56
Chapter 6 Problems 6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24
Chapter 6 Problems 6.6, 6.9, 6.15, 6.16, 6.19, 6.21, 6.24
Chapter 6
Chemical Equilibrium
Chemical Equilibrium Equilibrium Constant
Solubility product (Ksp) Common Ion Effect Separation by precipitation Complex formation
Solubility Product
Introduction to Ksp
Solubility Product
solubility-productthe product of the solubilities
solubility-product constant => Ksp
constant that is equal to the solubilities of the ions produced when a substance dissolves
Solubility Product
For silver sulfateAg2SO4 (s) 2 Ag+
(aq) + SO4-2
(aq)
Ksp = [Ag+]2[SO4-2]
Solubility of a Precipitatein Pure WaterEXAMPLE: How many grams of
AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25oC?AgCl <=> Ag+ + Cl-
Ksp = [Ag+][Cl-] = 1.82 X 10-10 (Appen. F)
EXAMPLE: How many grams of AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25oC?
AgCl(s) Ag+ (aq) + Cl- (aq)
Initial Some - -Change -x +x +xEquilibrium -x +x +x
(x)(x) = Ksp = [Ag+][Cl-] = 1.82 X 10-10
x = 1.35 X 10-5M
EXAMPLE: How many grams of AgCl (fw = 143.32) can be dissolved in 100. mL of water at 25oC?
How many grams is that in 100 ml?
# grams = (M.W.) (Volume) (Molarity) = 143.32 g mol-1 (.100 L) (1.35 x 10-5 mol
L-1) = 1.93X10-4 g = 0.193 mg
x = 1.35 X 10-5M
Solubility Product
For silver sulfateAg2SO4 (s) 2 Ag+
(aq) + SO4-2
(aq)
Ksp = [Ag+]2[SO4-2]
The Common Ion Effect
Ag2CO3 2 Ag+ + CO3-2
What is the effect on solubility of AgWhat is the effect on solubility of Ag22COCO33 if I add CO if I add CO33-2-2??
The Common Ion Effect
Add common ion effect a salt will be less soluble if one of
its constituent ions is already present in the solution
The Common Ion EffectEXAMPLE: Calculate the molar
solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3.
Ag2CO3 2 Ag+ + CO3-2
Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12
EXAMPLE: Calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3.
Ag2CO3 2 Ag+ + CO3-2
Initial Solid - 0.0200MChange -x +2x +xEquilibrium Solid +2x 0.0200+
x
Ksp=(2x)2(0.0200M + x) = 8.1 X 10-12
Ksp = [Ag+]2[CO3-2] = 8.1 X 10-12
4x2(0.0200M + x) = 8.1 X 10-12
EXAMPLE: Calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3.
4x2(0.0200M + x) = 8.1 X 10-12
no exact solution to a 3rd order equation, need to make some approximationfirst, assume the X is very small
compared to 0.0200 M4X2(0.0200M) = 8.1 X 10-12
4X2(0.0200M) = 8.1 X 10-12
X= 1.0 X 10-5 M
EXAMPLE: Calculate the molar solubility of Ag2CO3 in a solution that is 0.0200 M in Na2CO3.
X = 1.0 X 10-5 M(1.3 X 10-4 M in pure water)
Second, check assumption[CO3
-2] = 0.0200 M + X ~ 0.0200 M0.0200 M + 0.00001M ~ 0.0200M
Assumption is ok!
Separation by Precipitation
Separation by PrecipitationComplete separation can mean a lot
… we should define complete.
Complete means that the concentration of the less soluble material has decreased to 1 X 10-6M or lower before the more soluble material begins to precipitate
Separation by PrecipitationEXAMPLE: Can Fe+3 and Mg+2 be separated
quantitatively as hydroxides from a solution that is 0.10 M in each cation? If the separation is possible, what range of OH- concentrations is permissible.
Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Fe3+
Mg2+
Mg2+
Mg2+
Mg2+Mg2+
Mg2+Mg2+
Fe3+
Mg2+
Mg2+
Fe3+Fe3+
Mg2+
Mg2+
Add OHAdd OH--
Fe3+
Mg2+
Mg2+
Mg2+
Mg2+Mg2+
Mg2+Mg2+
Mg2+
Mg2+ Mg2+
Mg2+
Fe(OH)Fe(OH)33(s)(s)
What is the [OH-] when this happens^
@ equilibrium
Is this [OH-] (that is in solution) great enough to start precipitating Mg2+?
Separation by PrecipitationEXAMPLE: Can Fe+3 and Mg+2 be separated
quantitatively as hydroxides from a solution that is 0.10 M in each cation? If the separation is possible, what range of OH- concentrations is permissible.
Two competing reactionsFe(OH)3(s) Fe3+ + 3OH-
Mg(OH)2(s) Mg2+ + 2OH-
EXAMPLE: Separate Iron and Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
Ksp = [Mg+2][OH-]2 = 7.1 X 10-12
Assume quantitative separation requires that the concentration of the less soluble material to have decreased to < 1 X 10-6M before the more soluble material begins to precipitate.
EXAMPLE: Separate Iron and Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
Ksp = [Mg+2][OH-]2 = 7.1 X 10-12
Assume [Fe+3] = 1.0 X 10-6MWhat will be the [OH-] @ equilibrium required to reduce the [Fe+3] to [Fe+3] = 1.0 X 10-6M ?Ksp = [Fe+3][OH-]3 = 2 X 10-39
EXAMPLE: Separate Iron and Magnesium?
Ksp = [Fe+3][OH-]3 = 2 X 10-39
(1.0 X 10-6M)*[OH-]3 = 2 X 10-39
333 102][ OH11103.1][ OH
EXAMPLE: Separate Iron and Magnesium?
