MBBS Chapter 3 and Chapter 4 Electrolyte Solutions and Buffers

Solution

strong acid:

strong base:salt: NaCl

weak acid

weak base

Strong electrolyte

Weak electrolyte

Nonelectrolyte

Electrolyte

NaOH KOH

HCl HC1O4 H2S

O4 HNO

3

NH3

Monoprotic acid: HAc

Diprotic acid: H2CO3

Triprotic acid: H3PO4

completely ionized

partially ionized

Chapter 3 and ChapteChapter 3 and Chapter 4 Electrolyte Solutior 4 Electrolyte Solutions and Buffers (11h)ns and Buffers (11h)

Chapter 3 Electrolyte Solutions Chapter 3 Electrolyte Solutions (8h)(8h)

3.1 Strong Electrolytic Solutions (cancel)3.1 Strong Electrolytic Solutions (cancel)

3.2 Brønsted-Lowry Acid-Base Theory3.2 Brønsted-Lowry Acid-Base Theory

3.3 Solving Problems Involving Acid-Base Equilibri3.3 Solving Problems Involving Acid-Base Equilibria (monoprotic acids or bases) a (monoprotic acids or bases)

3.4 Electron-Pair Donation and the Lewis Acid-Bas3.4 Electron-Pair Donation and the Lewis Acid-Base Definition (cancel)e Definition (cancel)

3.5 Equilibria of Slightly Soluble Ionic (cancel)3.5 Equilibria of Slightly Soluble Ionic (cancel)

Chapter 4 BuffersChapter 4 Buffers (4h)(4h)

Key Points:Key Points: Identify conjugate acid-base pairs Identify conjugate acid-base pairs Calculate the concentrations of the species presCalculate the concentrations of the species pres

ent in a solution of a weak acid or base ent in a solution of a weak acid or base Determine the relationship between KDetermine the relationship between Kbb for a bas for a bas

e and the Ke and the Kaa of its conjugate acid of its conjugate acid Determine the pH of a buffer solution containinDetermine the pH of a buffer solution containin

g a weak acid and its conjugate baseg a weak acid and its conjugate base Determine how the addition of small amounts oDetermine how the addition of small amounts o

f acids or bases changes the pH of a buffer solutf acids or bases changes the pH of a buffer solution ion

KKnow how to prepare a buffer of a given pHnow how to prepare a buffer of a given pH

3.2 Brønsted-Lowry Acid-Base 3.2 Brønsted-Lowry Acid-Base TheoryTheory

1.1. History of the Acid-Base ConceptHistory of the Acid-Base Concept In In 16801680, Boyle noted that acids , Boyle noted that acids change the cchange the c

olorolor of certain natural dyes. of certain natural dyes. In the eighteenth centuryIn the eighteenth century it was recognized it was recognized

that acids have a that acids have a sour taste.sour taste.


Lavoisier, in Lavoisier, in 17871787, proposed that acids , proposed that acids are binary compounds of are binary compounds of oxygen oxygen . .

Davy, in Davy, in 18111811, proposed that acids are , proposed that acids are compounds of compounds of hydrogenhydrogen. .

Arrhenius concept, the Brønsted-LowrArrhenius concept, the Brønsted-Lowry protonic concept, the Lewis theoryy protonic concept, the Lewis theory. .


2.2. Arrhenius Concept of Acids and BasesArrhenius Concept of Acids and Bases Definition: Definition: AcidsAcids are substances that prod are substances that prod

uce uce HH33OO++ and and basesbases are substances that pro are substances that produce duce OHOH －－ in aqueous solutions.in aqueous solutions.

HC1O4(aq)+H2O(l)→H3O+(aq)+ClO4-(aq)

NaOH(s) + H2O(l)→ Na+(aq) + OH-(aq) CH3COOH + H2O H3O+ + CH3COO- Reactions: This neutralization of a acid by Reactions: This neutralization of a acid by

a base is essentially the reaction of Ha base is essentially the reaction of H33OO++ (a (aq) with OHq) with OH--(aq) and always give the water. (aq) and always give the water.


Advantage and Disadvantage : Advantage and Disadvantage : The Arrhenius definition applies to aThe Arrhenius definition applies to a

queous solutions, and does not scribe queous solutions, and does not scribe the behavior of substances in other sthe behavior of substances in other solvents or in gas-phase. olvents or in gas-phase.

It singles out the OHIt singles out the OH-- ion as the sourc ion as the source of base character. e of base character.


3.3. Brønsted -Lowry Concept of Acids and BasesBrønsted -Lowry Concept of Acids and Bases Definition: An acid is a proton donorDefinition: An acid is a proton donor ，， and and

a base is a proton acceptor. a base is a proton acceptor. HAc HHAc H++ + Ac + Ac －－ NHNH44

++ H H++ + NH + NH33

HH33POPO44 H H++ + H + H22POPO44 －－ HH22POPO44

－－ HH++ + HPO + HPO4422 －－

HPOHPO4422 －－ HH++ + PO + PO44

33 －－

Acid Proton + Conjugate baseAcid Proton + Conjugate base


Notice:Notice: AcidsAcids contain contain hydrogenhydrogen and all and all negative ionsnegative ions

must be considered asmust be considered as bases bases. When a acids l. When a acids loses a proton, the species formed is base. Thoses a proton, the species formed is base. The two species are conjugate acid-base pair. Ee two species are conjugate acid-base pair. Every Brønsted -Lowry acid has a conjugate bvery Brønsted -Lowry acid has a conjugate base, and every base has a conjugate acid.ase, and every base has a conjugate acid.

Write the formula of the conjugate base of aWrite the formula of the conjugate base of any acid simply by removing a proton . ny acid simply by removing a proton .

conjugate acid-H+

conjugate base+H+


Acids and bases can be ions as well as Acids and bases can be ions as well as molecular substances. molecular substances.

Some species that can act as either aciSome species that can act as either acids or bases are called ds or bases are called amphiproticamphiprotic. Th. The e proton-containing negative ionsproton-containing negative ions are are amphiprotic. amphiprotic.

A salt may be regarded as an ionic comA salt may be regarded as an ionic compound which can act as acids or bases. pound which can act as acids or bases.


Reactions: An acid-base reaction is the Reactions: An acid-base reaction is the trantransfer of a protonsfer of a proton from a proton donor (acid) from a proton donor (acid) to a proton acceptor (base). to a proton acceptor (base).

