chapter fifteenprentice-hall ©2002slide 1 of 31 920203 http:\\asadipour.kmu.ac.ir........57 slides...

$: Chapter FifteenPrentice-Hall ©2002Slide 1 of 31 920203 http:\\asadipour.kmu.ac.ir........57 slides 1$
Chapter FifteenPrentice-Hall ©2002 Slide 1 of 31• http:\\asadipour.kmu.ac.ir........57 slid

es• 1• 920203

Equilibrium In Solutions Of Weak Acids And Weak Bases

weak acid: HA + H2O H3O+ + A-

[H3O+][A-]

Ka =

[HA]

weak base: B + H2O HB+ + OH-

[HB+][OH-]

Kb =

[B]

• 920203

Some Acid-Base Equilibrium Calculations

• cHA≈[HA]

[H3O+][A-] [H3O+][A-]

Ka = --------------------= ----------------

cHA –[H3O+] cHA

• cHA > [HA] Analytical C> Equilibrium C

• - the calculations can be simplified.

• - When Macid/Ka or Mbase/Kb > 100,

• - When Ka or Kb<1×10-4 (In usual Conc.)

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An Example

1.Determine the concentrations of H3O+, CH3COOH and CH3COO-, and the pH of 1.00 M CH3COOH solution. Ka = 1.8 x 10-5.

2. What is the pH of a solution that is 0.200 M in methylamine, CH3NH2? Kb = 4.2 x 10-4.

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Are Salts Neutral, Acidic or Basic?

Salts are ionic compounds formed in the reaction between an acid and a base.

1. NaClNa+ is from NaOH , a strong baseCl- is from HCl, a strong acid

H2ONaCl (s) → Na+ (aq) + Cl- (aq)Na+ and Cl- ions do not react with water.The solution is neutral.

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2.KCN

K+ is from KOH , a strong base

CN- is from HCN, a weak acid

H2O

KCN (s) → K+ (aq) + CN- (aq)

K+ ions do not react with water, but CN- ions do.

CN- + H2O HCN + OH- hydrolysis

The OH- ions are produced, so the solution is basic.

• 920203


3.NH4Cl

NH4+ is from NH3 , a weak base

Cl- is from HCl, a strong acid

H2O

NH4Cl (s) → NH4+ (aq) + Cl- (aq)

Cl- ions do not react with water, but NH4+ ions do.

NH4+ + H2O H3O+ + NH3 hydrolysis

The H3O+ ions are produced, so the solution is acdic.

• 920203


3.NH4CN

NH4+ is from NH3 , a weak base

CN- is from HCN, a weak acid

H2O

NH4CN (s) → NH4+ (aq) + CN- (aq)

NH4+ + H2O H3O+ + NH3 Ka hydrolysis

CN- + H2O HCN + OH- Kb hydrolysis

(Ka>Kb ,Acidic)’’’(Ka< Kb,Basic)‘’’ (Ka= Kb,Nutral)• 920203

Ions As Acids And Bases

Certain and anion ions can cause an aqueous solution to become acidic or basic due to hydrolysis.

• Salts of strong acids and strong bases form neutral solutions.• Salts of weak acids and strong bases form basic solutions.• Salts of strong acids and weak bases form acidic solutions.• Salts of weak acids and weak bases form solutions that are

acidic in some cases, neutral or basic in others.

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Strong Acids And Strong Bases

Strong acids:

HCl, HBr, HI, HNO3, H2SO4, HClO4

Strong bases:

Group IA and IIA hydroxides

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An Example

Indicate whether the solutions

(a) Na2S and (b) KClO4 are acidic, basic or neutral.

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The pH Of a Salt Solution

What is the pH of 0.1M NaCN solution?

What is the pH of 0.1M NH4Cl solution?

What is the pH of 0.1M NH4CN solution?

Ka of HCN=1.0×10-9.

Kb for NH3=1.0×10-5

Ka x Kb = Kw

so, Kb = Kw/Ka

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es• 14• 920203

Common Ion Effect Illustrated

((1.00 M CH3COOH)) ((1.00 M CH3COOH + 1.00 M CH3COONa))

yellow:pH < 3.0

blue-violet:

pH > 4.6

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• CH3COOH CH3COO- + H+

The Common Ion Effect

Calculate the pH of 0.10 M CH3COOH solution.

