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CHAPTER 2CHAPTER 2

ELECTROLYTE SOLUTIONELECTROLYTE SOLUTION

2-1 Strong and Weak Electrolyte Solution

2-2 Theory of Acid-base

2-3 Acidity and Calculation of Solution

2-4 Equilibrium Between Dissolution

and Precipitation

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2-1 2-1 Strong and Weak Electrolyte Strong and Weak Electrolyte SolutionSolution

2-1.1 Theory of Strong Electrolyte Solution

Ion-ion Interaction Theory

Figure 2-1 Ion atmosphere

of strong electrolyte solution

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Ion Activity and Activity Coefficient

Activity(a):

Ion concentration, which can play a real action in

solution is ionic effective concentration, is called ion

activity.

actual concentration of ion (c) multiply a correction factor - activity coefficient ( f ).

a = f ·c (2-1)

Generally,

a ＜ c, 0 ＜ f ＜ 1

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Activity coefficient are influenced by

ion concentration

the electric-charge number of ion

has nothing to do with the nature of ion.

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Ionic Strength ( I )

Where, I is ionic strength;

c is the amount-of-substance

concentration of the ion i;

z i is the charge number of the ion i.

Note that the activity is for an ion; the ion

strength is for a solution.

2ii

222

211 zczczcI 2

1 ......

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Table 2-1 Ion activity coefficient and ion strength of solution

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Example 2-1

25ml 0.02mol /L HCl mixed with 25ml 0.18mol /L KCl, calculate activity of H+ ?

Solution:

(1) calculate the ion strength of the solution:

I =( 0.01×12+0.01×12+0.09×12+0.09×12)/2 =0.1

(2) look up the activity coefficient of ion has one charge:

when I = 0.1, Z =1, f = 0.78

(3) calculate the activity of H+ (aH+)

cH+ =0.02/2 =0.01mol /L, f = 0.78

So, aH+ = f ·c = 0.78×0.01 = 0.0078 (mol /L)

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2-1.2 2-1.2 Ionization Equilibrium of

Weak Electrolyte Solution

The law of Chemical Equilibrium

(Equilibrium Constant)

a A + b B → c C + d D

[C]c[D]d

K = ------------ (2-3) [A]a[B]b

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Ionization Constant (KKi i ))

HAc + H2O H3O+ + Ac -

or simply

HAc H+ + Ac -

The corresponding equilibrium-constant expression is

[H+][Ac -] K i = ------------ (2-4) [HAc]

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Degree of Ionization (αα)

1. Definition:

Number of ionized molecules α= ----------------------------×100% total number of solute molecules

Number of ionized molecules = -----------------------------------×100% ionized molecules + non- ionized molecules

concentration of ionized weak electrolyte = ----------------------------------×100% initial concentration of weak electrolyte

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2. 2. The factor of influencing degree ofThe factor of influencing degree of ionization ionization

① the nature of solute:

18℃， 0.1mol/L, αHAc= 1.33%, αH2S= 0.07%,

αHCN= 0.007%

② the initial concentration of solute:

(the more dilute the solution, the greater the

degree of ionization).

③ temperature:

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Dilution Law

HA H+ + A - c α

Initial c c 0 0

Equilibrium c – cα cα cα concentration

[H+][A -] cα·cα cα2

KHA = ------------ = ---------- = ------- [HA] c- cα 1-α

For Ka is very small, α is very small, 1-α≈ 1

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KHA = cα2

c

KHAor

Physical meaning:

Note that above dilute law only is for some given

conditions : (1) the weak electrolyte must be monoprotic

(2) α≤ 5%

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The Common Ion Effect and Salt Effect

1. Common ion effect

The ionization of a weak electrolyte is markedly decreased by

the adding to the solution an ionic compound containing

one of the ion of the weak electrolyte, this effect is called

the common ion effect.

For example, HAc Ac - + H+

NaAc → Ac – + Na+

shift the equilibrium from right to left, decreasing the [H+].

