(aq) + Cl

-(aq)

2 conjugate acid-base pairs : &

4. Which of the following is a conjugate acid-base pair ? A. NH4

+(aq) and NH3(aq)

B. H2SO4(aq) and SO42-

(aq) C. H3O

+(aq) and OH-

(aq) D. HCl(aq) and NaOH(aq)

(1 mark, 06, I, 3c) 1.2.3 Amphoteric Substances Some substances can both act as an acid to donate a proton, or act as a base to accept a proton. Such behaviour

is known as amphoteric. Example : 1. water

+±±− →← →←++

OHOHOHHH

32

2. HSO4-(aq). 424

24 SOHHSOSO

HH →← →←++ ±−±−

As an acid : HSO4-(aq) + OH

-(aq) SO4

2-(aq) + H2O(l)

As a base : HSO4-(aq) + H3O

+(aq) H2SO4(aq) + H2O(l)

For an amphoteric substance, whether it behaves as an acid or a base depends on the other substance present. If a stronger acid is present, the amphoteric species is forced to act as acid/base to accept/donate a proton.

If a stronger base is present, the amphoteric species is forced to act as acid/base to accept/donate a proton. Example : Give the equation for the reaction between HCO3

-(aq) and

(a) NaOH(aq) (b) HCl(aq)


1.2.4 Acid Strength & Base Strength The strength of an acid is a measure of its tendency to donate a proton. The strength of a base is a measure of its tendency to accept a proton. Both are related to the extent of equilibrium.

Example : HCl(aq) + H2O(l) Cl-(aq) + H3O

+(aq) complete ionization/dissociation

This equilibrium lies very far to the right

⇒ most HCl(aq) dissociate

⇒ HCl(aq) has a very high tendency to lose proton

⇒ HCl(aq) is a strong acid

Example : HCN(aq) + H2O(l) H3O+(aq) + CN

-(aq) incomplete ionization/dissociation

This equilibrium lies very far to the left

⇒ few HCN(aq) dissociate

⇒ HCN(aq) has a low tendency to lose proton

⇒ HCN(aq) is a weak acid The strength of a base can be defined accordingly. Example :

Strong base : equilibrium favours the gain of proton

CN-(aq) + H2O(l) HCN(aq) + OH

-(aq) equilibrium far to right

Weak base : equilibrium not favoured by gain of proton


-(aq) equilibrium far to left

1.2.5 Strength within Conjugate Acid-Base Pair

Fact : The conjugate base of a strong acid is a weak base. Reason :

For a strong acid, its tendency to lose proton must be high/low

⇒ the ability for its conjugate base to accept a proton to form back the strong acid must be high/low,

⇒ its conjugate base must be strong/weak Example :

HCl(aq) + H2O(l) Cl-(aq) + H3O

+(aq)

HCl(aq) is a strong acid

⇒ the above equilibrium lies far to left/right

⇒ Cl- has little/strong tendency to accept H

+ to form back HCl

⇒ Cl- is a strong/weak base

Exercise : Convince yourself with the following statement.

The conjugate base of a weak acid is a strong base. Example : HCN vs CN-

The conjugate acid of a strong base is a weak acid. Example : CN- vs HCN

The conjugate acid of a weak base is a strong acid. Example : Cl- and HCl

Remark :

Strength is a relative term. It takes two acids or two bases to tell which is stronger and which is weaker.


1.3 Hydrogen Ion

Fact : In aqueous solution, isolated hydrogen ion H+(aq) never exists.

Reason :

Hydrogen ion is just a bare proton with extremely high/low charge density (due to exceptionally )

⇒ H+ seeks out positive/negative centre immediately and will bond to in water

⇒ H+ is therefore hydrated with molecules of water

Remark :

1. The most predominant hydrated species is believed to be hydroxonium ion (or hydronium ion) H3O+(aq), with

the lone pair on H2O forming a bond with H+. Other higher hydrated ions like

[H(H2O)2]+(aq) and [H(H2O)3]

+(aq) etc. are also believed to exist.

O

H H

H+ H+

+

O

H H

2. For convenience, it is understood that H+(aq) or H3O

+(aq) refers to hydrated H

+ in aqueous solution.


2. Dissociation of Water

2.1 Ionic Product of Water The ionization of water can be represented by the following equilibrium :

H2O(l) H+(aq) + OH

-(aq) ∆H = +57.3 kJ mol

-1

with equilibrium constant, K, at a given temperature as :

eqm

eqmeqm

OH

OHHK

][

][][

2

−+

=

As the degree of ionization is extremely small

⇒ [H2O]eqm is virtually constant

⇒ the equation can be rewritten as

Kw = K [H2O] = [H+]eqm [OH

-]eqm where Kw is called the ionic product of water

At 25oC (298 K), Kw = 1.0 x 10

-14 mol

2 dm

-6.

Remarks :

1. In pure water at 25oC, [H

+] = [OH

-] = 1 x 10

-7 mol dm

-3.

This corresponds to approximately 1 in every 5 x 108 molecules of water being split up into ions at 298 K,

confirming that the degree of ionization is extremely small. 2. This equilibrium always holds in solution, no matter the solution is acidic or alkaline. In acidic solution, [H

+] increases and [OH

-] decreases accordingly. Similarly, in alkaline solutions, [OH

-]

increases and [H+] decreases accordingly. In any case, at 25

oC, [H

+] [OH

-] = 1.0 x 10

-14 mol

2 dm

-6.

3. Kw is like any equilibrium constant, its value only changes with .


2.2 Kw and Temperature Values of KW vs T are given below :

KW / mol2 dm

-6 Temperature / K

0.11 x 10-14 273

0.30 x 10-14 283

0.68 x 10-14 293

1.00 x 10-14 298

5.47 x 10-14 323

51.3 x 10-14 373

Trend : As temperature increases, Kw increases/decreases.

Reason : H2O(l) H+(aq) + OH

-(aq) ∆H = +57.3 kJ mol

-1

Either : Q ∆H is positive, by CRT

HK +

∆−=ln

As T ↑

⇒ T

1 ↑↑↑↑/↓↓↓↓

⇒ RT

H∆− ↑↑↑↑/↓↓↓↓

⇒ ln K ↑↑↑↑/↓↓↓↓

⇒ K ↑↑↑↑

Or : By Le Chatelier’s Principle, As temperature increases, equilibrium position shifts to endothermic/exothermic reaction

⇒ shifts to the left/right

⇒ more/less water ionized

⇒ K ↑↑↑↑ Exercise : 1. What is the concentration of H

+(aq) and OH

-(aq) if 0.01 mole of solid NaOH is dissolved in a 1.0 dm

3 of pure water

at 25oC ?

[OH

-] =

[H

+] =

2. What are the concentrations of H

+(aq) and OH

-(aq) in pure water at 373 K ?

In pure water, [H

+] [OH

-] =

⇒ [H+] =


3. pH and its Measurement

3.1 The ‘p’ Notation

Definition : p value of a number is the negative of the logarithm (base 10) of that number. Reason of using ‘p’ notation : [H+] & [OH

-] are usually very small, 10

-something

⇒ inconvenience to handle p values can avoid the use of negative indices in small numerical values.

3.2 pH

pH = -log [H+] pOH = -log [OH

-]

For a neutral solution at 25

oC, [H

+] = [OH

-] = 1 x 10

-7 mol dm

-3

pH = pOH = -log (1 x 10-7) = - (-7) = 7 ← no unit

If the pH/pOH of the solution is given, its [H

+]/[OH

-] can be found easily.

pH = -log [H+] pOH = -log [OH

-]

log [H+] = -pH log [OH

-] = -pOH

[H+] = 10

-pH [OH

-] = 10

-pOH

Example : For a solution has pH = 3.25, [H

+] = = mol dm

-3

Remarks :

1. At 25oC, [H

+] [OH

-] = 1 x 10

-14

log ([H

+] [OH

-]) = log (1 x 10

-14)

log [H

+] + log [OH

-] = -14

-log [H

+] - log [OH

-] = 14

pH + pOH = 14

2. A change of 1 unit in pH/pOH represents a change in [H+]/[OH

-] by a factor of .

Exercise : 1. What is the pH of 0.02 M HCl ? [H

+] =

2. What is the pH of 10

-3 M H2SO4 ? [H

+] =

3. What is the pH of 3 M HX which is only 50% dissociated ? [H

+] =

4. If a solution contains a concentration of H

+(aq) equal to 8.69 x 10

-9 mol dm

-3, what is the pH of the solution ?

pH = 5. If the pH of a solution is 2.73, what is the concentration of H

+(aq) ions ?

[H

+] =


6. Account for the fact that at 373K, the pH of pure water is less than 7.0. (94, II) Solution : From Le Chatelier’s Principle,

As T ↑, equilibrium position shifts to exothermic/endothermic reaction

⇒ more/less water is ionized

⇒ [H+] ↑↑↑↑/↓↓↓↓, higher than that at 25

oC

⇒ [H+] >/< 1 x 10

-7 mol dm

-3

⇒ pH more/less than 7 7. At 37

oC, Kw = 2.42 x 10

-14 mol

2 dm

-6. The pH of a solution is 6.808. Is the solution acidic ?