Ka of CH3COOH=1.0×10-5

Calculate the pH of 0.10 M CH3COONa solution.

Calculate the pH of 0.10 M CH3COOH/

0.10 M CH3COONa solution.

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Depicting Buffer Action

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Buffer Solutions

• A buffer solution is a solution that changes pH only slightly when small amounts of a strong acid or a strong base are added.

• A buffer contains

CH3COOH CH3COO- Acidic buffer

NH3 NH4+ Alkalin buffer

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How A Buffer Solution Works

• The acid component of the buffer can neutralize small added amounts of OH-, and the basic component can neutralize small added amounts of H3O+.

• CH3COOH CH3COO- + H+

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es• 20

eqb

eqbeqb

a HA

AOHK

3

•Ionization constant of an acid

• Taking log of the equation on both sides,

eqb

eqbeqb

a HA

AOHLogLogK

3

][][

3 HAA

LogOHLogLogKa

cb

LogaLogaswrittenbecancab

LogSince )()(

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es• 21

•Ionization constant of an acid

][][

3 HAA

LogOHLogLogKa

• Multiplying both sides of the equation by -1

][][

3 HAA

LogOHLogLogKa

pHOHLogandpKLogKbut aa 3,

][][

HAA

LogpHpKa

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][][

HAA

LogpKpH a

• Henderson-Hasselbach equation

Henderson-Hasselbalch Equation For Buff Solutions

[conjugate base]

pH = pKa + log

[conjugate acid]

If [conjugate acid] = [conjugate base], pH = pKa

Requirement:- [B] / [A] between 0.10 and 10

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Buffer Capacity

• There is a limit to the capacity of a buffer solution to neutralize added acid or base, and this limit is reached before all of one of the buffer components has been consumed.

• In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize.

• As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal. [B]=[A]

• [Buffer]=[Acid]+[Base]

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Buffer Capacity

• [Buffer]=[Acid]+[Base]

• [Acid]↑ & [Base]↑ Capacity ↑

• In equimolar buffersis is important

• Capacity ↑

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30

8

9

20

10

10

][

][

Acid

Base

1][

][

Acid

Base

15

5

5

15

8

12

10

10

Calculations in Buffer Solutions

1) A buffer solution is 0. 1 M NH3 (pKb=5)and 1.0 M NH4Cl.

(a) What is the pH of this buffer?

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95 ab pKpK 8190.1

1.0log9 pH

][

][

A

BLogpKpH a



(a) What is the pH of this buffer?(b) If 5 mmol NaOH is added to 0.500 L of this solution,

what will be the pH?

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046.8954.09495

55log9

51500

51.0500log9

pH



(a) What is the pH of this buffer?(b) If 5 mmol NaOH is added to 0.500 L of this solution,

what will be the pH?(c) If 5 mmol HCl is added to 0.500 L of this solution,

what will be the pH?

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950.7050.19505

45log9

51500

51.0500log9

pH



(a) What is the pH of this buffer?(b) If 5 mmol NaOH is added to 0.500 L of this solution, what

will be the pH?(c) If 5 mmol HCl is added to 0.500 L of this solution, what

will be the pH?

(d) If 5 mmol NH4Cl is added to 0.500 L of this solution, what will be the pH?

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996.7004.19505

50log9

51500

1.0500log9

pH


es• 29


2) What concentration of acetate ion in

500 ml of 0.500 M CH3COOH (pKa=5) produces a buffer solution with pH = 4.00?

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MBB

A

B05.0][1.0

5.0log1

A

B

A

BpKpH a log54log


2) What concentration of acetate ion in500 ml of 0.500 M

CH3COOH (pKa=5) produces a buffer solution with pH = 4.00?

• 920203

• How many mg?

205082

05.0500 mgmg

Acid-Base Indicators

• An acid-base indicator is a weak acid having one color and the conjugate base of the acid having a different color. One of the “colors” may be colorless.

HIn + H2O H3O+ + In-

• Acid-base indicators are often used for

applications in which a precise pH reading isn’t necessary.

• A common indicator used in chemistry laboratories is Phenolphetalein.

• 920203


es• 32• 920203


es• 33• 920203

Neutralization Reactions

• Neutralization is the reaction of an acid and a base.• Titration is a common technique for conducting a

neutralization.• At the equivalence point in a titration, the acid and base

have been brought together in exact stoichiometric proportions.

