Chemical Equilibrium

THE NATURE OF CHEMICAL EQUILIBRIUM

Reversible Reactions• In theory, every reaction can continue

in two directions, forward and reverse

• Reversible reaction ! chemical reaction in which the products can react to re-form the reactants

Mercury (II) oxide• Decomposes when heated

2HgO(s) ! 2Hg(l) + O2(g)

• Mercury and oxygen combine to form mercury (II) oxide when heated gently

2Hg(l) + O2(g) ! 2HgO(s)

• Suppose HgO is heated in closed container where mercury and oxygen can’t escape

• Once decomposition begins, mercury and oxygen released can recombine to form HgO again

• Both reactions happen at same time

• Rate of composition will eventually equal rate of decomposition• At equilibrium, mercury and oxygen will

combine at same rate that mercury (II) oxide decomposes

• Amounts of mercury(II) oxide, mercury and oxygen are expected to remain constant as long as conditions continue

• At this point, a state of dynamic equilibrium has been reached between two reactions

• No net change in composition of system

• A reversible reaction is in chemical equilibrium when the rate of its forward reaction equals the rate of its reverse reaction and the concentrations of its products and reactants remain unchanged

• Equation for chemical equilibrium written using double arrows

2HgO(s) ⇄ 2Hg(l) + O2(g)

Equilibrium, a Dynamic State• In some cases, the forward

reaction is almost done before the rate of the reverse reaction is high enough for equilibrium

•Here, products of forward reaction are favored (at equilibrium, there is higher concentration of products than reactants)

HBr(aq) + H2O(l) H3O+(aq) + Br−

(aq)

• Other cases, reverse reaction is favored• More reactants than products

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3−(aq)

• Sometimes forward and reverse happen almost same amount before equilibrium

H2SO3(aq) + H2O(l) ⇄ H3O+(aq) + HSO3−(aq)

The Equilibrium Expression• Let’s say 2 substances, A and B, react

to make products C and D• C and D react to make A and B

nA + mB ⇄ xC + yD

• At first, concentrations of C and D are zero and concentrations of A and B are max

• Over time, rate of forward reaction decreases as A and B are used up

• Rate of reverse increases as C and D are formed

•When two reaction rates become equal, equilibrium is created• Individual concentrations of A, B, C, and D don’t change if conditions stay the same

• After EQ, concentrations of products and reactants remain constant, so the ratio of their concentrations should also stay the same

• Ratio of product [C]x x [D]y to product [A]n x [B]m for this reaction has definite value at given temp• Definite value is EQ constant of the

reaction and is given letter K

• Following equation describes equilibrium constant for theoretical equilibrium system• [C] = concentration of C in

moles/liter

• Concentrations of products in chemical reaction appear in numerator of equilibrium ratio

• Each concentration to the power of the coefficient

• These are products of forward reaction

• Concentrations of reactants found in denominator (risen to power of coefficient)• These are reactants of forward reaction• K independent of original

concentrations, but dependent on temperature

The Equilibrium Constant• Value of K for particular equilibrium system is found

experimentally• Chemist analyzes equilibrium mixture and determines

concentrations of all substances• Value of K for given reaction at given temp shows degree

to which reactants converted to products

• If K = 1, products of concentrations raised to appropriate power in numerator and denominator have the same value• Therefore, at equilibrium,

there are equal concentrations of reactants and products

• If K is low, forward reactions happens before equilibrium, and reactants are favored• If K is high, the original

reactants are largely converted to products

•Only concentrations of substances that can actually change are included in K

• This means PURE solids and liquids are left out because their concentrations cannot change

Chemical-equilibrium Expression• In general, the equilibrium constant,

K, ! is the ratio of the mathematical product of the concentrations of the products at equilibrium to the mathematical products of the concentrations of reactants

• Each concentration raised to power equal to coefficient• Chemical-equilibrium expression !

the equation for K

The H2, I2, HI Equilibrium System• Consider reaction between H2 and I2 vapor in

sealed flask at elevated temp• Rate of reaction followed by observing rate at

which the violet color of iodine vapor lessens

• Suppose hydrogen gas is present in excess and reaction continues until all iodine is used up• Color (from iodine) gradually disappears

because the product, HI and hydrogen gas are both colorless

• In reality, color fades to constant intensity but does not disappear completely• The reaction is reversible•HI decomposes to form

hydrogen and iodine

• Rate of reverse reaction increases as concentration of HI increases• Rate of forward reaction decreases

accordingly

• Concentrations of hydrogen and iodine decrease as they are used up• As rates of opposing

reactions become equal, equilibrium is established

• Constant color indicates that equilibrium exists• Net chemical equation:

