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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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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