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General Chemistry IIGeneral Chemistry IICHEM 152 Unit 2CHEM 152 Unit 2

Week 9

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Week 9 Reading Assignment

Chapter 16 – Sections 16.2 (buffers), 16.3 (buffers), 16.4 (titrations), 16.5 (Ksp)

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Equilibrium in Equilibrium in Aqueous Solutions Aqueous Solutions

Acids and BasesAcids and Bases

We have explored the properties of acids and bases as we add them

independently to water. But what

happens when you mix them?

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TitrationsTitrations

You have covered the topic of mixing acids and bases for

titrations in lab. We will assume that

you understand this topic and cover

some other topics of mixing acids and

bases.

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What is the effect on the pH of adding What is the effect on the pH of adding NHNH44Cl to 0.25 M NHCl to 0.25 M NH33(aq)?(aq)?

NHNH33(aq) + H(aq) + H22O O NH NH44++(aq) + OH(aq) + OH--(aq)(aq)

Here we are adding an ion Here we are adding an ion COMMONCOMMON to the to the equilibrium (equilibrium (COMMON ION EFFECTCOMMON ION EFFECT))

Let us first calculate the pH of a Let us first calculate the pH of a 0.25 M NH0.25 M NH33 solution: solution:

[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial 0.250.25 0 0 00changechange -x-x +x +x +x +xequilibequilib 0.25 - x0.25 - x x x xx

Mixing Conjugate PairsMixing Conjugate Pairs

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NHNH33(aq) + H(aq) + H22O O NH NH44++(aq) + OH(aq) + OH--(aq)(aq)

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x2

0.25 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.25 - x

pH of Aqueous NHpH of Aqueous NH33

Assuming x is << 0.25, we haveAssuming x is << 0.25, we have

[OH[OH--] = x = [K] = x = [Kbb(0.25)](0.25)]1/21/2 = 0.0021 M = 0.0021 M

This gives pOH = 2.67This gives pOH = 2.67

and so pH = 14.00 - 2.67and so pH = 14.00 - 2.67

= = 11.3311.33 for 0.25 M NH for 0.25 M NH33

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What is the pH of a solution with 0.25 M What is the pH of a solution with 0.25 M NHNH44Cl and 0.25 M NHCl and 0.25 M NH33(aq)?(aq)?

NHNH33(aq) + H(aq) + H22O O NHNH44++(aq) + OH(aq) + OH--(aq)(aq)

We expect that the pH will decline on We expect that the pH will decline on adding NHadding NH44Cl. Let’s test that!Cl. Let’s test that!

pH of NHpH of NH33/NH/NH44++ Mixture Mixture

[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial 0.250.25 0.250.25 00

changechange -x-x +x+x +x+x

equilibequilib 0.25 - x0.25 - x 0.25 + x 0.25 + x xx

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Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x(0.25 + x)

0.25 - x

pH of NHpH of NH33/NH/NH44++ Mixture Mixture

[[OHOH--] = x = (0.25 / 0.25) K] = x = (0.25 / 0.25) Kbb = 1.8 x 10 = 1.8 x 10-5-5

MM

This gives pOH = 4.74 and pH = 9.26This gives pOH = 4.74 and pH = 9.26

pH drops from 11.33 to 9.26 on pH drops from 11.33 to 9.26 on

adding a common ion.adding a common ion.

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pH of NHpH of NH33/NH/NH44++ Mixture Mixture

What is the pH when 1.00 mL of 1.00 M HCl is What is the pH when 1.00 mL of 1.00 M HCl is added to 1.00 L of a mixture that has [NHadded to 1.00 L of a mixture that has [NH33]= ]=

0.25 M and [NH0.25 M and [NH44++]= 0.25 M ]= 0.25 M

(pH = 9.26)(pH = 9.26)

HH33OO++ + NH + NH33 NH NH44++ + H + H22OO

The reaction occurs completely because K is The reaction occurs completely because K is large.large.

HH33OO++ + NH + NH33 NH NH44++ + H + H22OO

Before rxnBefore rxn 0.001 0.250.001 0.25 0.250.25

Change -0.001 –0.001 +0.001Change -0.001 –0.001 +0.001

After rxn ~0 0.249 +0.251After rxn ~0 0.249 +0.251

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pH of NHpH of NH33/NH/NH44++ Mixture Mixture

Equilibrium will be re-established:

Ka=5.6 x 10-10 NH4+ + H2O H3O+ + NH3

Equilibrium 0.251-x x 0.249+x

[H3O ] [NH4+]

[NH3]

• Ka 0.2510.249

• (5.6 x 10-10 )

[H[H33OO++] = 5.64 x 10] = 5.64 x 10-10-10 M M pH = 9.25pH = 9.25

The pH has not changed much (pH = 9.26The pH has not changed much (pH = 9.26 to 9.25) on adding HCl to the mixture! on adding HCl to the mixture!