H2(g) + I2(g) ⇄ 2HI(g)

H2(g) + I2(g) ⇄ 2HI(g)• From this chemical equation we can

write chemical-equilibrium expression

Sample Problem• An equilibrium mixture of N2, O2 , and

NO gases at 1500 K is determined to consist of 6.4 x 10–3 mol/L of N2, 1.7 x 10–3 mol/L of O2, and 1.1 x 10–5 mol/L of NO. What is the equilibrium constant for the system at this temperature?

1. Analyze• Given: • [N2] = 6.4 × 10−3 mol/L• [O2] = 1.7 × 10−3 mol/L• [NO] = 1.1 × 10−5 mol/L

• Unknown: K

2. Plan• The balanced chemical equation is • N2(g) + O2(g) ⇄ 2NO(g)

• The chemical equilibrium expression is

3. Compute

• = 1.1 x 10-5

Practice Problems• At equilibrium a mixture of N2, H2, and

NH3 gas at 500°C is determined to consist of 0.602 mol/L of N2, 0.420 mol/L of H2, and 0.113 mol/L of NH3.What is the equilibrium constant for the reaction N2(g) + 3H2(g) ⇄ 2NH3(g) at this temperature?• 0.286

• The reaction AB2C(g) ⇄ B2(g) + AC(g) reached equilibrium at 900 K in a 5.00 L vessel. At equilibrium 0.084 mol of AB2C, 0.035 mol of B2, and 0.059 mol of AC were detected. What is the equilibrium constant at this temperature for this system?• 4.9 × 10−3

• At equilibrium a 1.0 L vessel contains 20.0 mol of H2, 18.0 mol of CO2, 12.0 mol of H2O, and 5.9 mol of CO at 427°C. What is the value of K at this temperature for the following reaction? CO2(g) + H2(g) ⇄ CO(g) + H2O(g)

• 0.20

• A reaction between gaseous sulfur dioxide and oxygen gas to produce gaseous sulfur trioxide takes place at 600°C. At that temperature, the concentration of SO2 is found to be 1.50 mol/L, the concentration of O2 is 1.25 mol/L, and the concentration of SO3 is 3.50 mol/L. Using the balanced chemical equation, calculate the equilibrium constant for this system.

• 4.36

Shifting Equilibrium

• In chemical equilibrium, forward and reverse reactions happen at same rate• Any change alters the rates

and disturbs equilibrium

• System will try to find a new equilibrium state• By shifting equilibrium in desired

direction, chemists can improve the yield of product they want

Predicting the Direction of Shift• Le Chatelier’s principle provides a way

of predicting the influence of stress factors on equilibrium systems

• Le Chatelier’s principle ! if a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that relieves the stress

• Principle true for all dynamic equilibria•Changes in pressure,

concentration, and temperature demonstrate the application of Le Chatelier’s principle to chemical equilibrium

Changes in Pressure• Change in pressure only affects equilibrium

systems where gases are involved• For changes to affect system, the total number of

moles of gas on left side must be different from total moles on right side

CaCO3(s) ⇄CaO(s) + CO2(g)

1 mol !1 mol + 1 mol

CaCO3(s) ⇄CaO(s) + CO2(g)• High pressure favors reverse reaction

b/c fewer CO2 molecules are produced

• Lower pressure favors increased production of CO2

Haber process• For synthesis of ammonia

N2(g) + 3H2(g) ⇄ 2NH3(g)

• Consider an increase in pressure which causes increase in concentration of all species• System reduces number of molecules (total

pressure) by shifting equilibrium to right

N2(g) + 3H2(g) ⇄ 2NH3(g)• For each four

molecules of reactants there are two molecules of products• By making more NH3,

and using up N2 and H2 the system reduces the number of molecules

• This decreases pressure• New equilibrium pressure higher than

before, but not as high as original stress

• Ammonia made in Haber process is continuously removed by condensation to liquid• This removes most of

the product from gas phase where reaction happens

• Result: decrease in partial pressure of NH3 gas• Same as decrease in product

concentration • Shifts equilibrium to the right

Changes in Concentration• Increase in concentration of reactant is a stress on

equilibrium system• Causes increase in collision frequency, and

increase in reaction rate

• Consider the following reaction:

A + B ⇄ C + D

A + B ⇄ C + D• Increase in [A] shift equilibrium to right• Both A and B used up faster, more of

C and D are formed• Equilibrium reestablished with lower

[B]

• Increase in [B] drives reaction to right• Increase in [C] or [D] increases rate of

reverse reaction (shift left)• Decrease in [C] or [D] equilibrium

shifts right

A + B ⇄ C + D

Changes in concentration have no effect on the value of the equilibrium constant

• This is b/c these changes have effect on numerator and denominator of chemical equilibrium expression

• So, all concentrations give same value or ratio for equilibrium constant when equilibrium is reestablished

•Many chemical reactions involve heterogeneous reaction where reactants and products are in different phases• Concentrations of solids

and liquids not changed by adding or removing it

• This is b/c concentration is density-dependent• Density of solids/liquids is constant,

regardless of amount present• A pure substance in a condensed

phase (solid, liquid) can be eliminated from the expression for the equilibrium constant

• Concentration of pure solid/liquid is set to equal 1 in equilibrium expression

CaCO3(s) ⇄ CaO(s) + CO2(g)

• Products are solid and gas so expression is

Changes in Temperature•Reversible reactions are exothermic in one direction and endothermic in the other

• Effect of changing temp depends on which direction is exothermic or endothermic

• According to Le Chatelier’s principle, addition of heat shifts equilibrium so heat is absorbed• This favors endothermic rxn•Removal of heat favors

exothermic rxn

• Rise in temp increase rate of any rxn• In equilibrium system, rates of 2 rxns

not equal• So, value of equilibrium constant is

affected by temp

• Haber process is exothermic

N2(g) + 3H2(g) ⇄ 2NH3(g) + 92 kJ

• High temp not desirable b/c it favors decomposition of NH3 (endothermic)• Low temp – forward rxn too slow to

be useful (doesn’t make enough NH3)

• Temp used in Haber process is balance between kinetic and eq requirements

• High enough that eq is created quickly• Low enough that eq concentration of

NH3 is large

• Average temp (500℃) and high pressure (700-1000 atm) makes satisfactory amount of NH3

• Production of colorless dinitrogen tetroxide gas, N2O4, from dark brown NO2 gas is exothermic

All 3 flasks contain same total mass of gas (mixture of NO2 and N2O4)

• Temp of system lowered to 0℃

• This causes eq to shift right (exothermic rxn, makes more heat) to balance for lower temp• Allows more colorless N2O4 to form• Makes light brown color

2NO2(g) ⇄N2O4(g)

• System is at equilibrium at 25℃

• At eq, system has equal amount of NO2 and N2O4

• Produces medium brown color

2NO2(g) ⇄N2O4(g)

• Raise temp to 100℃, equilibrium shifts to left

• Forward rxn = exothermic• Reverse rxn = endothermic (absorbs

excess heat)• Causes more dark brown NO2 gas to

form

2NO2(g) ⇄N2O4(g)

Endothermic Rxn• For endothermic, heat shows up on reactant side

556 kJ + CaCO3(s) ⇄ CaO(s) + CO2(g)

• Increase in temp causes value of K to increase

• Eq shifts right

External Stress Predicted Equilibrium Shift

Change in pressure Affects gases only.

Concentration of reactant(s) increased

Equilibrium shifts toward products (to the right).

Concentration of product(s) increased

Equilibrium shifts toward reactants (to the left).

Temperature increased Equilibrium shifts toward endothermic rxn

Temperature decreased Equilibrium shifts toward exothermic rxn

Rxns That Go to Completion• Some rxns involving compounds

formed by interaction of ions in solutions appear to go to completion (ions almost completely removed from solution)

• Degree to which reacting ions removed depends on solubility of compound formed and if compound formed is soluble, the degree of ionization

• A product that escapes as gas, precipitates as solid, or is only slightly ionized removes most of reacting ions from solution

1. Formation of a gas

2. Formation of precipitate

3. Formation of slightly ionized product

1. Formation of a Gas• Unstable substances formed as products of ionic

reactions decompose spontaneously• Ex. Carbonic acid, H2CO3, acid in carbonated water

(Pepsi, Coke, etc.)