• The point in the titration at which the indicator changes color is called the end point.

• The indicator endpoint and the equivalence point for a neutralization reaction can be best matched by plotting a titration curve, a graph of pH versus volume of titrant.

• In a typical titration, 50 mL or less of titrant that is 1 M or less is used.

• 920203

Drawing titration Curve ForStrong Acid - Strong Base

HCl + NaOH → NaCl +H2O

Calculate the pH at the

some points

and draw the curve.

4 essential points.

1)initial point

2)equivalence point

3)before the equivalence point

4)beyond the equivalence point

Ml تیتران ت

pH محی ط

0151919.519.92020.120.5212540

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4 questions.

1)What are the present compounds?

2)Which of them is effective on pH?

3)How much are the concentrations?

4)What is the relationship between their Conc. And pH?

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Calculate the pH at the following points in the

titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH.

(a) initial pH. (Before the addition of any NaOH) .

Answer Q1. There are:HCl & H2O

Answer Q2. HCl

Answer Q3. [HCl]

Answer Q4. pH=-log[H+]

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b)equivalence point.

Answer Q1. There are:NaCl & H2O

Answer Q2. H2O

Answer Q3.

Answer Q4. pH=7

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c)before the equivalence point.

Answer Q1. There are:HCl,NaCl & H2O

Answer Q2. HCl

Answer Q3.

Answer Q4. [H+]=N pH=-log[H+]21

2211

VV

VNVNN HCl

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d)after the equivalence point.

Answer Q1. There are:NaOH,NaCl & H2O

Answer Q2. NaOH

Answer Q3.

Answer Q4. [OH-]=N pOH=-log[OH-] pH=14-pOH

21

1122

VV

VNVNN OH

• 920203

Titration Curve ForStrong Acid - Strong Base

• pH is low at the beginning.• pH changes slowly until just before equivalence point.• pH changes sharply around equivalence point.• pH = 7.0 at equivalence

point.• Further beyond equivalence point, pH changes slowly.• Any indicator whose color changes in pH range of 4 – 10 can be used in titration.

• 920203


es• 42

Drawing titration Curve Forweak acid- Strong Base

CH3COOH + NaOH → CH3COO- + Na+ +H2O

Calculate the pH at the some points and draw the curve.

Ka=1×10-5

5 essential points.

1)initial point

2)equivalence point

3)beyond the initial point

4)before the equivalence point

5)beyond the equivalence point

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es• 43



4 questions.

1)What are the present compounds?

2)Which of them is effective on pH?

3)How much are the concentrations?

4)What is the relationship between their Conc. And pH?

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es• 44




titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH.

(a) initial pH. (Before the addition of any NaOH) .

Answer Q1. There are: CH3COOH & H2O

Answer Q2. CH3OOH

Answer Q3. CH3OOH

Answer Q4. pH=-log[H+]CKH a ][

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es• 45





b)equivalence point.

Answer Q1. There are: CH3COO- , Na+ & H2O

Answer Q2. CH3COO-

Answer Q3.

Answer Q4. pOH=-log[OH-] Ka×Kb=Kw

21

2211

VV

VNVNN

CKOH b ][• 920203


es• 46





c)beyond the initial point.

Answer Q1. There are: CH3COOH, CH3COO- ,Na+ & H2O

Answer Q2. CH3COOH, CH3COO-

Answer Q3.

Answer Q4.

21

2211

VV

VNVNN a

21

22

VV

VNN b

][

][log

A

BpKpH a

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es• 47





d)before the equivalence point.

Answer Q1. There are: CH3COOH, CH3COO- ,Na+ & H2O

Answer Q2. CH3COOH, CH3COO-

Answer Q3.

Answer Q4.

21

2211

VV

VNVNN a

21

22

VV

VNN b

][

][log

A

BpKpH a

• 920203


es• 48





e)after the equivalence point.

Answer Q1. There are:NaOH, CH3COO- , Na+ & H2O

Answer Q2. NaOH

Answer Q3.

Answer Q4. [OH-]=N pOH=-log[OH-] pH=14-pOH

21

1122

VV

VNVNN OH

• 920203

Titration Curve ForWeak Acid - Strong Base

• The initial pH is higher because weak acid is partially ionized.• At the half-neutralization point, pH = pKa.• pH >7 at equivalence point because the anion of the weak acid hydrolyzes.• The steep portion of titration curve around equivalence point has a smaller pH range.• The choice of indicators for the titration is more limited.

