Chemical Equilibrium Chapter 5

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Chemical Equilibrium 1. Introduction 2. Equilibrium Constant 3. Reaction Quotient 4. Calculations involving Equilibrium Constant and Reaction

Quotients. 5. Equilibrium Constant in Pressure 6. Le Chateliers Principle

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Kinetics applies to:

the speed of a reaction, the concentration of product that appears (or of reactant that disappears) per unit time. Rate laws can have a complicated dependence on concentrations,

unrelated to the reaction stoichiometry and depend on reaction mechanism

Equilibrium applies to:

the extent (yield) of a reaction, the concentration of reactant and product that has appeared after the reaction is set to run at a given temperature for an unlimited time until no further change occurs Equilibrium constant expressions can be deduced directly from the reaction stoichiometry and do not depend on the reaction mechanism.

k

K

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At equilibrium: rateforward = ratereverse

A system at equilibrium is dynamic on the molecular level; no further net change is observed because changes in one direction are balanced by changes in the other.

Concentration of reactants and products not necessarily equal.

Reactants A

Products B

2 moles of A is converted into B per second, 2 moles of B is converted into A per second.

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N2O4 (g, colourless) 2NO2 (g, brown)

Reaching equilibrium on the macroscopic and molecular levels.

A system is at equilibrium when its composition doesn't change any more over time.

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Approach Equilibrium from different Sides

H2 + I2 2HI From reactants

From products

*

pure H2 + pure I2

(1:1)pure HI

pure H2 + pure I2 + pure HI(1:1:1)

equilibrium mixture ofH2, I2, HI

= same value = Kc

eqilibriumn

eqilibriumm

][reactants[products]

The equilibrium was approached from both the reactant and the product side and a mixture of both, all three resulted in the same relative composition.

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Equilibrium constant (Kc): For a reaction at equilibrium aA + bB cC + dD

ba

dc

[B][A][D][C]

The Equilibrium Constant

Kc =

Kc = eqilibrium

n

eqilibriumm

][reactants[products]

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Do Equilibrium Constants have Units?

Equilibrium constants have no real units, because the equilibrium constant for aA + bB cC + dD is exactly expressed as: where [A], etc are the concentrations at an agreed reference state (which is 1 molL-1 for solutes). Because of the divisions, units (molL-1) of [A] and [A], etc cancel and Kc is truly dimensionless.

boao

doco

c )([B]/[B])([A]/[A])([D]/[D])([C]/[C]K =

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Writing an Equilibrium Constant

NH3 + O2 N2 + 1H2O 4NH3 + 3O2 2N2 + 6H2O KC(b) = (KC(a))4 An equilibrium constant belongs to the specific chemical equation with its specific stoichiometry.

0.7523

1.52

0.52

C ]][O[NHO][H][N(a)K =

32

43

62

22

C ][O][NHO][H][N(b)K =

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Combining Equilibrium Constants

Multiple equilibria N2(g) + O2(g) 2NO(g) 2NO(g) + O2(g) 2NO2(g)

N2(g) + 2O2(g) 2NO2(g)

Kc(1) = [N2][O2]

[NO]2 = 2.3 x 10-19

Kc(2) = [NO]2[O2]

[NO2]2 = 3 x 106

Kc(overall) = [N2][O2]2

[NO2]2

= 6.9 x 10-13

= Kc(1) x Kc(2)

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Write their Kc Expressions

1. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) 2. Hg(l) + Cl2(g) HgCl2(s) 3. CaCO3(s) CaO(s) + CO2(g)

COOH][CH]O][HCOO[CHK

3

33C

+

=

][Cl1K

2C =

][COK2C

=

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Solids and liquids are not included in Kc.

Reaction Quotient (Qc)

]COO][CH[NHCOOH]][CH[NHQ

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33c +=

The ratio of concentration terms written for a given reaction (non equilibrium) is called the reaction quotient (Qc)

NH4+ + CH3COO- NH3 + CH3COOH (in water)

]COO][CH[NHCOOH]][CH[NHK

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33c +=

(non equilibrium system)

(equilibrium system)

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Figure 17.5 Reaction direction and the relative sizes of Q and K.