H2CO3(aq) → H2O(l) + CO2(g)

• Reaction practically goes to completion b/c CO2 escapes as gas if container is open to air

2. Formation of Precipitate• When solutions of NaCl and AgNO3 are

mixed, white precipitate of AgCl forms

Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) ! Na+

(aq) + NO3−(aq) + AgCl(s)

• If equal amounts of reactants used, only Na+ and NO3

- ions stay in solution• Rxn goes to completion b/c an insoluble

product is formed

3. Formation of Slightly Ionized Product

• Neutralization rxns between H3O+ ions from aqueous acids and OH- ions from aqueous bases result in formation of H2O molecules

• H2O only slightly ionized

H3O+(aq) + Cl-−(aq) + Na+(aq) + OH-−(aq)! Na+(aq) + Cl-−(aq) + 2H2O(l)

• Get rid of spectator ions

H3O+(aq) + OH-−(aq) ! 2H2O(l)

H3O+(aq) + OH−(aq) → 2H2O(l)• b/c water only slightly ionized, it exists almost

entirely as covalently bonded molecules

• So, as long as they are originally in equal amounts, hydronium and hydroxide ions are almost entirely removed from solution

Common-Ion Effect• Eq rxn can be pushed in desired direction by

applying Le Chatelier’s principle

• Ex. HCl gas bubbled into saturated solution of NaCl• HCl is extremely soluble in water, and is almost

completely ionized

HCl(g) + H2O(l) ! H3O+(aq) + Cl−(aq)

HCl(g) + H2O(l) → H3O+(aq) + Cl−(aq)

• Eq for a saturated solution of NaCl is

NaCl(s) ⇄ Na+(aq) + Cl−(aq)

• As HCl dissolves in high enough amount, it increases [Cl-]• This is stress on eq system

• System balances, according to Le Chatelier’s principle, by combining some Cl- ions with equal number of Na+ ions• This causes some solid

NaCl to precipitate out• Relieves stress of added

chlorine

• New eq has higher [Cl-] but lower [Na+]• BUT product of [Na+] and [Cl-] still have

same value as before• Common-ion effect ! phenomenon, in

which the addition of an ion common to two solutes brings about precipitation or reduced ionization

• Common-ion effect also seen when one ion species of weak electrolyte is added in excess to solution

• A 0.1M CH3COOH (acetic acid) solution is only about 1.4% ionized to make hydronium ions (H3O+) and acetate ions (CH3COO-)

CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO−(aq)

CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO−(aq)

• Small additions of sodium acetate, NaCH3COO, to solution of acetic acid greatly increases acetate-ion concentration• Eq shifts to use up some acetate ions

• More molecules of acetic acid are formed and concentration of H3O+ reduced

Equilibria of Acids, Bases, and Salts

Ionization Constant of a Weak Acid• About 1.4% solute molecules in 0.1M acetic acid

solution are ionized at room temp• Remaining 98.6% of acetic acid molecules stay

unionized

• So, solution contains 3 species of particles in eq: CH3COOH molecules, H3O+ ions, and acetate ions,CH3COO−

• Eq constant for this system expresses eq ratio of ions to molecules

CH3COOH + H2O ⇄ H3O+ + CH3COO−

• At 0.1M concentration, water molecules are much more than the number of acetic acid molecules•One can assume the molar

concentration of H2O molecules stays constant• So, b/c both K and H2O are constant,

the product K[H2O] is constant

• Left side can be simplified by setting K[H2O] = Ka

• Acid-ionization constant ! Ka

• Ka, like K, is constant for specified temp and has new value for each new temp

• Ka for weak acid is a small value• To determine actual value for Ka

for acetic acid at specific temp, eq concentrations of H3O+ ions, CH3COO− ions, and CH3COOH molecules must be known

• The ionization of a molecule of CH3COOH in water yields one H3O+ ion and one CH3COO− ion • These concentrations can be found

experimentally by measuring the pH of the solution

Ionization for some dilute acetic acid solutions at 25℃

• Notice Ka is almost identical for every solution molarity shown

Molarity % Ionized [H3O+] [CH3COOH] Ka

0.100 1.33 0.00133 0.0987 1.79 x 10-5

0.0500 1.89 0.000945 0.0491 1.82 × 10−5

0.0100 4.17 0.000417 0.00958 1.81 × 10−5

0.00500 5.86 0.000293 0.00471 1.82 × 10−5

0.00100 12.6 0.000126 0.000874 1.82 × 10−5

• At constant temp, increase in [CH3COO-] through addition of NaCH3COO disturbs eq