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2. Salt effect

The ionization of a weak electrolyte is increased by

adding to the solution an soluble strong electrolyte

which not contains the common ion with the weak

electrolyte. This effect is called salt effect.

For example:

0.1mol/L [H Ac] α= 1.33%, adding NaCl,

[NaCl]= 0.1mol/L, α= 1. 68%

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Example: There is a solution of c(HAc) =0.1mol/L, if we add NaAc, when c(NaAc)=0.1mol/L. Calculate the α of HAc.

Solution: HAc H+ + Ac -

Initial c 0.1 0 0

Equilibrium c 0.1-[H+ ] [H+ ] 0.1+ [H+ ]

≈0.1 ≈0.1

[H+][Ac-] [H+]×0.1 Ka= ------------- = ------------ = [H+] = 1.8×10-5

[H Ac] 0.1 α= [H+]/[HAc] = 0.018% ＜＜ 1.33%

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2-2.2 Bronsted-Lowry Acids and Bases 1. Definition of acid and base

Acid - is a substance capable of donating a proton.

HCl, NH4+, HSO4

-, H2O

Base - is a substance capable of accepting a proton.

Cl-, NH3, HSO4-, OH-

2-2 2-2 Theory of Acid-base

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AA

HCl

baseacid

Conjugate acid-base pair

HH++ + B + B

H+ + Cl-

H2CO3 H+ + HCO3-

HCO3- H+ + CO3

2-

NH4+ H+ + NH3

H3O+ H+ + H2OH2O H+ + OH-

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Conclusion:Conclusion:

Acid or base may be a molecule, atom, or ion.

Some molecules or ions are capable of

donating a proton, and also accepting a proton,

which named ampholyte.

There are no concepts of salt in acid-base

proton theory.

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2. 2. Essence of Acid and Base ReactionEssence of Acid and Base Reaction

Essence: proton transfer reaction.

For example:

HCl(g) + NH3(g)H+

conjugate

NH4+ + Cl-

A1 + B2 A2 + B1

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(1) Compare by K a or Kb

HAc + H2O H3O+ + Ac-

[H3O+ ][Ac-] K a = ────── [HAc]

H+

The smaller the value for K a ， the weaker the acid ；

The greater the value for K a ， the stronger the acid.

3. 3. Relative Strength of Acids and BasesRelative Strength of Acids and Bases

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Ac- + H2O HAc + OH-

[HAc][OH-]K b = ────── [Ac -]

H+

The smaller the value for Kb ， the weaker the base ；

The greater the value for Kb ， the stronger the

base.

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The relationship between Ka and Kb:

Ka × Kb = K w

(2) The relationship between acid-base strength

and solvent

H2O weak acid

NH3 strong acid

HNO3

H2O strong acid

HAc weak acid

H2SO4 base substance

HAc

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4. The Leveling Effect and

Differentiating EffectThe leveling effect: The inability of a solvent to

differentiate among the relative strengths of all acids

stronger than the solvent’s conjugate acid is known

as the leveling effect.

Because the solvent is said to level the strengths of

these acids, making them seen identical.leveling solvent:

Strong acid such as HClO4, HCl, HNO3, H2SO4 will

appear to be of equal strength in aqueous solution.

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Strong acids--hydrochloric acid (HCl), nitric acid (HNO3),

perchloric acid (HClO4), and sulfuric acid (H2SO4), for

example- are all strong electrolytes. They may be

assumed to be completely ionized in water.

HCl(a q) + H2O(1) → H3O+(aq) + Cl - (aq)

HNO3(a q) + H2O(1) → H3O+(aq) + NO3-(aq)

HClO4(a q) + H2O(1) → H3O+(aq) + ClO4 - (aq)

H2SO4(a q) + H2O(1) → H3O+(aq) + HSO4 - (aq)we can not determine their strength, because H3O+ is

the strongest acid that can exist in aqueous solution.

• For strong acid or baseFor strong acid or base

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The differentiating effect:

The ability of a solvent to differentiate among the

relative strengths of all acids stronger than the solvent’s

conjugate acid is known as the differentiating effect.