Equilibrium: no net change

reactants products

Reaction Progress

reactants products

Reaction Progress

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H2 + I2 2HI Begin with 1 molL-1 of HI, and let reaction reach equilibrium with H2 and I2 at 25C (Kc = 25). What will be the final concentrations?

H2 I2 HI

initial 0 0 1

change +x +x -2x

equilibrium x x 1-2x

Forward reaction

Calculations involving Kc

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Insert into Kc equation: Final concentrations: [HI] = 0.72, [H2] = [I2] = 0.14. Put into Kc expression for final check.

0.14x17x5x2x15x2x1

25x2x)(1

(x)(x)2x)(1

]][I[H[HI]K 2

22

22

2

c

====

=

=

==

4H2 + 4I2 8HI Begin with 1 molL-1 of HI, and let reaction reach equilibrium with H2 and I2 at 25C (Kc = 254 = 390625). What will be the final concentrations?

H2 I2 HI initial 0 0 1 change +4x +4x -8x equilibrium 4x 4x 1-8x

0.14 ][I ][H and 0.72 [HI]

0.0357x20x8x 154x

8x13906254x

8x1

390625(4x)

8x)(1(4x)(4x)8x)(1

][I][H[HI]K

22

8

8

8

44

8

42

42

8

c

===

===

=

=

=

==

Typically, equations are written with smallest integer values.

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N2O4 2NO2 Begin with 1 molL-1 of N2O4, and let reaction reach equilibrium with NO2 at 500K (Kc = 46). What will be the final concentrations?

N2O4 NO2

initial 1 0

change -x +2x

equilibrium 1-x 2x

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Insert into Kc equation: (the other solution, x = -12.425, is unphysical)

Final concentrations: [N2O4] = 0.075, [NO2] = 1.85. Put into Kc expression for final check.

0.9258

53.4468

46*4*44646x

04646x4x46x464x

46x1

(2x)]O[N

][NOK

2

22

2

42

22

c

=+

=+

=

=+=

=

==

Quadratic Equation

2a4acbbx

2 =

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KP and Kc For gases, instead of Kc one usually uses KP: For any component X, with ideal-gas law PV = nRT, one gets For the above example:

ba

dc

c [B][A][D][C]K = b

Ba

A

dD

cC

P PPPPK =

[X]RTRTVnP XX ==

cb)](a-d)[(c

bbaa

ddcc

bB

aA

dD

cC

P KRT)((RT)[B](RT)[A](RT)[D](RT)[C]

PPPPK ++===

cn K(RT)=

Gas constant = R = 0.0821 L atm mol-1 K-1

KP

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KP has units? Equilibrium constants have no real units, because the equilibrium constant for aA + bB cC + dD is exactly expressed as: where PA, etc are the concentrations at an agreed reference state (which is in 1 bar (= 1 atm) (NOT Pa) for gases). Because of the divisions, units (bar) of PA and PA, etc cancel and KP is truly dimensionless.

bo

BB

ao

AA

do

DD

co

CC

P /P(P/P(P/P(P/P(PK

))))

=

*

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Le Chatelier's Principle If a system in a state of equilibrium is disturbed, it will undergo a change that shifts its equilibrium position in a direction that reduces the effect of the disturbance.

There are 4 types of stress/changes: - amount of reactants and/or products - pressure - temperature - catalyst

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Le Chateliers Principle

Q depends on the amount of the reactants and products, the pressure and the temperature.

The Q is always the same for a reaction, whether a catalyst is present or not.

K is only dependent on T.

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1. Change Amounts of Reactants and/or Products

Adding product makes Q > K

Removing reactant makes Q > K

Adding reactant makes Q < K

Removing product makes Q < K

Reactants Products

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2. Change Pressure

Only affects gases

- System will move in the direction that has the least moles of gas when P is increased.

- Because partial pressures (and concentrations)

change, new equilibrium pressures (and concentrations) must be reached.

- K remains the same.