• Disturbance causes decreases in [H3O+] and increase in [CH3COOH]

• Eventually, eq is reestablished with same value of Ka

• But higher [CH3COOH] and lower [H3O+] than before extra CH3COO- was added

• Changes in [H3O+] affect pH• Decrease in [H3O+] means

increase in pH

Buffers• Previous solution contains

weak acid (CH3COOH) and salt of weak acid (NaCH3COO)• Solution can react with acid

or base• When small amounts of

weak acid or bases added, pH remains nearly constant

• Weak acid and common ion, CH3COO-, act as “buffer” against large changes in pH

• Buffered solution ! solution that can resist changes in pH

• Let’s say a small amount of acid is added to acetic acid-sodium acetate solution• Acetate ions react with most of added

hydronium ions to form nonionized acetic acid molecules

CH3COO−(aq) + H3O+(aq) → CH3COOH(aq) + H2O(l)

• [H3O+] and pH remain amost unchaged

From added acid

• Now let’s say a small amount of base is added to original solution• The OH- ions of base react with and remove

H3O+ ions to form nonionized water molecules• Acetic acid molecules then ionize and

restore eq concentration of hydronium ions

CH3COOH(aq) + H2O(l) → H3O+(aq) + CH3COO−(aq)From reaction of OH-

of base with H3O+

• A solution of a weak base containing a salt of the base behaves in a similar way• [OH-] and pH of solution stay

constant with small additions of acids or bases

• Suppose a base is added to aqueous NH3 solution that also contains NH4Cl• Ammonium ions donate proton to added OH-

ions to form nonionized water molecules

NH4+(aq) + OH−(aq) → NH3(aq) + H2O(l)

From added base

• If small amount of acid added to solution instead, OH- from solution accept protons from added H3O+ from acid to form water• NH3 in solution then ionize and

restore eq concentration of H3O+ and pH of solution

NH3(aq) + H2O(l) → NH4+(aq) + OH−(aq)

Ionization Constant of Water• The self-ionization of water is an eq reaction

H2O(l) + H2O(l) ⇄ H3O+(aq) + OH−(aq)

• Eq is created with very low [H3O+] and [OH-]

Kw = [H3O+][OH−] = 1.0 × 10−14

Hydrolysis of Salts• Salts are formed during

neutralization reaction between Br∅nsted acid and Br∅nsted base•When a salt dissolves in

water, it makes cations of the base and anions of the acid

• Solution might be expected to be neutral• Some salts (NaCl and KNO3) are neutral•When Na2CO3 dissolves in water,

solution turns red litmus paper blue• Ammonium chloride in water turns blue

litmus red

• Variation in pH can be explained by looking at the ions formed when each of these salts dissociates

• If ions formed are from weak acids or bases, they react with water solvent, and pH will have value higher than 7

• Hydrolysis ! reaction between water molecules and ions of a dissolved salt

• If anions react with water, process is anion hydrolysis and solution is more basic• If cations react with water,

process is cation hydrolysis and solution is more acidic

Anion Hydrolysis• The anion of the salt is the

conjugate base of the acid from which it was formed• It is also a proton acceptor

• If acid is weak, the conjugate base (the anion) will be strong enough to remove protons from some water molecules (proton donors) to form OH- ions• Eq is created where the net effect of anion

hydrolysis is increase in [OH-] of solution (more basic)

• Eq equation for typical weak acid in water, HA, forming hydronium and an anion, A- is as followsHA(aq) + H2O(l) ⇄ H3O+(aq) + A−(aq)

• From this equation, Ka can be written

• Hydrolysis reaction between water and anion, A-that is made by dissociation of weak acid, HA, is represented as

A−(aq) + H2O(l) ⇄ HA(aq) + OH−(aq)

• In forward reaction, anion gets proton from water to form HA (weak acid) and OH- left over from water molecule

• Extent of OH- ion formation and position of eq depends on strength of anion

• Lower Ka value of HA, stronger attraction for protons that anion (A-) will have compared with OH-

•Greater production of OH-

• As strength of A- increase, eq position is more to the right

• Aq solutions of sodium carbonate are strongly basic• Sodium ions do not undergo

hydrolysis in aq solution, but carbonate ions react as a Br∅nsted base

• Carbonate anion receives proton from H2O to form slightly ionized hydrogen carbonate ion, HCO3