Why?Why? What would happen if we add NaOH?What would happen if we add NaOH?

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Formation of a BufferFormation of a Buffer

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The The NHNH33/NH/NH44++ mixture is a typical example mixture is a typical example

of a of a BUFFERBUFFER. A buffer is a solution that . A buffer is a solution that resists changes in the pH when limited resists changes in the pH when limited

amounts of base or acid are added to it.amounts of base or acid are added to it.

Buffer CompositionBuffer CompositionWeak AcidWeak Acid + Conjugate Base+ Conjugate Base

CHCH33COOHCOOH ++ CH CH33COOCOO

HH22POPO44 ++ HPO HPO44

22

Weak BaseWeak Base ++ Conjugate AcidConjugate Acid

NHNH33 ++ NH NH44++

Buffer SolutionsBuffer Solutions

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H+? OH-?

Action of a BufferAction of a Buffer

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[CH[CH33COOH] + HCOOH] + H22O O [CH [CH33COOCOO--]+ [H]+ [H33OO++]]initialinitialchangechangeEquilibEquilib

What is the pH of a buffer that has [CHWhat is the pH of a buffer that has [CH33COOH] COOH] = 0.700 M and [CH= 0.700 M and [CH33COOCOO--] = 0.600 M?] = 0.600 M?

CHCH33COOH + HCOOH + H22O O CH CH33COOCOO-- + H + H33OO++

KKaa = 1.8 x 10 = 1.8 x 10-5-5

Buffer SolutionsBuffer Solutions

0.7000.700 0.6000.600 00

-x-x +x+x +x+x

0.700 - x0.700 - x 0.600 + x0.600 + x xx

Ka 1.8 x 10-5 = [H3O+](0.600)

0.700 Ka 1.8 x 10-5 =

[H3O+](0.600)0.700

[H[H33OO++] = 2.1 x 10] = 2.1 x 10--

55 pH = 4.68pH = 4.68

Assuming that Assuming that x << 0.700 and 0.600x << 0.700 and 0.600, we , we havehave

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Notice that the expression for calculating Notice that the expression for calculating the Hthe H33OO++ concentration of the buffer is concentration of the buffer is

Buffer SolutionsBuffer Solutions

[H3O] [Acid]

[Conj. base]•Ka[H3O]

[Acid][Conj. base]

•Ka

Take the Take the negative lognegative log of both sides of of both sides of this equationthis equation

pH pKa + log[Conj. base]

[Acid]pH pKa + log

[Conj. base][Acid]

pH pKa - log[Acid]

[Conj. base]

Henderson-Henderson-Hasselbalch Hasselbalch

EquationEquation

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Preparing a BufferPreparing a Buffer

Imagine that you want to prepare a Imagine that you want to prepare a buffer solution with a pH = 4.30. buffer solution with a pH = 4.30.

It is best to choose an acid such that It is best to choose an acid such that [H[H33OO++] is about equal to K] is about equal to Kaa (or pH ≈ (or pH ≈

pKpKaa).).

Then you get the exact pH by adjusting Then you get the exact pH by adjusting the ratio of acid to conjugate base.the ratio of acid to conjugate base.

pH pKa + log[Conj. base]

[Acid]pH pKa + log

[Conj. base][Acid]

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You want to buffer a solution at You want to buffer a solution at pH = pH = 4.304.30 and have access to the following and have access to the following

acids:acids:

POSSIBLE ACIDSPOSSIBLE ACIDS KKaa

HSOHSO44- - / SO/ SO44

2-2- 1.2 x 101.2 x 10-2-2

CHCH33COOH/CHCOOH/CH33COOCOO--1.8 x 101.8 x 10-5-5

HCN / CNHCN / CN-- 4.0 x 104.0 x 10-10-10

Preparing a BufferPreparing a Buffer

Which one would you choose? If you have a 0.100 M

concentration of the base, what concentration of the acid would

you need?

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If you have a 0.100 M concentration of the base, what concentration of the acid would you need?

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Your TurnThe blood of mammals is an aqueous solution that maintains a constant pH. The normal pH

of human blood is 7.40. Carbon dioxide provides the most important blood buffer. In

solution, CO2, reacts with water to form H2CO3, which ionizes to produce H3O+ and

HCO3- ions:

CO2(g) + H2O(l) H2CO3(aq)

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

Ka= 4.2 x 10-7Use this information to determine the concentration ratio [HCO3

-]/[H2CO3] in blood.

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Determine the concentration ratio [HCO3

-]/[H2CO3] in blood

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If the pH of human blood decreases below 7.35, a condition known as acidosis occurs;

increasing the pH above 7.45 causes alkalosis. Both of these conditions can be

life-threatening.