NeutralizationNeutralizationHCl(g) + NHHCl(g) + NH33(g) → NH(g) → NH44Cl(s) Cl(s)

Ionization or dissociationIonization or dissociationCHCH33COCO22H + HH + H22O HO H33OO++ + CH + CH33COCO22

-- NHNH33 + H + H22O NHO NH44

++ + OH + OH--

HydrolysisHydrolysisAcAc －－ + H+ H22O HAc + OHO HAc + OH －－ NHNH44

++ + H + H22O NHO NH33 + H + H33OO++


Notice:Notice: For a given pair of reactants, write the equation fFor a given pair of reactants, write the equation f

or the transfer of or the transfer of one protonone proton from one species to t from one species to the other. (Do not transfer two protons.) A Brønstehe other. (Do not transfer two protons.) A Brønsted -Lowry acid-base reaction produces a conjugate d -Lowry acid-base reaction produces a conjugate acid and base.acid and base.

The direction for an acid-base reaction is from thThe direction for an acid-base reaction is from the stronger acid and base to the weaker acid and be stronger acid and base to the weaker acid and base. ase.

HBHB11 + B + B22-- BB11

-- + HB + HB22

stronger stronger weaker weakerstronger stronger weaker weaker acid base base acid acid base base acid


The strongest acids have the weakest conjugThe strongest acids have the weakest conjugate bases, and the strongest bases have the ate bases, and the strongest bases have the weakest conjugate acids. weakest conjugate acids.

Water is amphiprotic. Water is amphiprotic. Acid-base reactions are not restricted to aquAcid-base reactions are not restricted to aqu

eous solution. eous solution. Example 1 What are the conjugate bases of Example 1 What are the conjugate bases of

(a) H(a) H22SOSO33 and (b) H and (b) H22POPO44－－ ? ?

Example 2 What are the conjugate acids of Example 2 What are the conjugate acids of (a) CN(a) CN －－ and (b) Hand (b) H22POPO44

－－ ? ?


Example 3 Write the equations illustrating the fExample 3 Write the equations illustrating the following Brønsted -Lowry acid-base reactions.ollowing Brønsted -Lowry acid-base reactions.

(a) H(a) H22S as an acid with HS as an acid with H22O O (b) H(b) H22POPO44

－－ as an acid with OHas an acid with OH －－ (c) H(c) H22POPO44

－－ as an base with Has an base with H33OO++ (d) CN(d) CN －－ as an base with Has an base with H22O O Example 4 In the following equations, label eacExample 4 In the following equations, label eac

h species as an acid or a base. Show the conjugh species as an acid or a base. Show the conjugate acid-base pairs. ate acid-base pairs.

HCOHCO33--+ HF H+ HF H22COCO33+ F+ F--


4.4. Relative Strengths of Acids and Bases Relative Strengths of Acids and Bases The terms stronger and weaker are used here oThe terms stronger and weaker are used here o

nly in a comparative sense. nly in a comparative sense.

HC1(aq)+H2O(l)→H3O+(aq)+Cl-(aq)

CH3CO2H + H2O H3O+ + CH3CO2

-

Notice It is important to understand that the terms It is important to understand that the terms

stronger and weaker are used here only in a stronger and weaker are used here only in a comparative sense. It relates to the types of comparative sense. It relates to the types of solvent system. solvent system.


HC1(aq)+H2O(l)→H3O+(aq)+Cl-(aq)


-

HC1+H2O→H3O++Cl-

HC1O4+H2O→H3O++ ClO4

-

H2SO4 +H2O→H3O++HSO4

-

HNO3+H2O→H3O++NO3

-

HNO3(aq)+H2O(l)→H3O+(aq)+NO3

-(aq)

HNO3 + HAc H2Ac+ +NO3-

HC1 + HNO3 H2NO3 + + Cl -


Notice The strength of the acid is a measure of its The strength of the acid is a measure of its

tendency to liberate proton. tendency to liberate proton. When you say an acid loses its proton When you say an acid loses its proton

readily, you can also say that its conjugate readily, you can also say that its conjugate base does not hold the proton very tightly. base does not hold the proton very tightly. The stronger the acid, the weaker is its The stronger the acid, the weaker is its conjugate base. conjugate base.

The direction for an acid-base reactionThe direction for an acid-base reaction is is from the stronger acid and base to the from the stronger acid and base to the weaker acid and base.weaker acid and base.


4.4. Relative Strengths of Acids and Bases Relative Strengths of Acids and Bases Leveling effect Leveling effect HCl+H2O→H3O

++Cl-

HClO4+H2O→H3O++ ClO4

-

H2SO4 +H2O→H3O++HSO4

-

HNO3+H2O→H3O++NO3

-

This phenomenon is called the This phenomenon is called the leveling effect-leveling effect-- the - the solvent makes the strong acids appear equal, or levesolvent makes the strong acids appear equal, or level, in acidity.l, in acidity.


4.4. Relative Strengths of Acids and Bases Relative Strengths of Acids and Bases H+ (aq) (or H3O+(aq)) is the strongest acid that can

exist in aqueous solution. Any acid stronger than H+(aq) reacts with water completely to produce H+

(aq) and the weak conjugate base.HC1(aq)+H2O(l)→H3O

+(aq)+Cl-(aq)


- The conjugate base of the solvent is the strongest ba

se that can be present. In water, OH- is the strongest base.


4.4. Relative Strengths of Acids and Bases Relative Strengths of Acids and Bases Differentiating Effect: How to differentiate Differentiating Effect: How to differentiate

among the four strong acids? among the four strong acids? HC1+HAc H2Ac+ + Cl- HC1O4+HAc H2Ac+ + ClO4

-

H2SO4 +HAc H2Ac+ + HSO4-

HNO3+HAc H2Ac+ + NO3-

We could choose a different solvent that is a We could choose a different solvent that is a more acidic. The acid strengths are in the order . The acid strengths are in the order

HC1O4> H2SO4 > HC1> HNO3


4.4. Relative Strengths of Acids and Bases Relative Strengths of Acids and Bases The relative strengths of acid stronger than hydromi

um ion can be determined by studying reaction in nonaqueous solutions, using solvent more acidic than water.

The relative strengths of base stronger than hydroxide ion can be determined by studying reaction in nonaqueous solutions, using solvent more basic than water.

Summary of Acid-Base Summary of Acid-Base DefinitionsDefinitions

Acid: proton donor (contain hydrogen ) Acid: proton donor (contain hydrogen ) Base: proton acceptor ( all negative ions ) a Base: proton acceptor ( all negative ions ) amphiprotic: (proton-containing negative ions )mphiprotic: (proton-containing negative ions )

Reaction: The Brønsted -Lowry concept considReaction: The Brønsted -Lowry concept considers an acid-base reaction as a proton-transfer rers an acid-base reaction as a proton-transfer reaction. eaction.