-, and the OH- ion

CO32−(aq) + H2O(l) ⇄ HCO3

−(aq) + OH−(aq)

CO32−(aq) + H2O(l) ⇄ HCO3

−(aq) +OH−(aq)

• [OH-] increases until eq is established• As a result the [H3O+] decreases so that

the product [H3O+][OH-] stays equal to Kw

• pH is higher than 7, and solution is basic

Cation Hydrolysis• Cation of salt is conjugate acid of base

from which it was formed• Also a proton donor

• If base is weak, cation strong enough to donate proton to water molecule to form H3O+

• Eq created when net effect of cation hydrolysis is increase in [H3O+] of solution (more acidic)

• A typical weak base, B, used to get general expression for Kb (base dissociation constant)

B(aq) + H2O(l) ⇄ BH+(aq) + OH−(aq)

• Hydrolysis rxn between water and cation, BH+, made by dissociation of weak base, B, is represented as

BH+(aq) + H2O(l) ⇄ H3O+(aq) + B(aq)

• In forward reaction, BH+ donates proton to water to form H3O+

• Extent of H3O+ formation and position of eq depend on strength of BH+

• Lower Kb value of B, the stronger the donation of protons that BH+ will have and more H3O+ ions made

• As strength of BH+ increases, eq position lies to the right

• Ammonium chloride, NH4Cl, dissociates in water to make NH4

+ and Cl- and an acidic solution•Cl- are conjugate base of a

strong acid, HCl, so they don’t have tendency to hydrolyze in aqueous solution

• NH4+ are conjugate acid of weak base,

NH3

• NH4+ donate protons to water molecules

• Eq created with increase in [H3O+] so pH is lower than 7

Hydrolysis in Acid-Base Reactions• Hydrolysis helps explain why end point of

neutralization can happen at a pH other than 7• Hydrolysis properties of salts determined by

relative strengths of acids and bases from which the salts were formed• Salts can be placed in 4 general categories,

depending on their hydrolysis properties

1.Strong acid-strong base

2.Strong acid-weak base

3.Weak acid-strong base

4.Weak acid-weak base

• Salts of strong acids and strong bases produce neutral solutions• Neither the cation of strong

base nor anion of strong acid hydrolyze much in aqueous solutions

•HCl is strong acid – Cl- doesn’t hydrolyze in water•NaOH is strong base – Na+ doesn’t

hydrolyze in water

• So, aq solutions of NaCl are neutral

• Similarly, KNO3 is salt of strong acid HNO3 and strong base KOH

• pH of aqueous KNO3 is always very close to 7

Strong acid-weak base• Aq solutions of salts formed from rxns between weak

acids and strong bases are basic

• Anions of dissolved salt hydrolyzed in water• pH raised, indicating [OH-] increase• Anions from weak base (strong proton acceptor)

1 2

3

4

Equivalence point

At point 1 on the titration curve, only acetic acid is present. The pH depends on the weak acid alone. At 2 there is a mixture of CH3COOH and CH3COO−. Adding NaOH changes the pH slowly. At point 3 all acid has been converted to CH3COO−.This hydrolyzes to produce a slightly basic solution. At 4 the pH is due to the excess OH− that has been added.

Strong acid-weak base• Salts of strong acids and weak

bases produce acidic aqueous solutions

• Cations of dissolved salt hydrolyzed in water

• pH is lowered, indicated [H3O+] has increased

• Cations are positive ions from a weak base (strong proton donor)

12

3Equivalence point

4

At point 1 on the titration curve, only aqueous ammonia is present. The pH is determined by the base alone. At 2 there is a mixture of NH3 and NH4

+. Adding HCl changes the pH slowly. At point 3 all aqueous ammonia has been converted to NH4+. At 4 the pH is determined by the excess H3O+ that is being added.