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es• 50• 920203


es• 51

Application of Ka

• The Ka of nicotinic acid, HNic, is 1.4e-5. A solution containing 0.22 M HNic. What is its pH? What is the degree of ionization?

• Solution: HNic H+ + Nic–

0.22-x x x

• x 2

Ka = ———— = 1.4e-50.22 – x (use approximation, small indeed)

• x = (0.22*1.4e-5) = 0.0018 pH = – log (0.0018) = 2.76

• Degree of ionization = 0.0018 / 0.22 = 0.0079 = 0.79%

•

• 920203


es• 52

Determine Ka and percent ionization

• Nicotinic acid, HNic, is a monoprotic acid. A solution containing 0.012 M HNic, has a pH of 3.39. What is its Ka? What is the percent of ionization?


0.012-x x x

• x = [H+] = 10–3.39 = 4.1e-4

[HNic] = 0.012 – 0.00041 = 0.012

• (4.1e-4)2

Ka = ————— = 1.4e-5 0.012

• Degree of ionization = 0.00041 / 0.012 = 0.034 = 3.4%• 920203


es• 53

Using the quadratic formula

• The Ka of nicotinic acid, HNic, is 1.4e-5. A solution containing 0.00100 M HNic. What is its pH? What is the degree of ionization?


0.001-x x x

• x2

Ka = —————— = 1.4e-5 x2 + 1.4e-5 x – 1.4e-8 = 00.00100 – x

• –1.4e–5 + (1.4e–5)2 + 4*1.4e-8x = —————————————————— = 0.000111 M

2pH = – log (0.000111) = 3.95

• Degree of ionization = 0.000111/ 0.001 = 0.111 = 11.1%

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es• 54

Degree of or percent ionization

• The degree or percent of ionization of a weak acid always decreases as its concentration increases, as shown from the table given earlier.

• Concentration of acid

• % io

niza

tion

• 920203

• Deg.’f ioniz’n

0.220 0.8%0.012 3.4 %0.001 11.1 %


es• 55

Polyprotic acids

• Polyprotic acids such as sulfuric and carbonic acids have more than one hydrogen to donate.

• H2SO4 → H+ + HSO4– Ka1 very large

completely ionized

•HSO4

– H+ + SO4

2– Ka2 = 0.012

• H2CO3 H+ + HCO3– Ka1 = 4.3e-7

HCO3–

H+ + CO32– Ka2 = 4.8e-11

•Ascorbic acid (vitamin C) is a diprotic acid, abundant in citrus fruit.

• Others:H2S, H2SO3, H3PO4, H2C2O4 (oxalic acid) …

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es• 56

Species concentrations of diprotic acids

• Evaluate concentrations of species in a 0.10 M H2SO4 solution.

• Solution:H2SO4 → H+ + HSO4

– completely ionized

(0.1–0.1) 0.10 0.10

• HSO4–

H+ + SO42– Ka2 = 0.012

0.10–y 0.10+y y Assume y = [SO42–]

• (0.10+y) y————— = 0.012(0.10-y)

• [SO42–] = y = 0.01M

[H+] = 0.10 + 0.01 = 0.11 M; [HSO4

–] = 0.10-0.01 = 0.09 M• 920203


es• 57

Species concentrations of weak diprotic acids• Evaluate concentrations of species in a 0.10 M H2S solution.

• Solution:H2S = H+ + HS– Ka1 = 1.02e-7(0.10–x) x+y x-y Assume x = [HS–]

• HS– = H+ + S2– Ka2 = 1.0e-13

x–y x+y y Assume y = [S2–]

• (x+y) (x-y) (x+y) y————— = 1.02e-7 ———— = 1.0e-13(0.10-x) (x-y)

• [H2S] = 0.10 – x = 0.10 M[HS–] = [H+] = x y = 1.0e–4 M;

[S2–] = y = 1.0e-13 M• 0.1>> x >> y:

x+ y = x-y = xx = 0.1*1.02e-7 = 1.00e-4y = 1e-13

• 920203

chapter fifteenprentice-hall ©2002slide 1 of 31 920203 http:\\asadipour.kmu.ac.ir........57 slides...

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