CO2(g) + H2O(l) H2CO3(aq)

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

Ka= 4.2 x 10-7

What is an easy way to treat alkalosis?

(Apply Le Chatelier’s Principle)

Your Turn

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More with Henderson-More with Henderson-HasselbalchHasselbalch

CH2

H2N C

O

OH

pKpKaa(carb. acid) = 2.0(carb. acid) = 2.0

pKpKaa(prot. amine) = (prot. amine) = 10.510.5

Draw the structure of the amino acid at Draw the structure of the amino acid at

a)a)pH = 1.0, b) pH = 7.0, and c) pH = 12.0pH = 1.0, b) pH = 7.0, and c) pH = 12.0

Start by using the Henderson-Hasselbalch Start by using the Henderson-Hasselbalch equation!equation!

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Functional group structuresFunctional group structuresAcidic formsAcidic forms Basic formsBasic forms

C

O

OH

C

O

O

N

H

H

N

H

H

H

Which version of each functional group should Which version of each functional group should exist at a particular pH?exist at a particular pH?

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@pH = 1.0@pH = 1.0

The acid form (-COOH) is more prevalent. The acid form (-COOH) is more prevalent.

For the carboxylic acid group:For the carboxylic acid group:

[acid][base]

log2.01.0 [acid][base]

0.10

The acid form (-NHThe acid form (-NH33++) is more prevalent. ) is more prevalent.

For the amine group:For the amine group:

[acid][base]

log10.51.0 [acid][base]

103.2 10

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@pH = 1.0@pH = 1.0

CH2

H3N C

O

OH

Now – in your groups – Draw the prevalent Now – in your groups – Draw the prevalent structures of the amino acid at pH=7 and at structures of the amino acid at pH=7 and at

pH = 12.pH = 12.

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What ishappenin

ghere?

solution

solution

Solubility

Equilibria

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Solubility of Salts Solubility of Salts

Not all salts are completely soluble in water. Many of them only dissolve to a

small extent.

When equilibrium has been established, no

more AgCl dissolves and the solution is

SATURATED.

When an insoluble salt is placed in water, chemical

equilibrium can be established:

AgCl(s) Ag+(aq) + Cl-(aq)

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Solubility of Salts

Consider: AgCl(s) Ag+(aq) + Cl-(aq)

Write the expression of the equilibrium constant Ksp (solubility product)

associated with this process.

Ksp = [Ag+][Cl-]/ [AgCl(s)]

Ksp = [Ag+][Cl-]

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Solubility of Salts

Consider: AgCl(s) Ag+(aq) + Cl-(aq)

The concentration of [Ag+] in a saturated solution of AgCl is

1.34x10-5 M. This number represents the SOLUBILITY of the salt.

What is the concentration of Cl¯ ions in the system?

What is the value of Ksp for this salt?

Ksp =[Ag+][Cl-] = (1.34 x10-5)2 = 1.80 x 10-10

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Estimate the solubility of the following salts:

PbSO4

Your Turn

Ksp= 1.8 x10-8

Ksp= (x)(x)

x = 1.3 x 10-4 M

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Determine the molar solubility of PbCl2? Ksp = 1.7 x 10-5

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Estimate the solubility of the following salt:

Your Turn

PbCl2

Ksp= 1.7 x 10-5

Ksp= (x)(2x)2

x =1.6 x 10-2 M

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PbIPbI22(s) (s) Pb Pb2+2+(aq) + 2I(aq) + 2I--(aq)(aq)

Calculate KCalculate Kspsp for PbI for PbI22 if if

[Pb[Pb2+2+ ] = 0.00130 M in a saturated ] = 0.00130 M in a saturated solution.solution.

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Consider PbIConsider PbI22 dissolving dissolving

in water in water

PbIPbI22(s) (s) Pb Pb2+2+(aq) + 2I(aq) + 2I--

(aq)(aq)

Calculate KCalculate Kspsp for PbI for PbI22 if if

[Pb[Pb2+2+ ] = 0.00130 M ] = 0.00130 M

PbIPbI22(s) (s) Pb Pb2+2+(aq) + 2I(aq) + 2I--

(aq)(aq)

KKspsp =[0.00130][0.00260] =[0.00130][0.00260]2 2 = =

8.79 x108.79 x10-9-9

0.00130 0.00260

Your Turn

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Solubility of Salts Consider the following insoluble salt:

CaCO3(s)

Will the solubility of this salt increase, decrease, or remain the same after

adding:

a)Sulfuric acid (H2SO4)

b)Hydrochloric acid (HCl)

c) Calcium Chloride (CaCl2)

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Solubility of Salts

Consider the following insoluble salt:

AgCl(s)

Will the solubility of these salts increase, decrease, or remain the

same after adding:

a)Sulfuric acid (H2SO4)

b)Hydrochloric acid (HCl)

c) Calcium Chloride (CaCl2)

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The Common Ion EffectThe Common Ion EffectAdding an ion “common” to an equilibrium causes the

equilibrium to shift back to reactant.