Weak acid-weak base• Salts of weak acids and weak bases can make

either acidic, neutral, or basic solutions, depending on salt dissolved

• b/c both ions of dissolved salt are hydrolyzed greatly

• If both ions hydrolyzed equally, solution stays neutral

• In cases where both acid and base are very weak, the salt can undergo complete decomposition to the products of hydrolysis

• Ex. When Al2S3 is placed in water, both precipitate and gas are formed

Al2S3(s) + 6H2O(l) → 2Al(OH)3(s) + 3H2S(g)

• Both products hardly soluble in water and removed from solution

Solubility Equilibrium

Solubility Product• Saturated solution contains maximum amount

of solute possible at given temp with an undissolved excess of substance

• Not necessarily concentrated

• Depends on solubility of solute

• General rule used to express solubilities

• Substance is said to be soluble if solubility is greater than 1 g per 100 g of water

• Insoluble ! less than 0.1 g in 100 g water

• Slightly soluble ! between 0.1-1.0 g per 100 g water

• Silver chloride is so slightly soluble in water that it’s sometimes called insoluble• Solution reaches saturation at

very low concentration of ions• All Ag+ and Cl- ions in excess

of this concentration will precipitate as AgCl

• Let’s look at eq system in saturated solution of AgCl containing excess of solid salt (AgCl)• System represented as:

AgCl(s) ⇄ Ag+(aq) + Cl−(aq)

• b/c concentration of pure substance as solid or liquid stays constant, adding more solid AgCl doesn’t change concentration of undissolved AgCl

• So, [AgCl] does not appear in final expression

• Rearrange equation so both constants on same side gives solubility-product constant, Ksp

K[AgCl] = [Ag+][Cl-]

• Solubility-product constant ! of a substance the product of molar concentrations of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the chemical equation

Ksp = [Ag+][Cl-]• This equation is solubility-equilibrium

expression for the reaction

• Expresses that Ksp of AgCl is product of molar concentrations of its ions in saturated solution

Calcium fluoride•CaF2 is another slightly soluble salt

CaF2(s) ⇄ Ca2+(aq) + 2F−(aq)

Ksp = [Ca2+][F−]2

• Ksp can be determined from solubility data

• Indicates a maximum of 8.9 x 10-5 g AgCl can dissolve in 100 g water at 10℃

• 1 mol AgCl has mass of 143.32 g

• Saturation concentration (solubility) of AgCl can be expressed in moles per liter of water

• = 6.2 x 10-6 mol/L

• AgCl dissociates in solution, giving equal numbers of Ag+ and Cl- ions

• So, concentrations of ions are• [Ag+] = 6.2 x 10-6 • [Cl-] = 6.2 x 10-6

• Ksp = [Ag+][Cl-]

• Ksp = (6.2 x 10-6 )(6.2 x 10-6 )

• Ksp = 3.8 x 10-11

• This is solubility-product constant of AgCl at 10℃

Practice Problem• Calculate the solubility-product

constant, Ksp, for copper(I) chloride, CuCl, given that the solubility of this compound at 25°C is 1.08 x 10–2 g/100. g H2O.

1. Analyze•Given: • solubility of CuCl = 1.08 × 10−2 g CuCl/

100. g H2O• Unknown: Ksp

2. Plan• Start by converting the solubility of CuCl in

g/100. g H2O to mol/L• You will need the molar mass of CuCl to get

moles CuCl from grams CuCl• Then use the solubility of the [Cu+] and [Cl−]

ions in the Ksp expression and solve for Ksp

CuCl(s) ⇄ Cu+(aq) + Cl−(aq)

Ksp = [Cu+][Cl−]

[Cu+] = [Cl−] = solubility in mol/L

3. Compute

= 1.09 x 10-3 mol/L CuCl

[Cu+] = [Cl-] = 1.09 x 10-3 mol/L

Ksp = (1.09 x 10-3) (1.09 x 10-3) = 1.19 x 10-6

Practice Problem• Calculate the solubility-product

constant, Ksp, of lead(II) chloride, PbCl2, which has a solubility of 1.0 g/100. g H2O at a temperature other than 25°C.

• 1.9 × 10−4

• Five grams of Ag2SO4 will dissolve in 1 L of water. Calculate the solubility product constant for this salt.