If we use a more weakly basic solvent like acetic acid,

acetic acid can function as a base by accepting a proton.

Since acetic acid is a much weaker base than water, it is

not as easily protonated. Thus there are appreciable

differences in the extent.

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HCl(aq) + CHHCl(aq) + CH33COOH(COOH(l) CH) CH33COOHCOOH22++(aq) + Cl (aq) + Cl - - (aq)(aq)

HNOHNO33(aq) + CH(aq) + CH33COOH (COOH (l) CH) CH33COOHCOOH22++ (aq) + NO (aq) + NO33

--(aq)(aq)

HClOHClO44(aq)+ CH(aq)+ CH33COOH (COOH (l) CH) CH33COOHCOOH22++ (aq) + ClO (aq) + ClO44 - - (aq)(aq)

HH22SOSO44(aq)+ CH(aq)+ CH33COOH (COOH (l) CH) CH33COOHCOOH22++ (aq) + HSO (aq) + HSO44 - - (aq)(aq)

In acetic acid solvent, their relative strength increase as follows:

HNO3 ＜ H2SO4 ＜ HCl ＜ HClO4

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2-3 Acidity and Calculation of Solution

2-3.1 Autoionization of Water

H2O(l) + H2O(l) H3O+ (aq) + OH-(aq)

Water is capable of acting as a proton donor and proton

acceptor toward itself. The process by which this occurs

is called autoionization of water.

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Kw = [H3O+][OH-]

Where Kw is the equilibrium constant for water (unitless) ,

is called ion product of water or autoionization

equilibrium constant.

At 25 , ℃ Kw = [H3O+][OH -] = 1.0 ×10 -14

[H+] ＞ [OH-] in acid solutions

[H+] ＜ [OH-] in basic solutions

[H+] = [OH-] in neutral solutions

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2-3.2 Acidity of solution

pH = - log aH+ = -log [H3O+]

pOH = - log aOH- = -log[OH-]

For, [H+][OH-] = Kw= 1×10-14

So, pH + pOH =pKw= 14.00

In neutral solutions, pH ＝ 7 ＝ pOH,

In acid solutions, pH ＜ 7 ＜ pOH

In basic solutions, pH ＞ 7 ＞ pOH

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2-3.3 Calculation of Acidity of Solution

● For Strong Acids and Bases

pH = - log[H+] = -log[acid]

pOH = - log[OH-] = -log[base]

Example:

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● ● Monoprotic Weak Acids and BasesBases

H Ac

Solve this equation:

[H+] = - Ka /2 + √ Ka2 /4 +Ka c

H+ + Ac – initial c 0 0

equilibrium c- [H+] [H+] [Ac -] = [H+]

[H+] [Ac -] [H+]2

Ka= -------------- = ----------- [H Ac] c-[H+]

[H+]2 + Ka[H+] – Kac = 0

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When: c/Ka ≥103,

or α≤5%, c - [H+] ≈ c

thus, cKH a ][

Similarly, for weak base , there is an equation:

cKOH b ][

Note that above equation is limited for some

given condition:

Example 4-5: P41

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2-4 Equilibrium between Dissolution

and Precipitation

2-4.1 Solubility Product Constant (Ksp)

Ksp = [Ag+][Cl-]

AgCl(s)

dissolution

precipitationAg+ + Cl-

solubility product constant

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● Mg(OH) 2 (s) Mg2+ + 2OH-

Ksp = [M g2+][OH-]2

● Ag2CrO4 (s)

Ksp = [A g+]2[CrO42-]

2Ag+ + CrO42-

● Fe(OH) 3 (s) Fe3+ + 3OH-

Ksp = [Fe3+][OH-]3

● AmBn(s) mAn+ + nBm-

Ksp = [An+]m [Bm-]n

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2-4.2 Exchange between 2-4.2 Exchange between SSolubility and olubility and KKspsp

● AB (s , ksp)

spKs

AB(s) A + B

In saturated solution: [A]=[B]=s (mol/L)

Ksp = [A] [B]= s2

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● ● ABAB22 (A (A22B)B)

AB2(s)

3

4spK

s

A + 2B

In saturated solution: [A]=s (mol/L)

[B]=2s(mol/L)

Ksp = [A] [B]2

= s×(2s)2= 4s3

or

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3

4spK

s

The Ksp for CaF2 is 3.9×10-11. What is its solubility in water, in

grams per liter?