AgCH3CO2(s) Ag+(aq) + CH3CO2-(aq)

AgNO3(aq) Ag+(aq) + NO3-(aq)

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Kidney StonesKidney stones are normally

composed of: calcium oxalate (CaC2O4), calcium

phosphate (Ca3(PO4)2), magnesium ammonium phosphate (MgNH4PO4).Precipitate formed from calcium ions

from food rich in calcium, dairy products, and oxalate ions from chocolate, spinach,

celery, black tea:Ca+2 + C2O4

2- CaC2O4 How would you try to dissolve the

CaC2O4 stones?

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Precipitating an Insoluble SaltPrecipitating an Insoluble SaltConsider the following problem:

Imagine you mix 500.0 mL of a solution of AgNO3 1.5 x 10-5 M with 500.0 mL of a solution of NaCl 1.5 x 10-5 M. Will AgCl

precipitate?AgCl(s) Ag+(aq) + Cl-(aq) Ksp= 1.8 x 10-10

Reactant Quotient, Q=[Ag+]o[Cl-]o

Q < Ksp - no precipitate forms

unsaturated solution

Q > Ksp - precipitate may form supersaturated solution

Q = Ksp - no precipitate forms saturated solution

DID YOU CONSIDER DILUTION?

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Precipitating an Insoluble SaltPrecipitating an Insoluble Salt

Consider the case: Consider the case:

HgHg22ClCl22(s) (s) Hg Hg222+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg222+2+] [Cl] [Cl--]]22

If [HgIf [Hg222+2+]=0.010 M, what [Cl]=0.010 M, what [Cl--] is required to ] is required to

just begin the precipitation of Hgjust begin the precipitation of Hg22ClCl22??

That is, what is the maximum [ClThat is, what is the maximum [Cl--] that can be ] that can be

in solution with 0.010 M Hgin solution with 0.010 M Hg222+2+ without without

forming Hgforming Hg22ClCl22??KKspsp = 1.1 x 10 = 1.1 x 10-18-18 = [Hg = [Hg22+2+] [Cl] [Cl--]]22 = [0.010][Cl = [0.010][Cl--]]2 2

[Cl[Cl--]] = 1.0 x 10 = 1.0 x 10-8 M

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Separating Salts by Differences in Separating Salts by Differences in KKspsp

If two or more substances can precipitate If two or more substances can precipitate in a solution, the substance whose in a solution, the substance whose KKspsp is is

first exceeded precipitates first. first exceeded precipitates first.

Consider a solution containing 0.020 Consider a solution containing 0.020 M ClM Cl-- and 0.010 M CrO and 0.010 M CrO44

2-2- ions in which ions in which Ag+ ions are added slowly. Ag+ ions are added slowly.

Which precipitates first, AgCl or Which precipitates first, AgCl or AgAg22CrOCrO44??

KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12

KKspsp for AgClfor AgCl = 1.8 x 10= 1.8 x 10-10-10

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Separating Salts by Differences in Separating Salts by Differences in KKspsp

Consider a solution containing 0.020 Consider a solution containing 0.020 M ClM Cl-- and 0.010 M CrO and 0.010 M CrO44

2-2- ions in which ions in which Ag+ ions are added slowly. Ag+ ions are added slowly.

KKspsp for Ag for Ag22CrOCrO44 = 9.0 x 10 = 9.0 x 10-12-12

KKspsp for AgClfor AgCl = 1.8 x 10= 1.8 x 10-10-10

What is [AgWhat is [Ag++] when the maximum AgCl ] when the maximum AgCl is precipitated, but none of the is precipitated, but none of the

AgAg22CrOCrO44 has precipitated? has precipitated?

What percent of ClWhat percent of Cl-- remains remains

in solution?in solution?

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Separating Salts by Differences in Separating Salts by Differences in KKspsp

If

PbCl2(s) Pb2+ (aq) + 2Cl-(aq) Ksp=1.7 x 10-5

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

Ksp=1.8 x 10-4

What is the equilibrium constant for the reaction:

PbCl2 + CrO42- PbCrO4 + 2Cl-?

What does this tell you?

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1. What is the pH of a solution of 0.150 M HIO3 (Ka = 0.17)

and 0.350 M NaIO3?

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2. What is the maximum mass of calcium fluoride that will

dissolve in 500. mL of aqueous solution?Ksp = 1.46 x 10-10 M = 78.08 g/mol

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3. What is the molar solubility of CaF2 in a solution of 0.10

M NaF?Ksp = 1.46 x 10-10