Relative Strengths of Acids and Bases: The stroRelative Strengths of Acids and Bases: The strongest acids have the weakest conjugate bases. ngest acids have the weakest conjugate bases. The conjugate acid of the solvent is the strongThe conjugate acid of the solvent is the strongest acid that can be present.est acid that can be present.

conjugate acid-H+

conjugate base+H+

3.4 Solving Problems Involving 3.4 Solving Problems Involving Acid-Base EquilibriaAcid-Base Equilibria

1.1. Self-Ionization of WaterSelf-Ionization of Water Ionization Ionization

CHCH33COCO22H + HH + H22O HO H33OO++ + CH + CH33COCO22--

NHNH33 + H + H22O NHO NH44++ + OH + OH--

Self-Ionization of Water Self-Ionization of Water HH22O(l)+ HO(l)+ H22O(l) HO(l) H33OO++ (aq) + OH (aq) + OH--(aq) (aq)

[H[H22O]O]2 2 Kc =[HKc =[H33OO++ ][OH ][OH--]=constant= Kw ]=constant= Kw

22

3c O][H

]][OHO[HK


Notice Notice KKww is the ion-product constant for water. At 25 is the ion-product constant for water. At 25

℃, the value of K℃, the value of Kww is 1.0 × 10 is 1.0 × 10-14-14. . Like any equilibrium constant, Like any equilibrium constant, KKww varies with varies with

temperaturetemperature. UNLESS OTHERWISE INDICAT. UNLESS OTHERWISE INDICATED, ALL CALCULATIONS WILL BE AT 298.15 ED, ALL CALCULATIONS WILL BE AT 298.15 KK

If you add an acid or a base to water, the concIf you add an acid or a base to water, the concentrations of Hentrations of H++ and OH and OH-- will no longer be equ will no longer be equal. The equilibrium-constant equation Kal. The equilibrium-constant equation Kww ==[H[H33OO++ ][OH ][OH--] ] will still ho1d. will still ho1d.


In an acidic solution, [HIn an acidic solution, [H++] > 1.0 × 10] > 1.0 × 10-7-7 M. M.In a neutral solution, [HIn a neutral solution, [H++] = 1.0 × 10] = 1.0 × 10-7-7 M. M. In a basic solution, [HIn a basic solution, [H++] < 1.0 × 10] < 1.0 × 10-7-7 M. M. Equation indicates an inverse relationship bEquation indicates an inverse relationship b

etween etween [H [H ++] and [OH] and [OH - -]; if we know either the hydrog]; if we know either the hydrog

en or hydroxide ion concentration, we can, en or hydroxide ion concentration, we can, by Equation, find the other. by Equation, find the other.


2.2. The pH of A Solution (important)The pH of A Solution (important) Definition: pH is the negative logarithm of tDefinition: pH is the negative logarithm of t

he molar hydrogen ion concentration. he molar hydrogen ion concentration. pH = -log[HpH = -log[H++] ]

Remember, a high pH (greater than 7) meanRemember, a high pH (greater than 7) means basic and a low pH (less than 7) means acis basic and a low pH (less than 7) means acidic. The higher the pH, the lower the acidity,dic. The higher the pH, the lower the acidity, and vice versa. and vice versa.

If [HIf [H++] = l0] = l0-x-x, then pH = x. , then pH = x. [H[H++] = antilog(-pH) = 10] = antilog(-pH) = 10-pH-pH


Relationship between pH and pOH Relationship between pH and pOH [H[H++][OH][OH--] = Kw ] = Kw (-log [H(-log [H++]) + (-log [OH]) + (-log [OH--]) = -log Kw ]) = -log Kw pH + pOH = pKw=14 pH + pOH = pKw=14 Just as we know that the product of the hydJust as we know that the product of the hyd

rogen ion and hydroxide ion concentrationrogen ion and hydroxide ion concentrations is 1.0× 10s is 1.0× 10-14-14 , the sum of the pH and pOH , the sum of the pH and pOH must always be 14.00 (at 25℃). must always be 14.00 (at 25℃).

The relationships among [HThe relationships among [H33OO++], [OH], [OH--], pH, ], pH, and pOH can be summarized as fol1ows: and pOH can be summarized as fol1ows:


The relationships among [HThe relationships among [H33OO++], [O], [OHH--], pH, and pOH can be summarized ], pH, and pOH can be summarized as fol1ows: as fol1ows:

pH+pOH=14pH+pOH=14 pH pOHpH pOH-log [H-log [H++] -log [OH] -log [OH --]] [H[H++] [OH] [OH--]] [H[H++][OH][OH--] = K] = Kww = 1.0×10 = 1.0×10-14-14


EXAMPLE 1 A sample of orange juice has a EXAMPLE 1 A sample of orange juice has a hydrogen-ion concentration of 2.9×10hydrogen-ion concentration of 2.9×10-4-4 M. What is the pH? Is the solution acidic? M. What is the pH? Is the solution acidic?

EXAMPLE 2 The pH of human arterial blood EXAMPLE 2 The pH of human arterial blood is 7.40. What is the hydrogen-ion concenis 7.40. What is the hydrogen-ion concentration? tration?

Measurement of pH Measurement of pH Using a pH meter Using a pH meter Using acid-base indicators Using acid-base indicators

QuestionsQuestions

What is the hydrogen-ion concentratioWhat is the hydrogen-ion concentration of 0.10M niacin (nicotinic acid)? n of 0.10M niacin (nicotinic acid)?

What is the hydrogen-ion concentratioWhat is the hydrogen-ion concentration of the solution obtained by dissolvinn of the solution obtained by dissolving one 5.00-gram tablet of aspirin (acetyg one 5.00-gram tablet of aspirin (acetylsalicylic acid) in 0.500 L of water? lsalicylic acid) in 0.500 L of water?

The pH of human blood remains very The pH of human blood remains very nearly 7.35. Why? nearly 7.35. Why?

How to prepare a buffer of a given pH? How to prepare a buffer of a given pH?


3. The pH of Monoprotic Acids or Bases Acid-ionization Constant and Base-ionization C

onstant acid: HA+H2O A- +H3O

+

base: A-+H2O HA +OH-

Ka is the acid-ionization constant. Like any equilibrium constant, Ka (or Kb) varies with temperature.

If Ka is large, the acid is strong. If Kb is large, the base is strong.

O][HA][H

]][AO[HK

2

3c

[HA]

]][AO[HO][HKK 3

2ca

[HA]

]][AO[H3

aK

][A

][HA][OH

bK


3. The pH of Monoprotic Acids or Bases Acid-ionization Constant and Base-ionizati

on Constant The Degree of Ionization, α

Acid:

Base:

a

3 ][H

c

Oα

b

][OH

cα

where ca is the the initial concentration of the weak acid.

where cb is the the initial concentration of the weak base.