Solution : At pH = 6.808, [H

+] =

[OH

-] =

Since [H

+] >/=/< [OH

-], solution is acidic/neutral/alkaline.

8. Explain why the mean [H3O

+] between pH 4 and pH 5 corresponds to pH 4.26 rather than pH 4.50.

Solution : At pH = 4, [H

+] = At pH = 5, [H

+] =

Mean [H

+] =

⇒ pH = Remarks : 1. Since Kw changes with temperature,

a. pH = 7 does not imply that the solution is neutral. pH = 7 only implies a neutral solution at 25oC.

b. pH of any neutral solution (e.g. pure water) also changes with temperature and is not necessarily 7. A neutral solution has pH = 7 only at 25

oC.

2. A solution is neutral when [H+] = [OH

-]; is acidic when [H

+] > [OH

-] and is alkaline when [H

+] < [OH

-].


3.3 Measurement of pH pH of a solution can be measured accurately by a or roughly by or . 3.3.1 pH Meters

The theory and instrumentation of pH meter are not required.

Set up : A pH meter is an electrochemical cell consisting of 2 electrodes :

1. a reference electrode with constant electrode potential (e.g. calomel electrode, a platinum wire dipping into mercury in a solution of mercury(I) chloride).

2. a glass electrode sensitive to hydrogen ion concentration. Principle : The voltage or e.m.f. of the cell depends on the H

+(aq) concentration.

Remark :

1. The pH meter has to be calibrated before use. It usually takes at least 2 pH values for the calibration. Procedure : 1. Rinse the glass electrode with distilled water. 2. Dip the glass electrode into a buffer solution of known pH value. 3. Adjust the pH meter so that its reading agrees with the pH of the buffer solution, i.e. calibration. 4. Rinse the electrode with distilled water again. 5. Dip the glass electrode into another buffer solution of known pH value. 6. Calibrate the pH meter again. 7. The whole scale of pH should be calibrated. 8. Once the calibration is finished, dip the glass electrode into the solution whose pH to be determined and the

pH can be read from display. 2. The glass electrode is very delicate and sensitive. It should be kept clean and never be dried.


4. pH Calculations with Strong Acids or Bases

A strong acid or base is one that ionizes or dissociates completely in water.

Example : HCl(aq) → H+(aq) + Cl

-(aq)

NaOH(aq) → Na+(aq) + OH

-(aq)

In strong acid or base, in principle there are two sources of H

+(aq) or OH

-(aq) in solution :

1. from the complete ionization of the strong acid/base 2. from the self-ionization of water Example 1 : What is the pH of 1.00 M HCl(aq) ?

HCl(aq) → H+(aq) + Cl

-(aq) H2O(l) H

+(aq) + OH

-(aq)

1.0 1.0 (y + 1.00) y

Using ionic product of water, (y + 1) y = 1. 0 x 10-14

⇒ y = 1.0 x 10-14

⇒ [H+] = 1.00 + 1.0 x 10

-14 = 1.00

⇒ pH = 0 Finding : This is essential the same as if there is no contribution from the self-ionization of water.

Reason : [H+] arising from self-ionization of water is negligible compared to the contribution from strong acid.

Example 2 : What is the pH of 0.10 M HCl(aq) ? [H

+] = 0.1 mol dm

-3

pH = 1 Finding : A dilution of 10 times will change pH by 1 in strong acid/base. Example 3 : What is the pH of 1.0 x 10

-8 HCl(aq) ?

Finding : Direct pH calculation will give pH = ⇒ acidic/alkaline ⇒ reasonable/unreasonable

Reason : At such a low concentration of acid, self-ionization of water is negligible/not negligible.

HCl(aq) → H+(aq) + Cl

-(aq) H2O(l) H

+(aq) + OH

-(aq)

1 x 10-8 1 x 10

-8 (y + 1.0 x 10

-8) y

y (y + 1.0 x 10

-8) = 1.0 x 10

-14

y = 9.51 x 10

-8 or -1.05 x 10

-7 (rejected)

[H

+] = 9.51 x 10

-8 + 1.0 x 10

-8 = 1.051 x 10

-7 mol dm

-3

pH = 6.98 ← slightly acidic, makes sense


Exercise : 1. Calculate the pH and pOH of 3.2 x 10

-4 M solution of Ba(OH)2 at 25

oC. (You may assume that Ba(OH)2 is a

strong electrolyte that completely dissociates in water.) [OH

-] =

pOH = pH = 2. Calculate the molarity of an aqueous solution of NaOH having a pH of 10.5 at 25

oC.

[H

+] =

[OH

-] =

[NaOH] = 3. The degree of dissociation of Ca(OH)2 in a 0.1 M solution is 0.05 at 25

oC.

(a) Calculate the hydroxide ion concentration. [OH

-] =

(b) Calculate the pH of the solution. [H

+] =

pH =

Summary :

1. The contribution of H+(aq) or OH

-(aq) from self-ionization of water can usually be neglected, except at very

low concentration of strong acid/base.

2. A dilution of 10 times changes pH by 1 unit in strong acid/base.


5. pH Calculations of Weak Acid/Base

5.1 Weak Acid

The dissociation of weak acid HA(aq) in solution to form H+(aq) (or more formally H3O

+(aq)) and its conjugate base

(A-(aq)) can be represented by an equilibrium :

HA(aq) H+(aq) + A

-(aq) or HA(aq) + H2O(l) H3O

+(aq) + A

-(aq)

with equilibrium constant KC

eqm

eqmeqm

CHA

AHK

][

][][ −+

= or eqmeqm

eqmeqm

CHAOH

AHK

][][

][][

2

−+

=

Since [H2O] is a constant, we can define acid dissociation constant Ka for any weak acid as

eqm

eqm3

acid] tedundissocia[

base] conjugate[][

][

][][

][

][][ eqm

eqm

eqmeqm

eqm

eqmeqm

a

H

HA

AOH

HA

AHK

+−+−+

===

Unit of Ka : Example : What is the concentration of H

+(aq), OH

-(aq), CH3COOH(aq), CH3COO

-(aq) and pH of 0.1 M ethanoic acid (Ka = 1.75 x

10-5 mol dm

-3) ? What is the degree of dissociation ?

CH3COOH(aq) CH3COO-(aq) + H

+(aq)

Initial 0.1 0 0

At eqm. 0.1 - y y y

52

3

3 1075.11.0][

]][[ −−+

=−

== xy

y

COOHCH

COOCHHKa

Solve with quadratic equation, Solve with assumption,

y = 1.31 x 10-3 Assume y << 0.1, (0.1 - y) ≈ 0.1 ⇒ 5

2

1075.11.0

−≈ xy

[H+] = 1.31 x 10

-3 mol dm

-3 ⇒ y

2 = 1.75 x 10

-6 or y = 1.32 x 10

-3

pH = 2.88 ⇒ pH = 2.88 Check of assumption (5% check) :

%5%32.1%1001.0

1032.1 3

<=−

xx

⇒ valid

[OH-] =

3

14

1031.1

100.1−

−

x

x = 7.63 x 10

-12 mol dm

-3

[CH3COO-] = [H

+] = 1.31 x 10

-3 mol dm

-3

[CH3COOH] = 0.1 - 1.31 x 10-3 = 0.0987 mol dm

-3

Degree of dissociation = 0131.01.0

1031.1

ionconcentrat totaloriginal

ddissociateion concentrat

moles of no. totaloriginal

ddissociate moles of no. 3

===−x

Remarks :

1. The use of assumption can always simplify the problem. However, the assumption must be checked (usually error < 5% is acceptable).

2. The contribution of H+ from self-ionization of water is assumed to be negligible. This is a valid assumption here.

Q [H+] from self-ionization of water << [H

+] from weak acid


Exercise : If now water is added to the above mixture to make the concentration of the ethanoic acid 0.01 M. Calculate the H+(aq), OH

-(aq), pH and degree of dissociation. How can you account for the change in degree of dissociation

(compared to 0.1 M solution) in terms of Le Chatelier’s Principle ? Solution :

Concentration CH3COOH(aq) H+(aq) + CH3COO

-(aq)

Initial 0.01 0 0

At eqm.

Ka = [H+] =

[OH

-] =

[CH3COO

-] =

[CH3COOH] = Degree of dissociation =

If the equilibrium is rewritten as CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O

+(aq)

Addition of water shifts equilibrium position to left/right ⇒ more/less acid molecules dissociate

⇒ degree of ionization ↑↑↑↑/↓↓↓↓ Remarks :

1. In weak acid, a dilution of 10 times does not change pH by 1 (contrast to strong acid).

2. Dilution in weak acid increases the degree of dissociation, this can be explained by Le Chatelier’s Principle. Exercise : The pH of 0.001 M solution of aminoethanoic acid (NH2CH2COOH) is 6.38 at 25

oC. Calculate the value of Ka at this

temperature. Solution :

Concentration

+

Initial

At eqm.