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The position of the equilibrium i.e. the concentrations of the reactants and products, can be manipulated.

N2O4 (g, colourless) 2NO2 (g, brown)

At equilibrium at room T Cool Heat

3. Change in Temperature

Only temperature changes can alter K. The direction of the shift depends on whether the

reaction is exothermic or endothermic. If forward reaction is exothermic (H -ve) then reverse

reaction is endothermic (H +ve).

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Exothermic When H < 0 (negative) The reaction releases heat Think of heat as a product Raising temperature,

increases the product, push towards reactants (left) and Q and K decrease.

Decreasing temperature, decreases the product, push towards products (right) and Q and K increase.

Endothermic When H > 0 (positive) The reaction absorbs heat Think of heat as a reactant Raising temperature,

increases the reactant, push towards products (right) and Q and K increase.

Decreasing temperature, decreases the reactant, push towards reactants (left) and Q and K decrease.

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Exothermic H < 0:

In (K2/K1) = (-(- ve)/R) - ve K2/K1 = exp(- ve) K2/K1 = < 1 K2 < K1 Increase T, K decreases.

(1/T2 1/T1) is - ve when T2 > T1

Endothermic H > 0:

In (K2/K1) = (-(+ ve)/R) - ve K2/K1 = exp(+ ve) K2/K1 = > 1 K2 > K1 Increase T, K increases.

The vant Hoff Equation-the effect of T on K

ln K2

K1 = - H

orxn

R

1

T2

1

T1 -

R = universal gas constant = 8.314 J/mol*K

K1 is the equilibrium constant at T1 K2 is the equilibrium constant at T2

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Acid-Base Equilibrium

1. Arrhenius theory 2. Brnsted-Lowry and Lewis 3. Acid base strength 4. Autoionisation of water 5. pH scale and pKw 6. Relationship between Ka and Kb 7. Measure pH 8. Calculate pH, Ka and Kb 9. Hydrolysis of salts

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Arrhenius Theory Acid: - Has H in its formula. - Substance that dissociates in water to produce hydronium

ions (H3O+). - Acid + water H3O+ Base: - Has OH in its formula. - Substance that dissociates in water to produce hydroxide

ions (OH-). - Base + water OH-

1859 1927

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Limitation of Arrhenius definition

- substance can act as acid or base without H or OH in

their formula e.g. NH3 common base. - is limited to aqueous systems (many substances act

as acids or bases in non-aqueous solvents) e.g. in organic solvent methanol.

*

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Brnsted-Lowry Acid: - a substance which can donate a proton (H+). - Proton donor - must have a removable (acidic) proton. Base: - a substance which accepts a proton (H+). - Proton acceptor. -must have a pair of non-bonding electrons (to accept proton)

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Essentially, it is proton transfer from the acid to the base. An acid-base reaction rather than an acid reaction or a base reaction.

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HCl H2O

+ Cl- H3O+

+

(acid, H+ donor) (base, H+ acceptor)

NH3 H2O +

NH4+ OH-

+

(base, H+ acceptor) (acid, H+ donor)

Conjugate acid Conjugate base

Conjugate acid Conjugate base

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If it can be either acid or base ...it is amphiprotic.

(amphoteric) Acid Base CO32- HCO3 H2CO3 SO42- HSO4 H2SO4 OH- H2O H3O+

conjugate base of the acid

conjugate acid of the base

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Lewis Acids

Lewis acids are defined as electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids. AlCl3 is another common Lewis acid.

Lewis bases are defined as electron-pair donors. CM 1502

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

K = = Ka [HA]]][AO[H -3

+Acid Strength

Acid dissociation constant (Ka) is the measure of acid strength (in H2O solution).

Strong acids ~100% dissociation, higher [H3O+], Ka>>1 Weak acids, lower [H3O+], Ka

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Polyprotic Acids Have more than one proton to donate. Polyprotic acids dissociate in a stepwise manner.