Example2-4:

s = 2.1× 10- 4 mol /L

Solution:

2.1× 10 - 4 mol /L × 78.1 g /mol = 1.6 × 10-2 g / L

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Qi (ion product quotient): the product of the ion

concentration in solution when the system is under

any situation ( at equilibrium or not ).

Qi (AgCl)=cAg+ cCl

-

Difference between Qi and Ksp:

2-4.3 Formation and dissolution of precipitation2-4.3 Formation and dissolution of precipitation

● ● Rule of Solubility ProductRule of Solubility Product

HAcAg+ + Cl-

H+ + Ac-

AgCl(s)

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The relationship between QThe relationship between Qii andand KKspsp

1. If Qi = Ksp, equilibrium is reached - no precipitate will

form. Saturated solution

2. If Qi > Ksp , a precipitate will form (until Q i decreases

to Ksp). Supersaturated solution

3. If Qi < Ksp, any precipitate in solution will dissolve until

Qi increases to Ksp. Unsaturated solution

The state above is called rule of solubility product.

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● ● Formation of precipitationFormation of precipitation condition: Qcondition: Qi i ＞＞ KKspsp

Example 2-5: Does a precipitate form if 0.100 L of 3.0 × 10 -3mol/L

Pb(NO3)2 is added to 0.400 L of 5.0 × 10 -3 mol /L Na2SO4?

possible precipitate form is PbSO4 ( Ksp = 1.6 × 10-8 )

[Pb2+] = 6.0 × 10-4 mol /L

[SO42-] =4.0 × 10-3 mol /L

Qi = [Pb2+][SO42-] = (6.0 × 10 - 4)(4.0 × 10-3)

= 2.4 ×10 - 6

Because Q i> Ksp, PbSO4 will precipitate!

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● ● Dissolution of precipitationDissolution of precipitation

condition: Qcondition: Qi i ＜ ＜ KKspsp

(ion product (ion product ＜ ＜ solubility product))

There are several methods to dissolve precipitation.

1. Forming weak electrolytes by adding some

compounds make precipitation dissolve.

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For example:For example:

Mg(OH) Mg(OH)22 not only dissolve in acid, not only dissolve in acid,

but also dissolve in NHbut also dissolve in NH44Cl solution.Cl solution.

Mg(OH)2 + 2HCl = MgCl2 + 2H2O

Mg(OH)2(s)

2H2O

Mg2+ + 2OH-

2HCl 2Cl- + 2H++

Because formed weak electrolyte H2O, [OH-]

decrease, shift the equilibrium from left to right.

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2.2. Forming coordination compounds by addingForming coordination compounds by adding

some agents make precipitation dissolve some agents make precipitation dissolve..

For example, AgCl precipitation dissolve in

NH3·H2O.

AgCl(s) + 2NH3 = [A g(NH3)2 ]+ + Cl-

AgCl(s) Ag+ + Cl-

+ 2NH3

[A g(NH3)2 ]+

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3.3. ProducinProducing oxidation-reduction reactions byg oxidation-reduction reactions by adding oxidizing agents or reducing agents adding oxidizing agents or reducing agents make precipitation dissolve make precipitation dissolve..

For example :

To the CuS precipitation add the dilution

HNO3, CuS might dissolve.

3CuS(s) + 8HNO3(dilution) = 3Cu(NO3)2 + 3S + 2NO + 4H2O

Cu S(s) S2- + Cu2+

+

HNO3 S↓+ NO↑ + H2O