Ka =


Exercise : The degree of dissociation of a monobasic acid in 0.1 M solution at a certain temperature is 0.014. Calculate the pH and the dissociation constant of the acid at this temperature. Solution :

Concentration

+

Initial

At eqm.

pH = Ka =

5.2 Weak Base

The dissociation of weak base B(aq) in solution to form OH-(aq) and its conjugate acid (HB

+(aq)) can be represented

by an equilibrium :

B(aq) + H2O(l) HB+(aq) + OH

-(aq)

with equilibrium constant eqmeqm

eqmeqm

CBOH

HBOHK

][][

][][

2

+−

=

Since [H2O] is a constant, we can define base dissociation constant Kb for any weak base as

eqm

eqm

base][

acid] conjugate[][

][

][][ eqm

eqm

eqmeqm

b

OH

B

HBOHK

−+−

==

Unit of Kb : Exercise : The pH of a 0.50 M solution of ammonia at 298 K is 11.48. Calculate the base dissociation constant at 298 K for the reaction :


-(aq)

(At 298 K, Kw = 1.0 x 10-14 mol

2 dm

-6) (89, I)

Solution : [H

+]eqm =

[OH

-]eqm =

Concentration NH3 + H2O NH4+ + OH

-

Initial

At eqm.

Kb =


Exercise : When excess solid Mg(OH)2 is shaken with 1.00 dm

3 of 1.00 M NH4Cl solution, the resulting saturated solution has

a pH of 9.0 at 298 K. (i) Calculate the concentration of the OH

- ion.

(ii) Given that Kb for ammonia is 1.80 x 10-5 mol dm

-3 at 298 K, calculate the concentrations of NH3 and the NH4

+

ion. (87, II) Solution : (i) [H

+] =

[OH

-] =

(ii)


-

Initial

At eqm.

Kb =

⇒ [NH3] = [NH4

+] =


5.3 Magnitude of Ka and Kb

Acid : HA(aq) + H2O(l) H3O+(aq) + A

-(aq)

eqm

eqmeqm

aHA

AOHK

][

][][ 3−+

=

Base : B(aq) + H2O(l) OH-(aq) + HB

+(aq)

eqm

eqmeqm

bB

HBOHK

][

][][ +−

=

Larger Ka/Kb ⇒ forward reaction is more/less complete ⇒ stronger/weaker acid/base

Smaller Ka/Kb ⇒ forward reaction is more/less complete ⇒ stronger/weaker acid/base

Larger pKa/pKb ⇒ larger/smaller Ka/Kb ⇒ stronger/weaker acid

Smaller pKa/pKb ⇒ larger/smaller Ka/Kb ⇒ stronger/weaker acid

5.4 Relationship between Ka and Kb For an acid (HA), the basic properties of its conjugate base (A

-) can also be expressed by an equilibrium :

A-(aq) + H2O(l) HA(aq) + OH

-(aq)

eqm

eqmeqm

bA

HAOHAK

][

][][)(

−

−

−=

This Kb of A- is related to Ka of HA.

eqmeqmeqm

eqmeqm

w

b

OHHx

A

HAOH

K

AK

][][

1

][

][][)(

−+−

−−

=

)(

1

][][

][)(

HAKHA

HA

K

AK

aeqmeqm

eqm

w

b ==+−

−

⇒ )(

)(HAK

KAK

a

wb =

−

Similarly Ka of HB+ is related to Kb of HB.

)()(

BK

KHBK

b

wa =

+ (Verify it yourself)

Thus for a conjugate acid-base pair, Ka Kb = Kw log Ka + log Kb = log Kw -log Ka -log Kb = -log Kw

pKa + pKb = pKw

pKa + pKb = 14 at 25oC


Exercise : Calculate the pH of a 0.10 M aqueous solution of sodium ethanoate. The Ka value for ethanoic acid is 1.80 x 10

-5

mol dm-3, and Kw for water at 298 K is 1.00 x 10

-14 mol

2 dm

-6.

Kb for CH3COO

- =

Concentration CH3COO- + H2O CH3COOH + OH

-

Initial

0 0

At eqm.

Kb = [OH

-] =

[H+] =

pH =


Example : A solution is prepared by dissolving potassium hydrogencarbonate in water at 298 K. Write chemical equations for four equilibrium reactions, each involving the hydrogencarbonate ion, that occur in the solution and calculate the value of the equilibrium constant for each reaction. Given that at 298 K : K1 = 4.3 x 10

-7 mol dm

-3; K2 = 4.8 x 10

-11 mol dm

-3 for H2CO3

Kw = 1.0 x 10-14 mol

2 dm

-6 for H2O (93, I)

Solution :

HCO3- is amphoteric. 323

23 COHHCOCO

HH →← →←++ ±−±−

Species in solution :

With H3O+ : HCO3

-(aq) + H3O

+(aq)

K =

With OH- : HCO3

-(aq) + OH

-(aq)

K =

With H2O : HCO3-(aq) + H2O(l)

K =

With H2O : HCO3-(aq) + H2O(l)

K = Is a solution of potassium hydrogencarbonate acidic or alkaline ? Ka(HCO3

-) = Kb(HCO3

-) =

Since Ka >/< Kb, the solution is acidic/alkaline.


5.5 Calculations Involving Polybasic Weak Acid

For a polybasic weak acid, each ionization step is governed by an equilibrium with a specific Ka.

H3PO4(aq) H2PO4-(aq) + H

+(aq) Ka1 =

H2PO4-(aq) HPO4

2-(aq) + H

+(aq) Ka2 =

HPO42-(aq) PO4

3-(aq) + H

+(aq) Ka3 =

After each removal of H

+, the resulting species has one more negative charge

⇒ further removal of H+ will be progressively more/less difficult due to increased/decreased electrostatic

attraction

⇒ Ka1 >/< Ka2 >/< Ka3 Together with the ionic product of water, in a solution of phosphoric acid, there are a total of 4 equilibria, all these relationships have to be satisfied. Example : Calculate the concentration of all ionic species present in 0.05 M phosphorus acid. Although three ionization processes produce H

+, as Ka1 > Ka2 > Ka3, it is reasonable to assume that all H

+ come

solely from the first ionization. The assumption will be checked.

Concentration H3PO4 H+ + H2PO4

-

Initial 0.05 0 0

At eqm. 0.05 - x x x

Ka1 = x

x

−05.0

2

= 7.1 x 10-3

Assume 0.05 – x ≈ 0.05 ⇒ 05.0

2x = 7.1 x 10

-3 ⇒ x = 0.0188

Assumption is not valid.

Solve by quadratic equation, ⇒ x = 0.0156 [H+] = [H2PO4

-] = 0.0156 mol dm

-3

This [H

+] can be used to calculate [HPO4

2-] and [PO4

3-].

Since Ka1 > Ka2, we may assume that the amount of H2PO4

- and H

+ after dissociation is essentially the same as the

one just calculated.

Concentration H2PO4- H

+ + HPO4

2-

Initial 0.0156 0.0156 0

At eqm. 0.0156 - y 0.0156 + y y

Ka2 = y

yy

−

+

0156.0

)0156.0( = 6.3 x 10

-8

Assume 0.0156 ± y ≈ 0.0156 ⇒ y = 6.3 x 10-8

[HPO42-] = 6.3 x 10

-8 mol dm

-3

Notice that [HPO42-] << [H

+], assumption is valid.


Again as Ka2 > Ka3, we may assume that the amount of HPO42- and H

+ after dissociation is essentially the same as

the one just calculated.

Concentration HPO42- H

+ + PO4

3-

Initial 6.3 x 10-8

0.0156 0

At eqm. 6.3 x 10-8 - z 0.0156 + z z

Ka3 = zx

zz

−

+−8

103.6

)0156.0( = 4.2 x 10

-13

Assume z << 6.3 x 10-8, ⇒

8103.6

)0156.0(−

x

z = 4.2 x 10

-13 ⇒ z = 1.70 x 10

-18

[PO43-] = 1.70 x 19

-18 mol dm

-3

Notice that [PO4

3-] << [H

+], assumption is valid.

Lastly [OH-] =

][ +H

Kw = 0156.0

101 14−x = 6.41 x 10

-13 mol dm

-3

Example :

The dissociation constant of H2PO4-(aq) in water is 6.2 x 10

-8 mol dm

-3 at 298 K, where

][

]][[

42

24

−

+−

=POH

HHPOK a .

Calculate the pH at 298 K of (a) an aqueous solution of 0.10 M NaH2PO4(aq). (b) an aqueous solution of 0.05 M NaH2PO4(aq) and 0.05 M Na2HPO4(aq). (The concentrations of H3PO4 and PO4

3- are assumed to be negligible in these solutions.)