Question Question Although Although Ka and K and Kbb are listed in some t are listed in some tables, more frequently they are not. How do ables, more frequently they are not. How do you determine you determine Ka or K or Kbb of an ionic acid, basic, of an ionic acid, basic, or amphiprotic substance? or amphiprotic substance?

The Relationship between KThe Relationship between Kaa and K and Kbb for Conj for Conjugate Acid-Base Pairs (important)ugate Acid-Base Pairs (important)

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

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

If we multiply If we multiply Ka(HA) by K(HA) by Kbb(A (A --), we obtain ), we obtain KKaaKKbb = K = Kww

[HA]

]][A[HKa

][A

][HA][OHK

-b


Notice Notice This relationship is general and shows that This relationship is general and shows that the the

product of acid- and base-ionization constantsproduct of acid- and base-ionization constants in aqueous solution for conjugate acid-base pain aqueous solution for conjugate acid-base pairs equals the irs equals the ion-product constant for waterion-product constant for water, , KKww. .

As As Ka decreases, K decreases, Kbb for the conjugate base mus for the conjugate base must increase, since Kt increase, since Kww is a constant. We see that t is a constant. We see that the he weaker anweaker an acidacid (the smaller K (the smaller Kaa ), the ), the strongstronger its conjugate baseer its conjugate base (the larger K (the larger Kbb ). ).

Obtain Obtain Ka from K from Kbb or K or Kbb from K from Kaa . .


ExampleExample Use Tables 3-3 to obtain the following at 25℃: (a) Kb for CN-; (b) Ka for NH4

+. Solution

(a) The conjugate acid of CN- is HCN, whose Ka is 6.2×10-10. Hence,

Kb=Kw/Ka=1.0×10–14/6.2×10-10=1.6×10–5

(b) The conjugate base of NH4+ is NH3, whose Kb

is 1.8×10-5. Hence,

Ka=Kw/Kb=1.0×10–14/1.8×10-5=5.6×10-10

3.4 Solving Problems Involving Ac3.4 Solving Problems Involving Acid-Base Equilibria (monoprotic aciid-Base Equilibria (monoprotic aci

d)d)Question:Question: Once you know the value of Once you know the value of Ka for an for an

acid HA, how can you calculate the equilibriacid HA, how can you calculate the equilibrium concentrations of species HA, Aum concentrations of species HA, A--, and H, and H++ for solutions of different molarities?for solutions of different molarities?

Calculating Concentrations of Species in a Calculating Concentrations of Species in a Weak Acid Solution (Approximation MethoWeak Acid Solution (Approximation Method) d)

ProcedureProcedure Write the equilibrium equation. Write the equilibrium equation. Set up a table in which you write the starting,Set up a table in which you write the starting,

change, and equilibrium values of each sub change, and equilibrium values of each substance under the balanced equation. stance under the balanced equation.


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

q)Starting Starting ccaa 0 0 0 0 Change - Change - xx + + xx + + xx Equilibrium Equilibrium ccaa- - xx xx xx Substitute the equilibrium-constant equation Substitute the equilibrium-constant equation

for the equilibrium concentrations.for the equilibrium concentrations.

Solve the equilibrium-constant equation for tSolve the equilibrium-constant equation for the equilibrium concentrations. he equilibrium concentrations.

ca/Ka≥500 (or α< 5%α< 5% ):):

aaa c

x

xc

xxK

2%5

[HA]

]O][H[A 3

aa3 ][A][H cKOx

[HA]≈ ccaa pH=-log[H+ ] =-log[H3O+ ]


d)d)Question Question How do you know when you can How do you know when you can

use the simplifying assumption, where you use the simplifying assumption, where you neglected neglected xx in the denominator of the in the denominator of the equilibrium equation?equilibrium equation?

Solution α<5% or Solution α<5% or ca/Ka ≥500≥500 Notice Notice Every time an approximation is introducEvery time an approximation is introduc

ed in a calculation, you must check to see if ted in a calculation, you must check to see if the approximation is valid he approximation is valid (α<5% or (α<5% or ca/Ka ≥500 ≥500 ).). If you find you have made an invalid assu If you find you have made an invalid assumption, you can always solve the equation usmption, you can always solve the equation using the ing the quadratic formula.quadratic formula. 4

4

2]H[

2aaaa cKKK


d)d)Example Example A 0.25 M solution of HCN has a pH of A 0.25 M solution of HCN has a pH of

5.00. What is 5.00. What is KKaa? What is α of Hydrocyanic a? What is α of Hydrocyanic acid in this solution? cid in this solution?

ProcedureProcedure Convert pH to [HConvert pH to [H++] . ] . Write the equilibrium equation. Write the equilibrium equation. Set up a table in which you write the starting,Set up a table in which you write the starting,

change, and equilibrium values of each sub change, and equilibrium values of each substance under the balanced equation. stance under the balanced equation.

Use these equilibrium values to calculate Ka Use these equilibrium values to calculate Ka and α. and α.


d)d)Solution Solution [H[H++] = antilog (-5.00) = 1.0×10] = antilog (-5.00) = 1.0×10-5-5 HCN (aq) HCN (aq) + H2O(l) H H33OO++ (aq)+ (aq)+

CN CN --(aq) (aq) Starting 0.25 0 0 Starting 0.25 0 0 Change -1.0×10Change -1.0×10-5-5 +1.0×10 +1.0×10-5-5 +1.0×10 +1.0×10--

55 Equilibrium 0.25 1.0×10Equilibrium 0.25 1.0×10-5-5 1.0×1 1.0×1

00-5-5 01

-5-53

a 104[0.25]

]10][110[1

[HCN]

]][CN [HK

O

0.004%1040.25

101][Hα 5

5

a

3

c

O


d)d)Example: Example: What are the concentrations of nicoti

nic acid, hydrogen ion, and nicotinate ion in a solution of 0.10M nicotinic acid, HC6H4NO

2, at 25 ? What is the pH of the solution? W℃hat is the degree of ionization of nicotinic acid? The acid-ionization constant, Kaa, is 1.4×10-5.