H3PO4(aq) + H2O(l) H2PO4-(aq) + H3O+(aq)

H2PO4-(aq) + H2O(l) HPO42-(aq) + H3O+(aq)

HPO42-(aq) + H2O(l) PO43-(aq) + H3O+(aq)

Ka1 = [H3O+][H2PO4-]

[H3PO4]

Ka2 = [H3O+][HPO42-]

[H2PO4-]

Ka3 = [H3O+][PO43-]

[HPO42-] Ka1 > Ka2 > Ka3 Generally, easiest to lose the first proton, then second, then third.

= 7.2x10-3

= 6.3x10-8

= 4.2x10-13

H3PO4: triprotic acid

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Polyprotic Acids If the difference between the Ka for the first

dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

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Base Strength

Standard: H2O is the acid Base dissociation constant (Kb) is the measure of base strength

(in H2O solution)

K = ][B

][HB][OH-

-

= Kb B-(aq) + H2O(l) HB(aq) + OH-(aq)

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Autoionization of Water

In pure water, same number of molecules act as bases and as acids. Hence, [H3O+] = [OH-]. This is referred to as autoionization of water.

Kw = [H3O+][OH-], called ion product constant for water.

At 25 0C, Kw = [H3O+][OH] = 1.0 x 10-14 Kw = [H3O+]2 = [OH-]2

[H3O+] = [OH-] = [H3O+] = [OH-] = = 1.0 107 M

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

wK1410 x 1.0

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Relationship between Ka and Kb of a conjugate Acid-Base Pair

Ka = [HA]]][AO[H -3

+

Kb =

Ka Kb =

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

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

][HA][OH-

-

[H3O+][OH] = Kw

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T (C) Kw 0 0.114 x 10-14

10 0.292 x 10-14 20 0.681 x 10-14 25 1.01 x 10-14 30 1.47 x 10-14 40 2.92 x 10-14 50 5.47 x 10-14 60 9.61 x 10-14

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pH scale The range of [H3O+] and [OH-] is from ~10 to 10-15 molL-1.

Thus, instead of having 10-x, it is convenient to use a logarithmic scale.

pH is defined as the negative base-10 logarithm of the hydronium ion [H3O+] concentration. (p -log10)

pH = log [H3O+]

Reverse operation [H3O+] = 10-pH

large x, -log x small opposite directions

Other p Scales pOH = log [OH] pKw = log Kw = 14 pKa = log Ka pKb = log Kb CM 1502

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Figure 18.6 The relations among [H3O+], pH, [OH-], and pOH.

Therefore, in pure water, pH = pOH = log (1.0 107) = 7.00 NEUTRAL

An ACID has a higher [H3O+] than pure water and [OH-], so its pH is < 7.

A BASE has a higher [OH-] than pure water and [H3O+], so its pH is > 7.

-log10 -log10

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Why does pH + pOH = 14?

Kw = [H3O+][OH] = 1.0 1014,

we know that (or if log each sides of equation), log [H3O+] + log [OH] = log Kw = 14.00

or, in other words,

pH + pOH = pKw = 14.00

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Acid base properties of salt solutions

Acid + Base Salt + Water e.g. NaOH(aq) + HCl(aq) NaCl(aq) + H2O Aqueous solutions of salts can be neutral, acidic or basic. SALTS ARE NOT ALWAYS NEUTRAL! Strong acid + Strong base Neutral solutions Weak acid + Weak base Neutral, Acidic, Basic solutions Strong acid + Weak base Acidic solutions Weak acid + Strong base Basic solutions

WHY?

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Hydrolysis of Salts Salt hydrolysis refers to the situation when a salt is

dissolved in water and its constituent ion reacts with water. When salts such as NH4Cl are dissolved in aqueous

solutions, an acidic solution is obtained i.e. pH < 7 Cation Hydrolysis: If an ion from the salt associates with the

OH- entity of the H2O => acidic solution When salts such as Na2CO3, NaAc (acetate) are dissolved

in aqueous solutions, a basic solution is obtained i.e. pH > 7 Anion Hydrolysis: If an ion from the salt removed a H+ from

the H2O => basic solution

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Cation Hydrolysis

Salts from strong acids and weak bases.