Solution :

The equation for equilibrium : H2PO4- HPO4

2- + H

+

(a)

Concentration H2PO4- HPO4

2- + H

+

Initial 0.10 0 0


822

42

24 102.6

10.010.0][

]][[ −

−

+−

=≈−

== xx

x

x

POH

HHPOK a ⇒ x = 7.874 x 10

-5

[H+] = 7.874 x 10

-5 M ⇒ pH = 4.10

(b)

Concentration H2PO4- HPO4

2- + H

+

Initial 0.05 0.05 0

At eqm. 0.05 - x 0.05 + x x

8

42

24 102.6

05.0

05.0

05.0

)05.0(

][

]][[ −

−

+−

==≈−

+== xx

x

x

xx

POH

HHPOK a

[H+] = 6.2 x 10

-8 M ⇒ pH = 7.21


Exercise : The following reversible reaction occurs in an aqueous solution of ammonium ethanoate :

CH3COO-(aq) + NH4

+(aq) CH3COOH(aq) + NH3(aq) (I)

(i) Write an expression for the dissociation constant, Ka, of ammonium ion. (ii) Calculate the equilibrium constant of reaction (I) at 298 K. (iii) For a 0.10 M solution of ammonium ethanoate at 298 K, calculate (1) the concentration of ammonia, and (2) the pH of the solution. (At 298 K, the dissociation constants of ethanoic acid and ammonium ion are 1.76 x 10

-5 mol dm

-3 and 5.59 x 10

-10

mol dm-3 respectively.) (95, II)

Solution : Exercise : The pH of a 0.02 M ethanedioic acid solution (H2C2O4) is 1.8. Calculate the concentration of each species present in the solution at 25

oC. (90, II)

[Dissociation constants of ethanedioic acid at 25oC are : K1 = 6.5 x 10

-2 mol dm

-3, K2 = 6.1 x 10

-5 mol dm

-3]

Solution :


6. Calculations on Acid-Base Equilibria

6.1 Procedure

1. Identify the conjugate acid/base pair.

2. Write appropriate equilibrium equation using the acid/base given and water.

3. Use the appropriate K value, Ka for acid; Kb for base.

4. Find initial concentrations of species before equilibrium is established.

5. Define the change required to achieve equilibrium; hence calculate the equilibrium concentrations.

6. Relate equilibrium concentrations to the appropriate K.

7. Solve the equation, making reasonable assumption if necessary.

8. Check any assumption to make sure it is valid. 9. Convert to quantities required in question. In general, acid/base calculations are of three general types.

A. Only acid is present initially.

B. Only base is present initially.

C. Both acid-base pair are present initially.

6.2 Acid Only (Initially)

Calculate the pH of a solution of 0.10 M CH3(CH2)2COOH, whose Ka = 1.5 x 10-5 mol dm

-3.

Step 1 Conjugate acid/base pair CH3(CH2)2COOH/CH3(CH2)2COO-

Step 2 Equilibrium equation Since acid is present initially, its reaction with water will be

CH3(CH2)2COOH + H2O CH3(CH2)2COO- + H3O

+

Step 3 Write K expression ])([

])(][[

223

2233

COOHCHCH

COOCHCHOHK a

−+

=

Step 4 Find initial conc. CH3(CH2)2COOH + H2O CH3(CH2)2COO- + H3O

+

init. conc. 0.10 0 0

Step 5 Define eqm. conc. eqm. conc. 0.10 - x x x

Step 6 Relate to K 52

223

2233 105.110.0])([

])(][[ −−+

=−

== xx

x

COOHCHCH

COOCHCHOHK a

Step 7 Solve eqn. Assume x << 0.10, i.e. 0.10 - x ≈≈≈≈ 0.10

52

105.110.0

−= xx

⇒⇒⇒⇒ x = 1.225 x 10-3

Step 8 Check assumpt. Since 1.225 x 10-3 << 0.10, assumption is valid.

Step 9 Calculate pH [H+] = x = 1.225 x 10

-3 M

pH = -log [H+] = -log (1.225 x 10

-3) = 2.91

In short, the steps will be :


+


eqm. conc. 0.10 - x x x

52

223

2233 105.110.0])([

])(][[ −−+

=−

== xx

x

COOHCHCH

COOCHCHOHK a

Assume x << 0.10, i.e. 0.10 - x ≈≈≈≈ 0.10

52

105.110.0

−= xx

⇒⇒⇒⇒ x = 1.225 x 10-3

Since 1.225 x 10-3 << 0.10, assumption is valid.

[H+] = x = 1.225 x 10

-3 M pH = -log [H

+] = -log (1.225 x 10

-3) = 2.91


6.3 Base Only (Initially)

Calculate the pH of a solution by dissolving 0.10 mole CH3(CH2)2COO-Na

+ in 1 dm

3, Ka of CH3(CH2)2COOH =

1.5 x 10-5 mol dm

-3.

Step 1 Conjugate acid/base pair CH3(CH2)2COOH/CH3(CH2)2COO-

Step 2 Equilibrium equation Since the salt ionized completely, CH3(CH2)2COO- is

present initially, its reaction with water will be

CH3(CH2)2COO- + H2O CH3(CH2)2COOH + OH

-

Step 3 Write K expression ])([

])(][[

223

223

−

−

=COOCHCH

COOHCHCHOHKb

Step 4 Find initial conc. CH3(CH2)2COO- + H2O CH3(CH2)2COOH + OH

-


Step 5 Define eqm. conc. eqm. conc. 0.10 - x x x

Step 6 Relate to K 102

223

223 10667.610.0])([

])(][[ −

−

−

==−

== xK

K

x

x

COOCHCH

COOHCHCHOHK

a

wb

Step 7 Solve eqn. Assume x << 0.10, i.e. 0.10 - x ≈≈≈≈ 0.10

102

10667.610.0

−= xx

⇒⇒⇒⇒ x = 8.165 x 10-6

Step 8 Check assumpt. Since 8.165 x 10-6 << 0.10, assumption is valid.

Step 9 Calculate pH [OH-] = x = 8.165 x 10

-6 M

pOH = -log [OH-] = -log (8.165 x 10

-6) = 5.09

pH = 14 - pOH = 8.91 In short, the steps will be :


-


eqm. conc. 0.10 - x x x

102

223

223 10667.610.0])([

])(][[ −

−

−

==−

== xK

K

x

x

COOCHCH

COOHCHCHOHK

a

wb

Assume x << 0.10, i.e. 0.10 - x ≈≈≈≈ 0.10

102

10667.610.0

−= xx

⇒⇒⇒⇒ x = 8.165 x 10-6


[OH-] = x = 8.165 x 10

-6 M

pOH = -log [OH-] = -log (8.165 x 10

-6) = 5.09

pH = 14 - pOH = 8.91


6.4 Both Conjugate Acid-Base Pair (Initially)

Calculate the pH of a solution made up by 500 cm3 of 0.10 M of CH3(CH2)2COOH and 500 cm

3 of 0.10 M

CH3(CH2)2COO-Na

+.

We may use acid dissociation or base dissociation, as long as the dissociation constant is appropriate.

(1) Use acid dissociation, use Ka, calculate H+ and get pH.


+

init. conc. 0.05 0.05 0

eqm. conc. 0.05 - x 0.05 + x x

5

223

2233 105.105.0

)05.0(

])([

])(][[ −−+

=−

+== x

x

xx

COOHCHCH

COOCHCHOHK a

Assume x << 0.05, i.e. 0.05 ±±±± x ≈≈≈≈ 0.05

5105.105.0

)05.0( −= xx

⇒⇒⇒⇒ x = 1.5 x 10-5


[H+] = x = 1.5 x 10

-5 M

pH = -log [H+] = -log (1.5 x 10

-5) = 4.82

(2) Use base dissociation, use Kb, calculate OH- and get pOH.


-

init. conc. 0.05 0.05 0

eqm. conc. 0.05 - x 0.05 + x x

10

223

223 10667.605.0

)05.0(

])([

])(][[ −

−

−

==−

+== x

K

K

x

xx

COOCHCH

COOHCHCHOHK

a

wb

Assume x << 0.05, i.e. 0.05 ±±±± x ≈≈≈≈ 0.05

1010667.605.0

05.0 −= xx

⇒⇒⇒⇒ x = 6.667 x 10-10


[OH-] = x = 6.667 x 10

-10 M

pOH = -log [OH-] = -log (6.667 x 10

-10) = 9.18

pH = 14 - pOH = 4.82

6.5 Reminders in Acid-Base Equilibrium Calculations

1. The ionic product of water [H+] [OH

-] = Kw always holds.

2. All equilibria hold, i.e. all equilibrium expressions must be satisfied. E.g. : For a tribasic acid, there are three dissociation as represented by three equilibria. All three equilibrium

expressions are satisfied by the concentrations of the species involved.

3. Calculations must be done using equilibrium concentrations.

4. In a solution, there is only one concentration for each species, which equals to the sum from all

contributions for that species. Equilibrium calculations are based on that concentration. E.g. : In dibasic acids, H

+ comes from ionization of water, first dissociation and second dissociation of acid.

However, there is only one [H+], which equals to sum of all of the above processes. All equilibria

involving H+ will make use of this concentration.

5. Acid-base equilibria proceed very fast and can be assumed to have established immediately (i.e. time

required for attainment of equilibrium is negligible).


7. Buffers

7.1 Introduction

Definition : Buffers are solutions that resist change in pH when small amounts of acid, base or water are added to it.