• 2 × 10−5

Calculating Solubilities• Once known, Ksp can be used to determine

solubility

• Suppose you want to know how much BaCO3 can be dissolved in 1 L of water at 25℃

• From table, Ksp for BaCO3 is 5.1 x 10-9

BaCO3(s) ⇄ Ba2+(aq) + CO32−(aq)

• Given value of Ksp, can write solubility-equilibrium expression as

Ksp = [Ba2+][CO32−] = 5.1 × 10−9

• BaCO3 dissolves until [Ba2+] and [CO32−] =

5.1 × 10−9

• Solubility-eq equation shows that Ba+2 and CO3

-2 enter solution in equal numbers• So, they have same concentration

• Let [Ba2+] = x, and [CO32−] = x

Let [Ba2+] = x, and [CO32−] = x

• [Ba2+][CO32−] = Ksp = 5.1 × 10−9

• (x)(x) = 5.1 × 10−9

• x =√5.1×10−9

• The solubility of BaCO3 is 7.14 × 10−5 mol/L• Solution concentration is 7.14 × 10−5 M for Ba2+ ions

and 7.14 × 10−5 M for CO32− ions

Sample Problem• Calculate the solubility of silver

acetate,AgCH3COO, in mol/L, given the Ksp value for this compound listed in Table 18-3.

1. Analyze• Given: • Ksp = 1.9 × 10−3

• Unknown: • solubility of AgCH3COO

2. Plan• AgCH3COO ⇄ Ag+(aq) + CH3COO−(aq)

• Ksp = [Ag+][CH3COO−]

• [Ag+] = [CH3COO−],

• so let [Ag+] = x and [CH3COO−] = x

3. Compute• Ksp = [Ag+][CH3COO−]

• Ksp = x2

• x2 = 1.9 × 10−3

• x =√1.9×10−3

• Solubility of AgCH3COO =√1.9×10−3 = 4.4 × 10−2 mol/L

Practice Problem• Calculate the solubility of cadmium

sulfide, CdS, in mol/L, given the Ksp value listed in Table 18-3.

• 8.9 × 10−14 mol/L

• Determine the concentration of strontium ions in a saturated solution of strontium sulfate, SrSO4, if the Ksp for SrSO4 is 3.2 × 10−7.

• 5.7 × 10−4 mol/L

Precipitation Calculations• In earlier example, BaCO3 served as source for Ba2+ and

CO3-2

• b/c each mole of BaCO3 yields 1 mol Ba+2 and 1 mol CO3

-2, concentrations were equal

• Equilibrium conditions doesn’t require that the two ion concentrations be equal

• Eq still established to that [Ba2+][CO32−] doesn’t go over

Ksp

• Suppose unequal amounts of BaCl2 and Na2CO3 are dissolved in water and solutions are mixed

• If ion product [Ba2+][CO32−] goes over Ksp for

BaCO3, a precipitate forms

• Precipitation continues until ion concentrations decrease to point where [Ba2+][CO3

2−] = Ksp

Sample Problem•Will a precipitate form if 20.0 mL of

0.010 M BaCl2 is mixed with 20.0 mL of 0.0050 M Na2SO4?

1. Analyze• Given:• concentration of BaCl2 = 0.010 M• volume of BaCl2 = 20.0 mL• concentration of Na2SO4 = 0.0050 M• volume of Na2SO4 = 20.0 mL

• Unknown: • whether a precipitate forms

2. Plan• The two possible new pairings of ions are NaCl

and BaSO4

• Of these, BaSO4 is a sparingly soluble salt• It will precipitate if the ion product [Ba2+][SO4

2−] in the combined solution exceeds Ksp for BaSO4• From the list of solubility products in Table 18-3,

the Ksp is found to be 1.1 × 10−10

BaSO4(s) ⇄ Ba2+(aq) + SO4 2−(aq)• The solubility-equilibrium expression is written as

Ksp = [Ba2+][SO42−] = 1.1 × 10−10

• First [Ba2+] and [SO42−] in the above solution

must be found• Then the ion product is calculated and compared

with the Ksp

3. Compute• Calculate the mole quantities of Ba2+

and SO42− ions

• Calculate the total volume of solution containing Ba2+ and SO4

2− ions• 0.020 L + 0.020 L = 0.040 L• Calculate the Ba2+ and SO4

2− ion concentrations in the combined solution

• Trial value of the ion product:

[Ba2+][SO42−] = (5.0 × 10−3)(2.5 × 10−3) = 1.2 ×

10−5

• The ion product is much greater than the value of Ksp, so precipitation occurs.

Practice Problem• Does a precipitate form when 100. mL

of 0.0025 M AgNO3 and 150. mL of 0.0020 M NaBr solutions are mixed?

• AgBr precipitates.

• Does a precipitate form when 20. mL of 0.038 M Pb(NO3)2 and 30. mL of 0.018 M KCl solutions are mixed?

• PbCl2 does not precipitate