E.g. AlCl3, the Cl- does not undergo anion hydrolysis but Al3+ has weak acid properties and react with the OH- entity of H2O => acidic solution!

General case: Mn+ + 2H2O H3O+ + MOH(n-1)+

[H3O+][MOH(n-1)+] [Mn+] Ka =

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Anion Hydrolysis

Salts from weak acids and strong bases.

E.g. NaAc, the Na+ does not undergo cation hydrolysis but Ac- has basic properties and react with the H+ entity of H2O => basic solution!

General case: X- + H2O HX + OH-

[HX][OH-] [X-] Kb =

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Exception Strong Acid and Strong Base

Salts formed from reactions of strong acid and strong base, such as NaCl, NaNO3, when dissolved in aqueous solutions, a neutral solution is obtained i.e. pH = 7.

Na+(aq) + H2O(l) NaOH(aq) + H+(aq) The ions do not undergo hydrolysis.

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

Cannot occur

X

X

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Exception Weak Acid and Weak Base (BUFFERS)

Aqueous solution of the salts may be basic, acidic or neutral. The pH depends on the relative Ka and Kb of the parent acid

and base. Consider the case when Ka = Kb with salt: NH4CH3COO,

ammonium acetate. NH4CH3COO(aq) NH4+(aq) + CH3COO-(aq) Both ions can undergo hydrolysis.

Ka = Kb => The same concentration of H3O+ as OH- is produced => solution is NEUTRAL!

Ka

Kb

NH4+(aq) + 2H2O(l) NH4OH(aq) + H3O+(aq) CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

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If the parent Ka > Kb: solution is acidic!

e.g. NH4F

Kb(F-): 1.5 x 10-11 < Ka(NH4+): 5.7 x 10-10

So NH4+ hydrolyses to a greater extent than F-

more H3O+ is produced than OH- => ACIDIC!

Ka Kb

NH4+(aq) + 2H2O(l) NH4OH(aq) + H3O+(aq) F-(aq) + H2O(l) HF(aq) + OH-(aq)

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If the parent Ka < Kb: solution is basic!

e.g. NH4CN

Kb(CN-): 1.6 x 10-5 > Ka(NH4+): 5.7 x 10-10

So CN- hydrolyzes to a greater extent than NH4+

more OH- is produced than H3O+ => BASIC!

Ka Kb

NH4+(aq) + 2H2O(l) NH4OH(aq) + H3O+(aq) CN-(aq) + H2O(l) HCN(aq) + OH-(aq)

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Ionic Equilibria in Aqueous systems

1. Buffer solutions

2. Indicators

3. Titrations

4. Solubility Equilibrium

5. Calculations involving Solubility and Solubility Product

6. The Common Ion Effect

7. Solubility and pH

8. Complex Ion Equilibrium

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Buffer Solutions A buffer solution is a solution which does not undergo

any significant change in pH when small amounts of acid or base are added to it.

Resists changes in pH.

In order to neutralize the acid or base added, the buffer solution must contain large and roughly equal concentrations of both a weak acid and a weak base (usually the acids conjugate base).

Weak acid +its salt, eg: HF+NaF Weak base +its salt, eg: NH3 + NH4Cl

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The Common Ion Effect

The shift of equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction is called the common ion effect.

This effect makes a solution of NaF and HF less acidic than a solution of HF alone.

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Buffer with equal concentrations of conjugate acid and base

OH- H3O+

Buffer after addition of H3O+

H2O + CH3COOH H3O+ + CH3COO-

Buffer after addition of OH-

CH3COOH + OH- H2O + CH3COO-

CH3COOH amount produced is much smaller compared to initial amount, so changes in pH is small.

CH3COO- amount produced is much smaller compared to initial amount, so changes in pH is small.