Importance of Buffers For many chemical reactions (especially those in living organisms with enzymes), they can only proceed within a narrow range of pH. Example : Normal human blood has a pH range of 7.3 - 7.5 and a variation of 0.4 unit could prove fatal. Types of Buffers It can be divided into 2 classes : 1. Acidic buffer : made by a solution of weak acid and its salt (which is a weak base) (e.g. CH3COOH & CH3COO

-Na

+)

2. Alkaline buffer : made by a solution of weak base and its salt (which is a weak acid) (e.g. NH3 & NH4

+Cl-)

Principle of Buffers In a buffer there are appreciable amounts of weak acid and weak base. In acidic buffer, they are the weak acid (CH3COOH) and its conjugate base (CH3COO

-).

In alkaline buffer, they are the weak base (NH3) and its conjugate acid (NH4+).

When a strong acid is added, it will react and thus be removed by the weak base in solution. The weak base is converted to its conjugate weak acid. When a strong base is added, it will react and thus be removed by the weak acid in solution. The weak acid is converted to its conjugate weak base.

Example : CH3COOH/CH3COO

- buffer

Addition of HCl : Addition of NaOH :

⇒ CH3COO- + HCl → CH3COOH + Cl

- ⇒ CH3COOH + NaOH → CH3COO

- + Na

+ + H2O

⇒ [weak acid] ↑, [weak base] ↓ ⇒ [weak acid] ↓, [weak base] ↑ The addition of strong acid/base will change the relative amount of weak acid and weak base in buffer. The pH change is then governed by the equilibrium of the conjugate acid/base pair.


7.2 Buffer Calculations

It is just a modification of the case where both conjugate acid/base pair are present initially. The added strong acid/base can be assumed to be removed completely and immediately upon addition, thus changing the initial concentrations of the conjugate acid/base pair. Example :

For a solution made up by 500 cm3 of 0.10 M of CH3(CH2)2COOH and 500 cm

3 of 0.10 M CH3(CH2)2COO

-Na

+,

(1) Calculate the pH of the solution after addition of 0.001 mole of gas HCl. The added strong acid can be assumed to be completely removed by the weak base present, thus changing [CH3(CH2)2COOH]init. and [CH3(CH2)2COO

-]init. accordingly.

HCl + CH3(CH2)2COO- → CH3(CH2)2COOH + Cl

-


+

init. conc. 0.05 + 0.001 0.05 - 0.001 0

= 0.051 = 0.049

eqm. conc. 0.051 - x 0.049 + x x

5

223

2233 105.1051.0

)049.0(

])([

])(][[ −−+

=−

+== x

x

xx

COOHCHCH

COOCHCHOHK a

Assume x << 0.049, i.e. 0.051 - x ≈≈≈≈ 0.051; 0.049 + x ≈≈≈≈ 0.049

5105.1051.0

)049.0( −= xx

⇒⇒⇒⇒ x = 1.561 x 10-5


[H+] = x = 1.561 x 10

-5 M

pH = -log [H+] = -log (1.561 x 10

-5) = 4.81 makes sense as this is lower than before

(2) Calculate the pH of the solution after addition of 0.001 mole of solid NaOH. The added strong base can be assumed to be completely removed by the weak acid present, thus changing [CH3(CH3)2COOH]init. and [CH3(CH2)2COO

-]init. accordingly.

NaOH + CH3(CH2)2COOH → CH3(CH2)2COO- + Na

+ + H2O


+

init. conc. 0.05 - 0.001 0.05 + 0.001 0

= 0.049 = 0.051

eqm. conc. 0.049 - x 0.051 + x x

5

223

2233 105.1049.0

)051.0(

])([

])(][[ −−+

=−

+== x

x

xx

COOHCHCH

COOCHCHOHK a

Assume x << 0.049, i.e. 0.051 + x ≈≈≈≈ 0.051; 0.049 - x ≈≈≈≈ 0.049

5105.1049.0

)051.0( −= xx

⇒⇒⇒⇒ x = 1.441 x 10-5


[H+] = x = 1.441 x 10

-5 M

pH = -log [H+] = -log (1.441 x 10

-5) = 4.84 makes sense as this is higher than before

Remarks :

1. The added strong acid/base will change the concentrations of the weak acid & base to by the same

amount.

If strong acid is added, [weak acid] ↑↑↑↑ and [weak base] ↓↓↓↓ by the same amount.

If strong base is added, [weak acid] ↓↓↓↓and [weak base] ↑↑↑↑ by the same amount.

2. Always check if the answer makes sense; the addition of strong acid should lower the pH whereas the

addition of strong base should raise the pH.


Exercise : Kb for NH3 = 1.8 x 10-5 mol dm

-3.

(1) Calculate the pH of a 0.20 M solution of NH3. (2) Calculate the pH of a 1 dm

3 solution with 0.40 mole NH4Cl.

(3) Calculate the pH of a solution by mixing 600 cm3 of solution in (1) and 400 cm

3 of solution in (2).

(4) Calculate the new pH of the solution in (3) when 0.01 mole of solid NaOH is added to it. (5) Calculate the new pH of the solution in (3) when 0.0005 cm

3 of 18M HCl is added to it. (You may neglect the

increase in volume after addition.) Solution : (1)


-

Initial 0.20 0 0

At eqm. 0.20 – x x x

52

3

4 108.120.0][

]][[ −−+

=−

== xx

x

NH

OHNHKb

Assume 0.20 >> x, i.e. 0.20 - x ≈ 0.20, 52

108.120.0

−= xx

⇒ x = 1.897 x 10-3


[OH-] = 1.897 x 10

-3 M ⇒ pOH = 2.72 ⇒ pH = 14 - 2.72 = 11.28

(2)

Concentration NH4+ + H2O NH3 + H3O

+

Initial 0.40 0 0


10

5

142

4

3 10556.5108.1

101

40.0][

]][[ −

−

−

+

+

===−

== xx

x

K

K

x

x

NH

HNHK

b

wa

Assume 0.40 >> x, i.e. 0.40 - x ≈ 0.40, 52

10556.540.0

−= xx

⇒ x = 1.491 x 10-5


[H+] = 1.491 x 10

-5 M ⇒ pH = 4.83

(3)


-

Initial 0.2x0.6/1 = 0.12 0.4x0.4/1 = 0.16 0

At eqm. 0.12 - x 0.16 + x x

5

3

4 108.112.0

)16.0(

][

]][[ −−+

=−

+== x

x

xx

NH

OHNHKb

Assume 0.12 >> x, i.e. 0.12 - x ≈ 0.12 and 0.16 + x ≈ 0.16, 5108.112.0

16.0 −= xx

⇒ x = 1.35 x 10-5


[OH-] = 1.35 x 10

-3 M ⇒ pOH = 4.87 ⇒ pH = 14 - 4.87 = 9.13


(4)


-

Initial 0.12 + 0.01 = 0.13 0.16 - 0.01 = 0.15 0

At eqm. 0.13 - x 0.15 + x x

5

3

4 108.113.0

)15.0(

][

]][[ −−+

=−

+== x

x

xx

NH

OHNHKb

Assume 0.13 >> x, i.e. 0.13 - x ≈ 0.13 and 0.15 + x ≈ 0.15, 5108.113.0

15.0 −= xx

⇒ x = 1.56 x 10-5


[OH-] = 1.56 x 10

-3 M ⇒ pOH = 4.81 ⇒ pH = 14 - 4.81 = 9.19

(5) No. of moles of HCl = 18 x 0.0005 = 9 x 10

-3


-

Initial 0.12 - 9x10-3 = 0.111 0.16 + 9x10

-3 = 0.169 0

At eqm. 0.111 - x 0.169 + x x

5

3

4 108.1111.0

)169.0(

][

]][[ −−+

=−

+== x

x

xx

NH

OHNHKb

Assume 0.111 >> x, i.e. 0.111 - x ≈ 0.111 and 0.169 + x ≈ 0.169,

5108.1111.0

169.0 −= xx

⇒ x = 1.182 x 10-5


[OH-] = 1.182 x 10

-5 M ⇒ pOH = 4.93 ⇒ pH = 14 - 4.93 = 9.07


7.3 Short Cut in Buffer Calculations

7.3.1 Acidic Buffer Take an acidic buffer as example.

HA + H2O H3O+ + A

-

eqm

eqmeqm

aHA

AHK

][

][][ −+

=

)][

][(][

eqm

eqm

aeqmA

HAKH

−

+= ←←←← [H

+] depends on ratio )

][

][(

eqm

eqm

A

HA

−

)][

][log(]log[

eqm

eqma

eqmA

HAKH

−

+=

)][

][log(log]log[

eqm

eqm

aeqmA

HAKH

−

++=

)][

][log(log]log[

eqm

eqm

aeqmA

HAKH

−

+−−=−

eqm

eqm

aHA

ApKpH

][

][log

−

+=

Henderson-Hasselbalch equation eqm

eqm

aacid

basepKpH

][

][log+=

Usually, as seen from above examples, [A-]eqm ≈≈≈≈ [A

-]init. and [HA]eqm ≈≈≈≈ [HA]init..