d)d)SolutionSolution HNic (aq) HNic (aq) + H2O(l) H H33OO++ (aq)+ (aq)+

Nic Nic --(aq) (aq)

mol/L0.0012101.40.1

][Nic][H

5

aa

cKx

[HNic]≈ ccaa =0.1 mol/L pH=-log[H+ ] = 2.92

1.2%0.0120.1

0.0012][H

a

cα

aa][A][H cKx [HA]≈ ccaa

pH=-log[H+ ] a

][Hα

c

3.4 Solving Problems Involving Ac3.4 Solving Problems Involving Acid-Base Equilibria (monoprotic baid-Base Equilibria (monoprotic ba

se)se)Monoprotic AcidMonoprotic Acid[[HH33OO

++]] Monoprotic BaseMonoprotic Base[[OHOH--]]HA+HHA+H

22O AO A--+H+H33OO

++ AA--+H+H22O HA +OHO HA +OH--

If cIf caa//KKaa≥500≥500 If cIf cbb//KKbb≥500≥500

[HA]

]][AO[H3

aK][A

][HA][OH

bK

ac

]O[H3

bc

][OH

aacK ]A[]OH[ 3 bbcK ]HA[]OH[

4

4

2]H[

2aaaa cKKK

4

4

2]OH[

2bbbb cKKK


se)se)ExampleExample Morphine, CMorphine, C1717HH1919NONO33, is administered medica, is administered medica

lly to relieve pain. It is a naturally occurring blly to relieve pain. It is a naturally occurring base, or alkaloid. What is the pH of a 0.0075 mase, or alkaloid. What is the pH of a 0.0075 mol·Lol·L-1-1 solution of morphine at 25 solution of morphine at 25 ℃℃? The base-i? The base-ionization constant, onization constant, KKbb, is 1.6×10, is 1.6×10-6-6 at 25 at 25 ℃℃..

SolutionSolution CC1717HH1919NONO33 (aq) + H (aq) + H22O(l) HO(l) HCC1717HH1919NONO33 ++(aq) + (aq) +

OHOH--(aq)(aq) Starting 0.0075 0 0 Starting 0.0075 0 0 Change - x + x + x Change - x + x + x Equilibrium 0.0075 - x ≈ 0.0075 x x Equilibrium 0.0075 - x ≈ 0.0075 x x

04.1096.3

/101.10075.0106.1][OH

500/or 5%Condition

46bb

bb

pHpOH

LmolcKx

Kc


se)se)ProcedureProcedure Identify a salt solution is acidic or basic. Identify a salt solution is acidic or basic. Obtain Obtain KKaa or or KKbb for the ion that hydrolyz for the ion that hydrolyz

es by using es by using KKaaKKbb = K = Kww . . Calculation of the concentrations of speCalculation of the concentrations of spe

cies present follows that for solutions of cies present follows that for solutions of weak acids or bases. weak acids or bases.

basic→basic→KKbb → → acidic→acidic→KKaa → →

bbCK][OH

aa3 CK][H O

3.4 Solving Problems Involving Ac3.4 Solving Problems Involving Acid-Base Equilibriaid-Base Equilibria

Shifting an Acid-Base Equilibrium Dilution Law HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

Starting ca 0 0 Change -caα caα caα Equilibrium ca - caα caα caα

ca/Ka≥500:

2%52

3

1[HA]

]][AO[H

aa

aa

aaa c

c

cc

ccK

aa / cK As ca gets smaller, αmust increas

e because Ka is a constant at a given temperature.

3.4 Solving Problems Involving Acid-Base Eq3.4 Solving Problems Involving Acid-Base Equilibria (uilibria (Common-Ion Effect and Salt Effect) Common-Ion Effect and Salt EffectCommon-Ion Effect and Salt EffectExamples Examples Question Question H H

Ac(aq) + HAc(aq) + H22O(l) HO(l) H33OO++(aq) + Ac(aq) + Ac--(aq) Suppose (aq) Suppose you add NaAc(aq) to this solution. What is the eyou add NaAc(aq) to this solution. What is the effect on the acid-ionization equilibrium? ffect on the acid-ionization equilibrium?

Solution Solution HAc(aq) + HHAc(aq) + H22O(l) HO(l) H33OO++(aq) + Ac(aq) + Ac--(aq) (aq) NaAc → NaNaAc → Na++ + Ac + Ac--

The equilibrium should shift to the left. The equilibrium should shift to the left. α of acetic acid is decreased. α of acetic acid is decreased. The NaAc has AcThe NaAc has Ac-- in common with HAc, so the i in common with HAc, so the i

nfluence is known as the common-ion effect. nfluence is known as the common-ion effect.

3.4 Solving Problems Involving Acid-Base Eq3.4 Solving Problems Involving Acid-Base Equilibria (uilibria (Common-Ion Effect and Salt Effect)

Common ion effectCommon ion effect Salt effectSalt effectHAc+ HHAc+ H22O HO H33OO++++AcAc-- NaAc→NaNaAc→Na+++ + AcAc--

EquilibriumEquilibrium←← ，， αα ↓ ↓

HAc+ HHAc+ H22O HO H33OO+++A+Acc--NaCl→NaNaCl→Na+++ Cl+ Cl--

EquilibriumEquilibrium→→ ，， αα ↑↑

The shift in equilibrium The shift in equilibrium caused by the addition caused by the addition of a compound having aof a compound having an ion in common with tn ion in common with the dissolved substance ihe dissolved substance is called the common-ios called the common-ion effect .n effect .

The shift in equilibriThe shift in equilibrium caused by the adum caused by the addition of a compound dition of a compound without a common iowithout a common ion is called the salt effn is called the salt effect .ect .


Calculating the Common-Ion Effect on Acid IonizatioCalculating the Common-Ion Effect on Acid Ionization n

Example The α of HAc in a 0.10M aqueous solution at 25℃ is 0.013. The pH of 0.10M acetic acid is 2.88. Ka at this temperature is 1.7×10–5. Calculate the α of HAc in a 0.10M solution at 25℃ to which sufficient NaAc is added to make it 0.10 M NaAc. How is the degree of ionization affected? What is the pH of this solution at 25℃? Calculate the change in pH that results when 0.01mol HCl is added to 1.00 L of a solution containing 0.10 M HAc and 0.10 M NaAc. Compare the pH change with what would occur if this amount of acid were added to 1 L of pure water.


Solution (1)Before addition of HCl HAc+ H2O H3O++Ac-

starting: 0.10 0 0.10 change: -x +x +xequilibrium: 0.10-x x 0.10 +x ≈0.10 ≈0.10

0.000170.10

101.7α

4.74)10lg(1.7]lg[HpH

Lmol101.7 101.70.10

0.10

0.10

0.10

5

5

155a

xx

x

xxK


(2)After addition of HCl HAc+ H2O H3O++Ac-

starting: 0.10 0.01 0.10 change: +0.01 -0.01 -0.01 F(S): 0.11 0 0.09 change: -x +x +xequilibrium: 0.11-x x 0.09 +x ≈0.11 ≈0.09

0.08pH 4.66)10lg(2.2]lg[HpH

Lmol10 2.2 101.70.11

0.09K

5

155a

xx


1 L of pure water pH = 7.00 0.010 mol of HCl was added [H+] = 0.10 mol·L-1 pH = 2.00pH changes 5.00 units.If 0.01 mol of hydrochloric acid is added to 1 L of p

ure water, the pH changes from 7.0 to 2.0 - a pH change of 5.0 units. By contrast, the addition of this amount of hydrochloric acid to 1 L of the solution containing weak acid and its conjugate base might change the pH by only 0.1 unit.