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Ka = 1.8 x 10-5 Kb = 5.6 x 10-10

How does a buffer work

-------------------------------------------------------------------------------------------------------

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HA(aq) + H2O(l) A-(aq) + H3O+(aq) [H3O+][A-] [HA] Ka =

Take the negative log of both sides:

Henderson Hasselbalch equation

[acid][base]log10+= apKpH[acid]

[base]log10+= apKpH

HelloSticky Noteuse for equilibrium state or buffer solution b/c buffer solution does not change much in components concentration

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Buffer Capacity Buffer capacity is the ability to resist pH change. It is the amount of H+ or OH- ions that a buffer can absorb without a significant change in pH. Determined by the absolute or relative magnitudes of [HA] and [A-]. The more concentrated the components of a buffer, the greater the buffer capacity. A buffer has the highest capacity when the component concentrations are equal.

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[X-] [HX]

[X-] [HX]

Buffer Range

The range of should be 0.1 10 (1:10 or 10:1).

pH = pKa + log10

*

Therefore, pH = pKa 1

Buffers have a usable range within 1 pH unit of the pKa of the acid component.

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Indicators Acid-base indicators are buffers a weak acid and its conjugate base, where each respectively have significantly different visible colors.

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

RED Solution acidic

BLUE Solution basic

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+ -3

,[H O ][In ]

[HIn]a a InK K= =

Taking logs of both sides and rearranging,

-

, 10[In ]log[HIn]a In

pH pK= +solution

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When an indicator is at the mid point of its color transition range, then

-

,

[HIn] [In ]

a InpH pK=

=

= 1

RED BLUE

-

, 10[In ]log[HIn]a In

pH pK= +solution

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For a colour change to be observed, (i) the acid colour of an indicator needs to be clearly developed which

occurs when ~90% or more of the indicator is in the HIn form, and (ii) the alkaline colour of an indicator needs to be clearly developed

which happens when ~90% or more of the indicator is in the In- form. Therefore range, Acid colour: log10 = -1 ( -0.95) Alkaline colour: log10 = 1 ( +0.95)

Acid colour ~90% HIn ~10% In-

Alkaline colour ~10% HIn ~90% In-

pH = pKa,In - 1 pH = pKa,In + 1

= 0.11

= 9

-

, 10[In ]log[HIn]a In

pH pK= +

* CM 1502

50% HIn 50% In-

67

pH Figure 19.5

A good indicator will change colour close to the end point of the titration.

pH = pKa,In 1 where an indicator will change colour

pH range of the solution where the equivalence point happens

=

Eq. point pH = pKa,In +1

pH = pKa,In -1

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Equivalence point: pH at which nacid = nbase End point: pH at which indicator changes colour.

Acid-Base Titrations

Buffer regions

Volume of base added

Volume of base added Volume of base added

Volume of base added

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Strong Acid-Strong Base Titrations A strong base added to strong acid :

Initially: calc. pH from [HA] = [H3O+] Before equiv. point: pH from [H3O+] = (nA nB)/(VA + VB) At equiv. point: nA = nB, [H3O+] = [OH-], pH = 7 After equiv. point: pH from [OH-] = (nB nA)/(VA + VB)

HA + H2O A- + H3O+

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Weak Acid-Strong Base Titrations

A strong base added to weak acid :

Initially: weak acid dissociation Ka = [H3O+]2/[HA] calc. pH from [H3O+] = Before equiv. point: Use buffer equation

.[HA]Ka

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10[base]log[acid]a

pH pK= +

HA + H2O A- + H3O+

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Weak Acid-Strong Base Titrations At equiv. point: nA = nB, salt formed undergoes anion hydrolysis Initial [A-] = nB/(VA + VB) Kb = [OH-]2/[A-] calc. pH from [OH-] = After equiv. point: pH from [OH-] = (nB nA)/(VA + VB)

].[AK -b

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A- + H2O HA + OH-

Initially: calc pH from [HA] = [H3O+] Before equiv. point: [H3O+] = (nA nB)/(VA + VB) At equiv. point: nA = nB, salt formed undergoes cation hydrolysis Initial [Mn+] = nA/(VA + VB) Ka = [H3O+]2/[Mn+] calc. pH from [H3O+] = After equiv. point: Use buffer equation for bases.