Thus .

.

][

][log

init

inita

HA

ApKpH

−

+≈ . This is very often a valid assumption.

After addition of strong acid/base, the ratio .

.

][

][

init

init

HA

A−

changes but usually only by a little.

⇒ the solution is resistant to pH change

From )][

][log(

eqm

eqm

aacid

basepKpH += , there are several results.

1. The pH of the buffer depends only on the ratio of the concentrations of acid and base, not on their actual values.

2. When [acid]eqm = [base]eqm, pH = pKa. Thus pKa equal to the pH of a solution made from a solution of the acid which has been half neutralized by a strong base.

3. Addition of water does not change the pH of the buffer.

Addition of water changes [acid] and [base] to the same extent, so the ratio ][

][

acid

base is not changed.

4. Once the weak acid/weak base in the buffer is completely used up, the buffering action will be gone. Example : In a solution with CH3COOH/CH3COO

-,

when enough strong base (e.g. NaOH) is added which reacts all CH3COOH, or when enough strong acid (e.g. HCl) is added which reacts all CH3COO

-, then

the solution is no longer a buffer and its resistance to pH change is gone.


5. For a buffer with larger values of [acid] and [base], when strong acid/base is added, change in ][

][

acid

base will

be smaller and thus the buffer is more resistant to change its pH. The buffer is said to have higher buffering capacity.

Example : For two buffers (1) [acid] = 0.1 M, [base] = 0.1 M (2) [acid] = 0.01 M, [base] = 0.01 M

][

][

acid

base= 1 for both buffers and thus pH = pKa.

Addition of 0.001 mole of NaOH to 1 dm3 of each buffer separately,

for (1), 020.1099.0

101.0

001.01.0

001.01.0

][

][==

−

+=

acid

base, pH pKa= + log( . )1020

for (2), 222.1009.0

011.0

001.001.0

001.001.0

][

][==

−

+=

acid

base, pH pKa= + log( . )1222

pH change in (2) will be larger than that in (1), buffer (1) is said to have a higher buffering capacity. Remarks : 1. There is a similar expression for pKb, pOH & [acid]eqm and [base]eqm.

B + H2O HB+ + OH

-

eqm

eqmeqm

aB

OHHBK

][

][][ −+

=

)][

][(][

eqm

eqm

beqmHB

BKOH

+

−= ←←←← [OH

-] depends on ratio )

][

][(

eqm

eqm

HB

B

+

)][

][log(]log[

eqm

eqmb

eqmHB

BKOH

+

−=

)][

][log(log]log[

eqm

eqm

beqmHB

BKOH

+

−+=

)][

][log(log]log[

eqm

eqm

beqmB

HBKOH

+

− −−=−

eqm

eqm

bB

HBpKpOH

][

][log

+

+=

Henderson-Hasselbalch equation eqm

eqm

bbase

acidpKpOH

][

][log+=

2. The expression )][

][log(

eqm

eqm

aacid

basepKpH += is also applicable to alkaline buffer. Make sure to use Ka

(usually Kb is given and it has to be converted to Ka first). Example : Consider an alkaline buffer NH3/NH4

+.

We may treat the problem

either as or

NH4+ + H2O NH3 + H3O

+ NH3 + H2O NH4

+ + OH

-

acid base base acid

eqm

eqm

aacid

basepKpH

][

][log+=

eqm

eqm

bbase

acidpKpOH

][

][log+=

⇒

eqm

eqm

aNH

NHNHpKpH

][

][log)(

4

3

4 +

++=

eqm

eqm

bNH

NHNHpKpOH

][

][log)(

3

4

3

+

+=


Example : A solution X is formed by mixing 0.50 dm

3 of 0.30 M methanoic acid with 0.50 dm

3 of 0.15 M sodium hydroxide.

(i) Given that Ka for methanoic acid is 1.75 x 10-4 mol dm

-3 at 298 K, calculate the pH of solution X.

(ii) What special property does solution X have ? What is it commonly known as ? (87, II) Solution : The substances present are methanoic acid (weak acid) and sodium methanoate (its conjugate base). (i)

HCOOH HCOO- + H

+

After mixing 075.0

1

15.05.03.05.0=

− xx

075.0

1

15.05.0=

x

0

At eqm. 0.075 - x 0.075 + x x

Kx x

x

xx xa =

+

−≈ = = −( . )

.

.

..

0 075

0 075

0 075

0 075175 10 4

⇒ [H+] = 1.75 x 10

-4 M ⇒ pH = 3.76

(ii) The solution has the ability to resist a change in pH when small amounts of either acid or base are added. The

solution is generally known as a buffer solution. Example : The following solutions can act as a buffer : (I) a solution with [CH3COOH] = 0.100 M and [CH3COO

-] = 0.075 M

(II) a solution with [CH3COOH] = 1.000 M and [CH3COO-] = 0.750 M

(i) Calculate the approximate pH value of (I) and (II). [Ka = 1.85 x 10

-5 mol dm

-3 for CH3COOH]

(ii) Which of the above solutions is the better buffer ? Explain your answer. (iii) Can a solution of ethanoic acid alone act as a buffer ? Explain your answer. (89, II) Solution :

(i) CH3COOH CH3COO- + H

+

)][

][log()

][

][log(

3

3

eqm

eqm

aeqm

eqm

aCOOHCH

COOCHpK

acid

basepKpH

−

+=+=

For I, [CH3COOH]eqm ≈ 0.100 M For II, [CH3COOH]eqm ≈ 1.000 M

[CH3COO-]eqm ≈ 0.075 M [CH3COO

-]eqm ≈ 0.750 M

pH = 4.61 pH = 4.61 (ii) Solution (II) is a better buffer than (I) because it has larger [CH3COOH] and [CH3COO

-] and thus has a larger

buffering capacity. (iii) An ethanoic acid solution alone cannot be a buffer because it contains little CH3COO

- ions (from the ionization

of ethanoic acid). Thus when a strong acid is added, the solution cannot effectively remove H+ ions and the pH

of the solution will change by a large amount.


Example : (i) At 298 K, the pH of 0.050 M CH3CH2COOH is 3.10. Calculate the Ka of CH3CH2COOH at 298 K. (ii) Calculate the pH at 298 K of a solution which is 0.050 M with respect to CH3CH2COOH and to CH3CH2COONa. (iii) 5.0 x 10

-4 mol of solid NaOH were added separately to

(1) 100 cm3 of 0.050 M CH3CH2COOH, and

(2) 100 cm3 of the solution in (ii) above.

Assuming that the change in volume upon the addition of NaOH is negligible, calculate the pH at 298 K of the solution in each case. Comment on the difference in the pH change of the two solutions. (96, II)

Solution : (i)

CH3CH2COOH CH3CH2COO- + H

+

Initial 0.050 0 0

At eqm. 0.050 – 10-3.1 10

-3.1 10

-3.1

35

1.3

1.31.3

dm mol1028.110050.0

1010 −−

−

−−

=−

= xx

K a

(ii) )][

][log()

][

][log(

23

23

eqm

eqm

aeqm

eqm

aCOOHCHCH

COOCHCHpK

acid

basepKpH

−

+=+=

[CH3CH2COOH]eqm ≈ 0.050 M, [CH3CH2COO-]eqm ≈ 0.050 M

89.4)050.0

050.0log()1028.1log( 5 =+−= −xpH

(iii) )][

][log(

23

23

eqm

eqm

aCOOHCHCH

COOCHCHpKpH

−

+=

For solution (1), Mxx

COOHCHCH eqm 045.0

1000100

105)1000

100(050.0][

4

23 =−

≈

−

Mx

COOCHCH eqm 005.0

1000100

105][

4

23 =≈−

−

pH = 3.93

For solution (2), Mxx

COOHCHCH eqm 045.0

1000100

105)1000

100(050.0][

4

23 =−

≈

−

Mxx

COOCHCH eqm 055.0

1000100

105)1000

100(050.0][

4

23 =+

≈

−

−

pH = 4.98 0.05 M CH3CH2COOH is not a buffer and thus pH changes significantly (0.83 unit) while the solution in (ii) is a

buffer and thus pH changes little (0.09 unit).


Exercise : 1. Calculate the volume of 0.1 M HCOO

-Na

+(aq) that must be added to 1 dm

3 of 0.1 M methanoic acid solution to

give a buffer solution of pH = 3.50. The pKa for methanoic acid is 3.75. Solution : Let V dm

3 be the volume of HCOO

-Na

+ required.