3.4 Solving Problems Involving Ac3.4 Solving Problems Involving Acid-Base Equilibria (polyprotic basid-Base Equilibria (polyprotic bas

e)e)4.4. Polyprotic Acids Polyprotic Acids Step- ionizationStep- ionization When polyprotic acids ionize, they do so by steWhen polyprotic acids ionize, they do so by ste

ps, releasing ps, releasing one hydrogen ion with each step. one hydrogen ion with each step. 1st stage H1st stage H22COCO3 3 ++ H2O H3O++ HCO HCO33

- - K Ka1a1 2st stage HCO2st stage HCO33

-- + + H2O H3O++ CO CO332-2- K Ka2a2

Each succeeding acid is weaker than the one bEach succeeding acid is weaker than the one before. For any diprotic acid, Kefore. For any diprotic acid, Ka2a2 < K < Ka1a1. This is al. This is always true because it is more difficult to removways true because it is more difficult to remove a proton from a negatively charged species e a proton from a negatively charged species (HA(HA--) than from an uncharged molecule (H) than from an uncharged molecule (H22A). A).


e)e)4.4. Polyprotic AcidsPolyprotic Acids Ionization Constants for Acids (Ionization Constants for Acids (KKaa) and ) and

Base (Base (KKbb) ) 1st H1st H22COCO3 3 + + H2O H3O++ HCO HCO33

- - K Ka1a1 = 4.3×10= 4.3×10-7 -7

2st HCO2st HCO33-- + + H2O H3O++ CO CO33

2-2- K Ka2a2 = 5.61×10= 5.61×10-11-11

COCO332- 2- + H+ H22O HCOO HCO33

- - + OH+ OH- - KKb1b1

HCOHCO33- - + H+ H22O HO H22COCO3 3 + OH+ OH- - KKb2b2

HH22COCO33 ～～ HCOHCO33- - KKa1 a1 KKb2 b2 ==KKW W

HCOHCO33-- ～～ COCO33

2- 2- KKa2 a2 KKb1 b1 ==KKWW


e)e)


e)e) Calculating the concentrations of species in a s

olution of a polyprotic acid (Approximation Method)

monoprotic acidmonoprotic acid[[HH33OO++]] polyprotic acidpolyprotic acid[[HH33OO

++]]HA+HHA+H

22O AO A--+H+H33OO

++ HH22A+HA+H22O HAO HA--+H+H

33OO++

HAHA--+ H+ H22O AO A2-2-+H+H

33OO++

ccaa//KKaa≥500(or≥500(orα<5%α<5%):):

[HA]≈ [HA]≈ ccaa

KKaa11//KKaa2 2 > 10> 1022, [H, [H33OO

++] ] is due is due

to the first dissociation to the first dissociation ONLYONLY . . ccaa//KKa1 a1 ≥ 500≥ 500 ：： [H2A

]≈ca

[[AA2-2-] ≈] ≈ KKaa22

aacK ]A[]OH[ 3

aa cK 13 ]HA[]OH[


e)e) AA2-2- ion ion is produced in the second is produced in the second

ionization; its concentration equals ionization; its concentration equals KKa2a2..

2st stage HA2st stage HA-- + + H2O H3O+ + A A2-2- K Ka2a2

Starting Starting x xx x 0 0

Change Change -y +y -y +y +y+y

Equilibrium Equilibrium x-y≈x x+y ≈x y x-y≈x x+y ≈x y

][A][HA

]][A[HK 2

23

a2

yx

yxO


e)e) Example 1Example 1 H2S is a diprotic acid. What is the pH o

f a 0.10 M solution? What is the concentration of [H+], [HS-] and [S2-]? （ Ka1=9.1×10-8 ， Ka2=1.1×10-12 )

Example 2 Example 2 What is the pH of a 0.10 M NaNa22COCO33 solution? （（ HH22COCO33::KKa1a1= 4.3×10= 4.3×10-7 -7 ，， KKa2a2= 5.61×10= 5.61×10-11-11 ））

Example 3 Example 3 How to calculate concentrations of species in a solution of H3PO4?

Monoprotic AcidMonoprotic Acid Polyprotic AcidPolyprotic Acid

★★

★★★★

★★

★★ When polyprotic acids ionize, When polyprotic acids ionize, they do so by steps, releasing they do so by steps, releasing one one hydrogen ion with each stephydrogen ion with each step. . For For any diprotic acid, any diprotic acid, KaKa22 < Ka < Ka11.. ★★ IfIf: : ca / /KKa1a1≥500 ≥500 aandnd KKa1a1//KKa2a2 > 10 > 1022

★★ AA2-2- ion ion is produced in the seco is produced in the second ionization; its concentration nd ionization; its concentration eequals Kaquals Ka22..

HA H++ A-

[HA]

]][A[HK a

a

][Hα

c

If:α<5% or ca /Ka≥500

aaCK][A][Hx

[HA]≈ ca

pH=-log[H+ ] 2aa cK

H2A H+ + HA- Ka1

HA- H+ + A2- Ka2

aa1][HA][Hx cK

[H2A]≈ca

pH=-log[H+ ]

[A2- ] = Ka2

3.4 Solving Problems Involving Ac3.4 Solving Problems Involving Acid-Base Equilibria (Amphoteric Suid-Base Equilibria (Amphoteric Su

bstances)bstances)5.5. The pH of Amphoteric Substances The pH of Amphoteric Substances Questions Consider NaHCO3, Since HCO3

- is an amphiprotic substance, will solutions of HCO3

- be acidic or basic? How to determine Ka or Kb for HCO3

- ？ What is the conjugate acid of HCO3

- ion? How do you predict the pH of HCO3

- ion? Solution If Ka > Kb, the solution is acidic. If Ka = Kb, the solution is neutral. If Ka < Kb, the solution is basic.


bstances)bstances)5.5. The pH of Amphoteric Substances The pH of Amphoteric Substances HCO3

- + H2O H3O+ + CO32-

Ka2= 4.8×10-11

HCO3- + H2O OH- + H2CO3

Kb2= Kw / Ka1 = 2.3×10-8

cKa2 ＞ 20Kw ，c ＞ 20Ka1 ,

cKa2 ＞ 20Kw, c ＞ 20Ka1

cK

KcKK

1a

w2a1a )(]H[

cK

cKK

1a

2a1a]H[

2a1a]H[ KK 2a1a]H[ KK


bstances)bstances)cKa2 ＞ 20Kw, c ＞ 20Ka1

Note that Ka2 is the acid-ionization constant when an amphoteric substance acts as a weak acid; Ka1 is the its conjugate acid-ionization constant when it acts as a weak base.