What [OH-] is required to begin the precipitation of Mg(OH)2?
[Mg+2] = 0.10 M
Ksp = (0.10 M)[OH-]2 = 7.1 X 10-12
[OH-] = 8.4 X 10-6M
EXAMPLE: Separate Iron and Magnesium?
[OH-] to ‘completely’ remove Fe3+ = 1.3 X 10-11 M
[OH-] to start removing Mg2+ = 8.4 X 10-6M
“All” of the Iron will be precipitated b/f any of the magnesium starts to precipitate!!
^̂@ equilibrium@ equilibrium
Complex Ion Formation
Complex Formation Consider Lead Iodide PbI2 (s) Pb2+ + 2I- Ksp
= 7.9 x 10-9
What should happen if I- is added to a solution?
Should the solubility go up or down?
Complex Formationcomplex ions (also called coordination ions)
Lewis Acids and Basesacid => electron pair acceptor (metal)base => electron pair donor (ligand)
Effects of Complex Ion Formation on SolubilityConsider the addition of I- to a
solution of Pb+2 ionsPb2+ + I- <=> PbI+
PbI+ + I- <=> PbI2 K2 = 1.4 x 101
PbI2 + I- <=> PbI3- K3 =5.9
PbI3+ I- <=> PbI42- K4 = 3.6
221 100.1
]][[][ xIPb
PbIK
Effects of Complex Ion Formation on SolubilityConsider the addition of I- to a
solution of Pb+2 ionsPb2+ + I- <=> PbI+
PbI+ + I- <=> PbI2 K2 = 1.4 x 101
Pb2+ + 2I- <=> PbI2 K’ =?
221 100.1
]][[][ xIPb
PbIK
Overall constants are designated with
This one is
Protic Acids and Bases
Section 6-7
Question Can you think of a salt that when
dissolved in water is not an acid nor a base?
Can you think of a salt that when dissolved in water IS an acid or base?
Protic Acids and Bases - Salts Consider Ammonium chloride
Can ‘generally be thought of as the product of an acid-base reaction.
NH4+Cl- (s) NH4
+ + Cl-
From general chemistry – single positive and single negative charges are STRONG ELECTROLYTES – they dissolve completely into ions in dilute aqueous solution
Protic Acids and BasesConjugate Acids and Bases in the B-L
conceptCH3COOH + H2O CH3COO- + H3O+ acid + base <=> conjugate base + conjugate acid
conjugate base => what remains after a B-L acid donates its protonconjugate acid => what is formed when a B-L base accepts a proton
Question: Question: Calculate the Concentration of H+ and OH- in Pure
water at 250C.
EXAMPLE: Calculate the Concentration of H+ and OH- in Pure water at 250C.
H2O H+ + OH-
Initial liquid - -Change -x +x +xEquilibrium Liquid-x +x +x
KW=(X)(X) = 1.01 X 10-14
Kw = [H+][OH-] = 1.01 X 10-14
(X) = 1.00 X 10-7
Example Concentration of OH-
if [H+] is 1.0 x 10-3 M @ 25 oC?
Kw = [H+][OH-]1 x 10-14 = [1 x 10-3][OH-]1 x 10-11 = [OH-]
“From now on, assume the temperature to be 25oC unless otherwise stated.”
pH
~ -3 -----> ~ +16pH + pOH = - log Kw = pKw = 14.00
Is there such a thing as Pure Water? In most labs the answer is NO Why?
A century ago, Kohlrausch and his students found it required to 42 consecutive distillations to reduce the conductivity to a limiting value.
CO2 + H2O HCO3- + H+
6-9 Strengths of Acids and 6-9 Strengths of Acids and BasesBases
Strong Bronsted-Lowry Acid A strong Bronsted-Lowry Acid is
one that donates all of its acidic protons to water molecules in aqueous solution. (Water is base – electron donor or the proton acceptor). HCl as example
Strong Bronsted-Lowry Base Accepts protons from water
molecules to form an amount of hydroxide ion, OH-, equivalent to the amount of base added.
Example: NH2- (the amide ion)
Weak Bronsted-Lowry acid One that DOES not donate all of its
acidic protons to water molecules in aqueous solution.
Example? Use of double arrows! Said to reach
equilibrium.
Weak Bronsted-Lowry base Does NOT accept an amount of
protons equivalent to the amount of base added, so the hydroxide ion in a weak base solution is not equivalent to the concentration of base added.
example: NH3
Common Classes of Weak Acids and BasesWeak Acids carboxylic acids ammonium ionsWeak Bases amines carboxylate anion
Weak Acids and Bases
HA H+ + A-
HA + H2O(l) H3O+ + A-
Ka
][]][[
HAAHKa
][]][[ 3
HAAOHKa
Ka’s ARE THE SAME
Weak Acids and Bases
B + H2O BH+ + OH-
][]][[
BOHBHKb
Kb
Relation Between KRelation Between Kaa and and KKbb
Relation between Ka and Kb Consider Ammonia and its
conjugate base.
][]][[
4
33
NH
OHNHKb
][]][[
3
4
NHOHNHKa
NH3 + H2O NH4+ + OH-
Ka
NH4+ + H2O NH3 + H3O+
Kb
H2O + H2O OH- + H3O+][]][[
][]][[
3
4
4
33
NHOHNH
NHOHNHK
][][ 3 OHOHKw
ExampleThe Ka for acetic acid is 1.75 x 10-5. Find
Kb for its conjugate base.Kw = Ka x Kb
a
wb KKK
105
14
107.51075.1100.1
bK
1st Insurance Problem
Challenge on page 107