)][

][()

][

][log(

eqm

eqm

aeqm

eqm

aHCOOH

HCOOpK

acid

basepKpH

−

+=+=

1

11.0][

+=V

xHCOOH ;

1

1.0][

+=−

V

xVHCOO

)1

1.0(

)1

1.0(log75.35.3

+

++=

V

VV

⇒ log .V = −0 25 ⇒ V = 0.562 dm3

2. 1 dm

3 of a solution is obtained by mixing 500 cm

3 of 0.200 M HCl and 500 cm

3 of 0.700 M NaCN. Given that

the dissociation constant of HCN is 4.0 x 10-10 mol dm

-3, calculate the pH of this solution. (88, II)

Solution :

)][

][log()

][

][log(

eqm

eqm

aeqm

eqm

aHCN

CNpK

acid

basepKpH

−

+=+=

Mx

HCN eqm 10.01

)1000

500(200.0][ =≈

Mxx

CN eqm 25.01

)1000

500(200.0)1000

500(7.0][ =

−≈−

)10.0

25.0log()100.4log( 10 +−= −xpH ⇒ pH = 9.80

3. Given : Ka for CH3(CH2)2COOH = 1.5 x 10

-5 mol dm

-3 at 298 K.

Calculate the pH of (i) an aqueous solution of 0.1 M CH3(CH2)2COOH ; (ii) an aqueous solution of 0.05 M CH3(CH2)2COONa and 0.05 M CH3(CH2)2COOH ; and (iii) 1.0 dm

3 of the solution in (ii) after the addition of 1.0 x 10

-3 mol of solid NaOH. (94, II)

Solution : (i)

CH3(CH2)2COOH CH3(CH2)2COO- + H

+

Initial 0.1 0 0

At eqm. 0.1 - x x X

522

105.11.01.0

−=≈−

= xx

x

xK a ⇒ x = 1.225 x 10

-3

⇒ [H+] = 1.225 x 10

-3 M ⇒ pH = 2.91


(ii) )][

][log()

][

][log(

23

23

eqm

eqm

aeqm

eqm

aCOOHCHCH

COOCHCHpK

acid

basepKpH

−

+=+=

[CH3(CH2)2COOH]eqm ≈ 0.05 M

[CH3(CH2)2COO-]eqm ≈ 0.05 M

)05.0

05.0log()105.1log( 5 +−= −xpH ⇒ pH = 4.82

(iii) [CH3(CH2)2COOH]eqm ≈ 0.05 - 1.0 x 10-3 = 0.049 M

[CH3(CH2)2COO-]eqm ≈ 0.05 + 1.0 x 10

-3 = 0.051 M

)049.0

051.0log()105.1log( 5 +−= −xpH ⇒ pH = 4.84

Exercise : A weak base MOH has an ionization constant Kb = 2.0 x 10

-5 mol dm

-3, where Kb = [M

+][OH

-]/[MOH].

Solution S is made up of 0.10 mol of MOH in 1.0 dm3, and a second solution T has 0.10 mol of MOH and 0.50 mol

of MCl in 1.0 dm3.

(i) Calculate the pH of solutions S and T.

(ii) 0.01 mol of a strong acid HX is added separately to the solutions S and T. The new pH value for solution S is

10.26; calculate that for solution T. What conclusion can you draw from this result ? (Kw = 1.0 x 10

-14 mol

2 dm

-6) (91, II)

Solution :


8. Indicators

8.1 Introduction

Acid-base indicators are usually coloured organic compounds. They are either weak acids or weak bases,

whose colour depend on pH of the solution.

Example : HIn(aq) + H2O(l) H3O+(aq) + In

-(aq)

colour A colour B

Colour A has to be distinct from colour B.

With Henderson-Hasselbalch equation )][

][log(

eqm

eqm

aacid

basepKpH += , and treating indicators as weak acids gives

)][

][log(

eqm

eqm

aHIn

InpKpH

−

+= or aeqm

eqmpKpH

HIn

In−=

−

)][

][log(

The colour of an indicator depends on the relative amount of the two coloured species, i.e. [ ]

[ ]

In

HIn

−

, which in

terms depends on the pH of the solution.

8.2 Working Range of Indicator

The pH range in which an indicator functions (i.e. changes colour) is known as its working range. Working range

of an indicator depends on its Ka. Example : Phenolphthalein (HPh) has a pKa = 9.

HPn(aq) + H2O(l) H3O+(aq) + Pn

-(aq)

colourless red

At pH < 7, 2)][

][log( −<

−

HPn

Pn ⇒

100

1

][

][<

−

HPn

Pn ⇒ [HPn] > 100 [Pn

-]

⇒ colour will be that of HPn, i.e. colourless

At pH = 8, 1)][

][log( −=

−

HPn

Pn ⇒

10

1

][

][=

−

HPn

Pn ⇒ [HPn] = 10 [Pn

-]

⇒ colour of Pn- (red) becomes detectable

At pH = 9, 0)][

][log( =

−

HPn

Pn ⇒ 1

][

][=

−

HPn

Pn ⇒ [HPn] = [Pn

-]

⇒ equal amount of HPn and Pn-, red colour darkens

At pH = 10, 1)][

][log( =

−

HPn

Pn ⇒ 10

][

][=

−

HPn

Pn ⇒ 10 [HPn] = [Pn

-]

⇒ red colour fully developed


Experimental working range of some indicators

Indicator pKa pH range Colour change

Acidic Colour Alkaline Colour

Thymol blue (1st) 1.7 1.2 - 2.8 Red Yellow

Methyl orange 3.7 3.2 - 4.2 Red Yellow

Methyl red 5.1 4.2 - 6.3 Red Yellow

Thymol blue (2nd) 8.9 8.0 - 9.6 Yellow Blue

Phenolphthalein 9.3 8.2 - 10.0 Colourless Red

Remarks :

1. Most indicators have a detectable colour change over a range of about 2 pH units. This is a consequence

of the sensitivity of our eyes to colour changes. Thus the working range of an indicator is usually pKa ±±±± 1.

2. Some pH ranges are not exactly pKa ±±±± 1. This is because the completion of colour change for different indicator is different. We have assumed above that completion occurs when there is a ten-fold ratio between the two colour forms. This ratio may be different for different indicator.

3. Thymol blue has two working ranges. This is because it is a dibasic acid, with three different coloured forms.

H2In(aq) + H2O(l) H3O+(aq) + HIn

-(aq) HIn

-(aq) + H2O(l) H3O

+(aq) + In

2-(aq)

red yellow yellow blue Exercise : 1. Account for the fact that at 298 K, in a solution of pH 7.0, the indicator methyl orange shows its alkaline colour

(yellow), while phenolphthalein shows its acidic colour (colourless). (94, II) Solution : 2. An acid-base indicator, HIn, is a weak acid with dissociation constant Ka.

HIn(aq) + H2O(l) H3O+(aq) + In

-(aq)

yellow blue

(i) Show that ][

][log

)(

)(

aq

aq

aHIn

InpKpH

−

+=

(ii) Given that Ka of HIn is 1.0 x 10-4 mol dm

-3, deduce the colour of a solution containing the indicator

(I) at pH 3.9. (II) at pH 10.0. (00, II)

Solution :


9. Acid-Base Titrations

Equivalence Point : Point in titration at which equivalent amounts of acid and alkali have reacted completely according to the stoichiometric equation.

End-Point : Point in titration at which there is a marked colour change in its indicator used. Remark : It is customary to put acid in the burette for the titration. This is because bases like NaOH can attack glass joint in stop-cock in burette, thus blocking the jet. The titration curves for the four possible combinations are shown below.

STRONG ACID vs STRONG BASE STRONG ACID vs WEAK BASE

WEAK ACID vs STRONG BASE WEAK ACID vs WEAK BASE


If base is used in the burette, the titration curve will just be the opposite of the above ones.

For a suitable indicator, its working range must lie in the steep region when the pH changes significantly.


In summary :

Strong Acid vs Strong Base

Strong Acid vs Weak Base

Weak Acid vs Strong Base

Weak Acid vs Weak Base

Example HCl vs NaOH HCl vs NH3 CH3COOH vs NaOH CH3COOH vs NH3

pH at equivalence point

7 < 7 > 7 depends on acid and base used

pH change near equivalence point

4-10 3-7 7-11 depends on acid and base used

Suitable indicators All indicators methyl orange and methyl red

phenolphthalein no indicator

Example : Calculate the pH at the equivalence point when 25 cm

3 of 0.1 M CH3COOH is neutralized by 0.1 M NaOH. Ka of

CH3COOH is 1.75 x 10-5 mol dm

-3.

CH3COOH + NaOH → CH3COO-Na

+ + H2O

At equivalence point, 25 cm3 of NaOH has to be added.

Total volume at equivalence point = 50 cm3.

At equivalence point, CH3COO-Na

+ ionizes completely to give CH3COO

- and Na

+. CH3COO

- is related to its

conjugate acid by the following equilibrium and is thus basic.

CH3COO-

+ H2O CH3COOH + OH-

Initial 0.05 0 0


5

1422

3

3

1075.1

100.1

05.005.0][

]][[

−

−

−

−

==≈−

==x

x

K

Kx

x

x

COOCH

OHCOOHCHK

a

wb ⇒ x = 5.34 x 10

-6 M

[OH-] = 5.34 x 10

-6 M ⇒ pOH = 5.27 ⇒ pH = 8.73

Exercise : Calculate the pH at the equivalence point when 25 cm

3 of 0.05 M NH3 is neutralized by 0.1 M HCl. Kb of NH3 is

1.80 x 10-5 mol dm

-3.