Example 1Example 1 What is the pH of a 0.1 mol·L0.1 mol·L-1-1 NaH NaHCOCO

3 3 solution? （（ HH22COCO33::KKa1a1= 4.3×10= 4.3×10-7 -7 ，， KKa2a2= 5.61×10= 5.61×10-11-11 ）） Example 2 Example 2 Identify a salt solution is acidic

or basic. (a)NaH(a)NaH22POPO44; (b)Na; (b)Na22HPOHPO44

2a1a]H[ KK

3.5 Buffers3.5 Buffers1.1. Definition, Definition, Composition and Action of BuComposition and Action of Bu

ffer Solutions ffer Solutions DefinitionDefinition:: A buffer is a solution A buffer is a solution

characterized by the ability to characterized by the ability to resist resist changes in pHchanges in pH when when limited limited amounts of acid amounts of acid or base are added to it.or base are added to it.

Composition Composition :: Buffers contain either a weak Buffers contain either a weak acid and its conjugate base or a weak base acid and its conjugate base or a weak base and its conjugate acid. and its conjugate acid.

HAcHAc ～～ NaAc; NHNaAc; NH44Cl Cl ～～ NHNH33 kH2PO

4

k2HPO

4

3.5 Buffers3.5 Buffers Question Which of the following solutions a

re buffer systems? (a) KH2PO4/H3PO4, (b) NaC1O4/HC1O4, Explain your answer.

How a Buffer Works How a Buffer Works HAc(aq) + HHAc(aq) + H22O(l) HO(l) H33OO++(aq) + Ac(aq) + Ac--(aq) (aq) large concentration large concentrationlarge concentration large concentration(added acid) Ac (added acid) Ac -- + H + H33OO++ HAc + H HAc + H22OO(added base) HAc + OH(added base) HAc + OH-- H H22O + AcO + Ac-- Antiacid component-base (Antiacid component-base (Ac Ac -- ))Antibase component- acid (Antibase component- acid (HAcHAc))

3.5 Buffers3.5 Buffers2.2. The pH of a Buffer-Henderson-HasselbalcThe pH of a Buffer-Henderson-Hasselbalc

h Equation h Equation HB+HHB+H22O BO B--+H+H33OO++

[HB]

]][BO[H3

aK

][B

[HB]]O[H

-3aK

cid][

ase][lgap

[HB]

][BlgppH a a

bKK

][B

[HB]loglogK]log[H a

3.5 Buffers3.5 Buffers NoticeNotice Influencing factors: pKa and [HA]/[A-]. Note Ka f

or the conjugate acids. [HA]≈Ca [A-] ≈Cb Another expressions

Mixing equal concentrations of HA and A-,

Buffer pairs that find extensive application are HAc and Ac-, NH3 and NH4

+, H2CO3 and HCO3-, H

2PO4- and HPO4

2-. Note Ka for the conjugate acids.

HB

B

n

nlgpKapH

HB

B

V

VlgpKapH

3.5 Buffers3.5 Buffers ApplicationApplication Calculating the pH of a BufferCalculating the pH of a BufferExample 1 Instructions for making up a buffer sExample 1 Instructions for making up a buffer s

ay to mix 60 mL of 0.100 M NHay to mix 60 mL of 0.100 M NH33 with 40 mL with 40 mL of 0.100 M NHof 0.100 M NH44C1. What is the pH of this bufC1. What is the pH of this buffer? Kfer? Kbb for NH for NH33 is 1.8 ×10 is 1.8 ×10 –5–5. .

Example 2 A 0.10 M solution of aqueous ammonExample 2 A 0.10 M solution of aqueous ammonia, also containing ammonium chloride, has ia, also containing ammonium chloride, has a hydroxide ion concentration of 5.6 ×10 a hydroxide ion concentration of 5.6 ×10 –10–10 M. What is the concentration of the ammoniM. What is the concentration of the ammonium ion in the solution?um ion in the solution?

3.5 Buffers3.5 Buffers ApplicationApplication Calculating the pH of a BufferCalculating the pH of a BufferExample 3 Ten milliliters of 4.0 M acetic acid is aExample 3 Ten milliliters of 4.0 M acetic acid is a

dded to 20 ml of 1.0 M sodium hydroxide. Caldded to 20 ml of 1.0 M sodium hydroxide. Calculate the hydrogen ion concentration of the culate the hydrogen ion concentration of the resulting solution. Kresulting solution. Kaa for HAc is 1.8 ×10 for HAc is 1.8 ×10 –5–5 . .

Example 4 Estimate the pH of a solution prepared by dissolving 0.03 mol of H3PO4 and 0.05 mol of K3PO4 in 1 liter of water. (H3PO4: pKa1= 2.16, pKa2= 7.21, pKa3= 12.67)

3.5 Buffers3.5 Buffers ApplicationApplication Calculating the change in pH of a buffer solutioCalculating the change in pH of a buffer solutio

n on addition of a small amount of strong base n on addition of a small amount of strong base Example 1 Calculate the change in pH that results Example 1 Calculate the change in pH that results

when 0.50 mL of 0.200 M NaOH is added to 10.0when 0.50 mL of 0.200 M NaOH is added to 10.00 mL of a buffer solution containing acetic acid 0 mL of a buffer solution containing acetic acid and potassium acetate, both at a concentration and potassium acetate, both at a concentration of 1.00 M. of 1.00 M.

Conclusion Conclusion (added acid)(added acid)

(added base)(added base)

)H(n(HB)

)n(H-)n(BlogpKpH a

n

)OH(n(HB)

)OH()n(BlogpKpH a

n

n

3.5 Buffers3.5 Buffers3.3. Buffer Capacity and Buffer Range Buffer Capacity and Buffer Range Buffer Range : Buffer Range : HB(aq) HHB(aq) H++ (aq)+ B (aq)+ B--(aq) (aq)

When the [base]/[acid] ratio is 1, pH = pKa, thWhen the [base]/[acid] ratio is 1, pH = pKa, the buffer efficiency is largest. e buffer efficiency is largest.