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Strong Acid-Weak Base Titrations A weak base added to strong acid :

].[MK na+

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Mn+ + 2H2O H3O+ + MOH(n-1)+

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Ksp is called the SOLUBILITY PRODUCT meaning product of soluble species

Solubility Equilibrium

BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+][SO42-]

MmXx (s) mMn+ (aq)+ xXn- (aq) Ksp = [Mn+]m[Xn-]x

The definition of solubility product is applied to any sparingly soluble ionic solid. Example:

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Solubility is the quantity of the ionic solid that dissolves in a given quantity of solvent to give a saturated solution (Units: gL-1 ; g / 100 cm3; mol dm-3; etc).

Solubility product is the equilibrium constant for the dissolution reaction (no units).

Solubility vs. Solubility Product

If the solubility of a compound is known, one can calculate the corresponding solubility product (and vice versa).

CM 1502 76

The main use of solubility products is to determine conditions under which an electrolyte will dissolve or precipitate from solution.

(i) [Mn+]m[Xn-]x (Qsp) < Ksp: soln. is unsaturated; MmXx(s) will dissolve.

(ii) [Mn+]m[Xn-]x (Qsp) = Ksp: soln. is saturated; no dissolution or precipitation of MmXx. (iii) [Mn+]m[Xn-]x (Qsp) > Ksp: soln. is supersaturated; MmXx(s) will precipitate.

Dissolve or Not?

CM 1502 77

The solubility of a sparingly soluble salt is decreased by the presence of a second solute having an ion in common with the sparingly soluble salt.

The Common Ion Effect on Solubility

PbCrO4(s) Pb2+(aq) + CrO42-(aq)

CrO42- added

By Le Chateliers Principle, the position of equilibrium will shift to the left, favouring the formation of the reactants, Q > K => solubility of PbCrO4 decreases!

CM 1502 78

Kstab is the stability constant of a complex ion (e.g. MX64-) used primarily in dilute aqueous solutions.

Kstab of a complex ion is the equilibrium constant for the formation of the complex ion in a solvent from its constituent ions. M2+(aq) + 6X-(aq) MX64-(aq)

Complex Ion Equilibrium

6-2

-46

stab ]][X[M][MX K +=

Cr(NH3)63+

Slide Number 1Chemical EquilibriumSlide Number 3Slide Number 4Slide Number 5Approach Equilibrium from different SidesSlide Number 7Do Equilibrium Constants have Units?Writing an Equilibrium ConstantCombining Equilibrium ConstantsWrite their Kc ExpressionsReaction Quotient (Qc)Slide Number 13Slide Number 14Calculations involving KcSlide Number 16Slide Number 17Slide Number 18Slide Number 19KP and KcKP has units?Le Chatelier's PrincipleLe Chateliers Principle1. Change Amounts of Reactants and/or Products2. Change Pressure3. Change in TemperatureExothermicSlide Number 28Slide Number 29Acid-Base EquilibriumArrhenius TheoryLimitation of Arrhenius definitionBrnsted-Lowry Slide Number 34If it can be either acid or baseLewis AcidsSlide Number 37Slide Number 38Polyprotic AcidsPolyprotic AcidsBase StrengthAutoionization of WaterRelationship between Ka and Kbof a conjugate Acid-Base PairSlide Number 44pH scaleSlide Number 46Why does pH + pOH = 14?Acid base properties of salt solutionsHydrolysis of SaltsCation Hydrolysis Anion HydrolysisException Strong Acid and Strong BaseException Weak Acid and Weak Base (BUFFERS)Slide Number 54Slide Number 55Slide Number 56Ionic Equilibria in Aqueous systemsBuffer SolutionsSlide Number 59Slide Number 60Slide Number 61Slide Number 62Buffer RangeIndicatorsSlide Number 65Slide Number 66Slide Number 67Slide Number 68Slide Number 69Strong Acid-Strong Base TitrationsSlide Number 71Slide Number 72Slide Number 73Slide Number 74Slide Number 75Slide Number 76Slide Number 77Slide Number 78