Solution :


5.6.1.3 By conductivity measurements Changing ions concentrations of a solution changes electrical conductivity of the solution, so acid-alkali (carbonate) titrations can be followed by conductivity measurements. Example : Referring to the graph, At A : Conductivity is high due to presence of

large number of mobile OH- and Na+ ions from NaOH.

A to E : HCl added ionizes in water, giving H+ and Cl-. HCl(aq) → H+

(aq) + Cl-(aq)

H+ added combines with OH- to form H2O, which undergoes essentially no ionization.

H+(aq) + OH-

(aq) → H2O(l)

Although the no. of ions is not changed in solution (the OH- lost is replaced by Cl- added), since Cl- is less mobile than OH-, conductivity decreases.

At E : The end point. Conductivity is at a minimum when neutralization is complete. The conductivity at the end point is not zero as the solution contains Na+ and Cl- ions which conduct

electricity. E to B : Conductivity increases due to the addition of excess mobile H+ and Cl- ions from HCl.


Example : Conductivity measurement is one of the principle ways of detecting the end-points of weak acid-weak base titrations. By plotting conductivity against volume of weak acid added, account for the change in conductivity in terms of the dissociation of the weak acid and weak base. Explain how this plot can be used to determine the end-point in such a titration. (88, II) Conductivity is directly proportional to the concentration of mobile ions in solution.

Initially, the conductivity is low as weak base only dissociates slightly. Addition of weak acid neutralizes the weak base, the resulting salt ionizes completely and conductivity increases. At equivalence point, the base is completely neutralized by the acid added. Only salt is present and it ionizes completely giving the highest conductivity. This is recorded as the end-point of the titration. After that, addition of further acid which ionizes slightly will barely change the conductivity. Exercise : Temperature change and conductivity measurement can also be used to determine end-points in titrations. Draw on the following graphs, when sodium hydroxide is added to (1) a strong acid, e.g. HCl, and (2) a weak acid, e.g. CH3COOH.

0 0

temperature conductivity

NaOH added NaOH added


The titration of sodium carbonate with hydrochloric acid consists of two stages.

Stage 1 Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq)

pH at this equivalence point ≈ 8.5 ⇒ phenolphthalein as indicator

Stage 2 NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)

pH at this equivalence point ≈ 4.0 ⇒ methyl orange as indicator

Example : 25.0 cm

3 of a solution containing sodium carbonate and sodium hydrogencarbonate required 27.5 cm

3 of 0.80 M

hydrochloric acid to decolorize phenolphthalein. On addition of methyl orange, a further 32.5 cm3 of the acid were

required to turn the indicator to its neutral colour. Determine the concentration of sodium carbonate and sodium hydrogencarbonate in the solution. The titration of sodium carbonate consists of two stages :

Stage 1 Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + NaCl(aq)

Stage 2 NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g) Stage 1 : No. of moles of HCl needed to convert CO3

2- to HCO3

- =

No. of moles of Na2CO3 = Concentration of Na2CO3 = Stage 2 : No. of moles of HCl needed to neutralize all HCO3

- =

No. of moles of HCl used to neutralize HCO3

- coming from Na2CO3 =

No. of moles of HCl used to neutralize HCO3

- coming from NaHCO3 =

No. of moles of NaHCO3 in solution = Concentration of NaHCO3 =


10. Solubility Product

10.1 Introduction Consider a saturated solution of silver chloride that is in contact with solid silver chloride. Since there is undissolved silver chloride solid present, the solution is a solution. Macroscopically, the amount of solid silver chloride present changes/does not change with respect to time. The system seemingly has attained a completely state. However, microscopically, a equilibrium has in fact been established, with the rate of exactly the same as the rate of . The solubility equilibrium can be represented as

AgCl(s) Ag+(aq) + Cl

-(aq)

dissolution

precipitation As for any heterogeneous reactions, the concentration of a solid is a constant; it does not appear in the equilibrium expression.

Therefore, equilibrium constant for the dissociation of AgCl is Ksp = [Ag+]eqm [Cl

-]eqm, where Ksp is called the

solubility product constant or simply the solubility product. Example : Write the solubility products and their units for the following compounds. 1. MgF2 Ksp = Unit : 2. Ag2CO3 Ksp = Unit : 3. Ca3(PO4)2 Ksp = Unit : Remarks :

1. The value of Ksp is related to the solubility of an ionic compound. A smaller Ksp implies a less/more soluble compound in water.

2. Such equilibrium only holds when both components are present, i.e. there is undissolved solid present in a saturated solution.

3. There is/is no fixed unit for Ksp. It depends on the produced from dissolution. 10.2 Solubility and Solubility Product Solubility of a solute is either expressed in g dm

-3 or mol dm

-3. Solubility and solubility product can be inter-

converted fairly easily. Example: The solubility of calcium sulphate is found to be 0.67 g dm

-3. Calculate the value of Ksp for calcium sulphate.

CaSO4(s) ∏

[Ca

2+] = [SO4

2-] =

Ksp = Exercise : The solubility of bismuth sulphide (Bi2S3) is 1.0 x 10

-15 g dm

-3. Calculate the solubility product of this compound.


Example: Calculate the solubility of copper(II) hydroxide, Cu(OH)2 in g dm

-3. [Ksp(Cu(OH)2 = 2.2 x 10

-20 mol

3 dm

-9.]

Cu(OH)2(s) Cu2+(aq) + 2OH

-(aq)

At eqm. x

Ksp = 2.2 x 10

-20 =

⇒ x = Solubility of Cu(OH)2 = Exercise : Calculate the solubility of silver chloride in g dm

-3 and mol dm

-3. [Ksp(AgCl) = 1.7 x 10

-10 mol

2 dm

-6.]

10.3 Ionic Product When a ionic solid is added to an aqueous solution, there can be three possibilities :

(1) the solution is unsaturated (2) the solution is saturated (3) the solution is supersaturated

For concentrations that do not correspond to equilibrium conditions, reaction quotient rather than solubility

product can be calculated. In solubility situation, we called the reaction quotient (Q) ionic product. For AgCl solution, ionic product = [Ag

+]0 [Cl

-]0

Remarks: Note that Q has the same form as Ksp except that the concentrations of ions are not equilibrium concentrations.

Ionic product is useful to predict whether a precipitate will form. The possible relationships between Q and Ksp are

Ionic product < Ksp ⇒ Unsaturated/Saturated/Supersaturated solution (no precipitation)

Ionic product = Ksp ⇒ Unsaturated/Saturated/Supersaturated solution (no precipitation yet)

Ionic product > Ksp ⇒ Unsaturated/Saturated/Supersaturated solution; AgCl will precipitate out until the ionic product = Example : Exactly 200 cm

3 of 0.0040 M BaCl2 are mixed with exactly 600 cm

3 of 0.0080 M K2SO4. Will a precipitate form ?

[Ksp(BaSO4) = 1.1 x 10-10 mol

2 dm

-6]

[Ba

2+] =

[SO4

2-] =

[Ba

2+] [SO4

2-] =

The solution is unsaturated/saturated/supersaturated. Some of the BaSO4 will out of the solution until


10.4 The Common Ion Effect and Solubility Suppose there is a saturated solution of AgCl(aq).


-(aq)

If AgNO3(aq) is added, since AgNO3 is essentially completely ionized,

AgNO3(s) →OH2 Ag

+(aq) + NO3

-(aq)

Since both species gives Ag

+(aq) upon dissolution, [Ag

+] = contribution from AgCl + contribution from AgNO3.

Such action has increased [Ag

+], making the ionic product greater than Ksp, and some AgCl has to precipitate out,

until the ionic product decreases to Ksp. This precipitation effect is known as the common ion effect. The phenomenon can also be explained by Le Chatelier’s principle :

An increase in [Ag+] ⇒ system responses by increasing/decreasing [Ag

+]

⇒ shifting equilibrium position to the left/right

⇒ more/less AgCl(s) is precipitated out Common Ion Effect is the effect of adding a common ion which causes a decrease in the solubility of the salt in the solution. Remarks : Notice that since [Ag

+] came from two sources, [Cl

-] no longer equals to [Ag

+].

In terms of the equilibrium law, the relationship Ksp = [Ag+] [Cl

-] still holds.

Example Calculate the solubility of silver chloride (in mol dm

-3) in a 6.5 x 10

-3 M silver nitrate solution.

[Ksp(AgCl) = 1.7 x 10-10 mol

2 dm

-6.]


-(aq)

At eqm. x

Ksp = =

⇒ x = Solubility of AgCl = Exercise :

Mercury(I) chloride, Hg2Cl2(s), is sparingly soluble in water, and gives Hg22+

(aq) and Cl-(aq) ions. The solubility product Ksp of Hg2Cl2(s) is 2.4 x10

-18 mol3 dm-9 at 298 K. Calculate, at 298K, the solubility of Hg2Cl2(s) (I) in pure water, and (II) in 0.10 mol dm-3 KCl(aq) (4 marks, 08)