If the [base]/[acid] ratio is larger than 10 or sIf the [base]/[acid] ratio is larger than 10 or smaller than 0.1, the solution will not be particmaller than 0.1, the solution will not be particularly effective at keeping the pH constant. ularly effective at keeping the pH constant.

cid][

ase][lgap

[HB]

][BlgppH a a

bKK

3.5 Buffers3.5 Buffers3.3. Buffer Capacity and Buffer Range Buffer Capacity and Buffer Range If this ratio is between 1:10 and 10:1, the soIf this ratio is between 1:10 and 10:1, the so

lution will be effective at keeping the pH clution will be effective at keeping the pH constant. onstant.

[base]/[acid] =10, pH=pKa + 1 [base]/[acid] =10, pH=pKa + 1 [base]/[acid] =0.1, pH=pKa -1 [base]/[acid] =0.1, pH=pKa -1

Buffer Range : pKa±1Buffer Range : pKa±1Buffers (Acid-Buffers (Acid-Base) Base)

pKapKa Buffer Buffer Range Range

HAc-AcHAc-Ac-- pKa =pKa = 4.764.76 3.76 3.76 ～～ 5.765.76

NHNH3 3 - NH- NH44++ pKa =pKa = 9.259.25 8.25 8.25 ～～

10.2510.25

HH22COCO33 -HCO -HCO33-- pKapKa11== 6.376.37 5.37 5.37 ～～ 7.377.37

HCOHCO33- - - CO- CO33

2-2- pKapKa22==

10.3210.329.32 9.32 ～～ 11.3211.32

3.5 Buffers3.5 Buffers Buffer Capacity Buffer Capacity Definition: the amount of strong acid or baDefinition: the amount of strong acid or ba

se needed to change the pH of one liter of se needed to change the pH of one liter of buffer by 1 unit. buffer by 1 unit.

Influencing factors: the total buffer concentrations c and [HB]/[B-]

ratio→same,β∝c (c (total)= [HB] + [B-]) c (total)→same, [base]/[acid] =1, βmax =0.576c

ccc

][A[HA]

2.303d[pH]

d[B]

3.5 Buffers3.5 Buffers4.4. Preparing a Buffer Preparing a Buffer PrinciplesPrinciples Most buffers are made from solutions with coMost buffers are made from solutions with co

ncentrations in the range from 0.05 to 0.2 M. ncentrations in the range from 0.05 to 0.2 M. The pKa value should be as close as possible tThe pKa value should be as close as possible t

o the desired pH .o the desired pH . MethodsMethods HB + B-

HB(excess) + NaOH→B-

B- (excess) + HCl→ HB x

xVK

2lgappH

xV

xK

2lgappH

xV

xK

-lgappH

3.5 Buffers3.5 Buffers4.4. Preparing a Buffer Preparing a Buffer ProcedureProcedure The first step in the process is to choose a suThe first step in the process is to choose a su

itable weak acid. The H-H equation is really itable weak acid. The H-H equation is really only valid in a pH range near the pKa of the only valid in a pH range near the pKa of the weak acid; therefore, we select a weak acid weak acid; therefore, we select a weak acid with a pKa near the pH range of interest with a pKa near the pH range of interest

Second Step Calculate the required ratio oSecond Step Calculate the required ratio of conjugate base to weak acid from the Hendf conjugate base to weak acid from the Henderson-Hasselbalch Equationerson-Hasselbalch Equation

3.5 Buffers3.5 Buffers Calculation Calculation HB + B-: HB + B-: Mixing equal concentrations of HB

and B-,

HB + NaOH→BHB + NaOH→B-- How to calculate the volum How to calculate the volume for preparing a buffer by mixing a strong e for preparing a buffer by mixing a strong base with a weak acid? The total volume of a base with a weak acid? The total volume of a buffer =100ml,V(NaOH)=x.buffer =100ml,V(NaOH)=x.

HB + NaOH→BHB + NaOH→B-- + H + H22OOStarting (100-x)c cx 0 Starting (100-x)c cx 0 Change -cx -cx +cx Change -cx -cx +cx Equilibrium (100-2x)c 0 cxEquilibrium (100-2x)c 0 cx

HB

B

V

VlgpKapH

2x-100

xlgpKapH

3.5 Buffers3.5 Buffers4.4. Preparing a Buffer Preparing a Buffer Example 1 A student wishes to prepare a buffer s

olution at pH=8.60.Which of the following weak acid should she choose and why? HA(Ka=2.7×10-3), HB(Ka=4.4×10-6), HC(Ka = 2.6× 10-9).

Example 2 What volume of 2.00 M NaOH must be added to 100.00 m L of 2.50 M HCOOH to prepare a buffer with pH = 4.00?

Example 3 How to prepare 30ml a phosphatic buffer with pH = 7.21 by using 0.1 mol/L NaOH and 0.1 mol/L KH2PO4?

3.5 Buffers3.5 Buffers5.5. Biological meaning Biological meaning Question: The pH of human blood remains very n

early 7.35. Why? Solution: Blood contains H2CO3 and HCO3

-, as well as other conjugate acid-base pairs: H2PO4

- - HPO42-

and proteins, HPr – Pr. Most of the buffering of blood is carried out by a carbonic acid-bicarbonate buffer.

H2CO3 + H2O HCO3 - + H3O+ (added acid) HCO3 - + H3O+ H2CO3 + H2O The extra carbon dioxide is released in the lungs

and exhaled

3.5 Buffers3.5 BuffersH2CO3 + H2O HCO3 - + H3O+ (added base) H2CO3 + OH- HCO3 - + H2O The extra bicarbonate ion formed is eventu

ally expelled by the kidneys. Terms If the pH of the blood drops below 7.2, a po

tentially fatal condition called acidosis results.

If the pH of the blood increases to over 7.5, another serious condition known as alkalosis results.

Summary Summary AcidAcid If: If: Ca/Ka≥100, Ca/Ka≥100,

BaseBase If: If: CCbb/K/Kbb≥100, ≥100,

Buffers Buffers (two (two forms)forms)

1.pH 1.pH 2. Buffer range: pKa±12. Buffer range: pKa±13.3.ββ: the total buffer concentratio: the total buffer concentrations and [B]/[A] ns and [B]/[A]

[HB]

]][B[HKa

aaCK][H

][B

][HB][OHK

-b

bC

][OHα

bbCK][OH

KaKb = Kw

[acid]

[base]logpK

[HB]

][BlogpKpH aa

aC

][Hα

MBBS Chapter 3 and Chapter 4 Electrolyte Solutions and Buffers

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Transcript of MBBS Chapter 3 and Chapter 4 Electrolyte Solutions and Buffers