Part II Applications of Acid-Base Equilibria. Common Ion Effect Calculate [H + ] and the percent...

Acids and Bases

Part II

Applications of Acid-Base Equilibria

Common Ion EffectCalculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2x10-4) and 1.0

M NaF.

FNaNaF

FHHFco

change

ceq

NaFHFsolution FFF ][][][

1.0 M 1.0 M

4102.7)1(

)1(

x

xxK a

4102.7 x

14.3)102.7log()log(]log[ 4 xHpH

Previous Example: Calculate [H+] and the percent dissociation of HF in a

solution containing 1.0 M HF (Ka = 7.2x10-4).

%072.0%1001

102.7%100%

4

oc

x

1 M HF 1 M HF + 1 M NaF

x 4.2x10-3 7.2x10-4

pH 1.57 3.14

a% 0.42% 0.072%

Le ChatelierImagine 1 M HF solution at equilibrium, then NaF is

added.

What happens? Common Ion Effect

Buffered Solution

• is a solution that persists the change in its pH when H+ or OH- is added.• Solution of weak acid and the salt of its conjugate

base: HAc/NaAc • Solution of weak base abd the salt of its

conjugate acid: NH3/NH4+

• Blood is a buffered solution.

A buffered solution contains 0.50 M acetic acid HC2H3O2, Ka=1.8x10-5) and 0.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution.

AcNaNaAc

AcHHAcco

change

ceq

0.5 M 0.5 M

5108.1)5.0(

)5.0(

][

][][

x

xx

HAc

AcHK a

5108.1][ xH 74.4pH

Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the above buffered solution. OHNaNaOH

0.01 mol 0.01 mol

OHAcOHHAc 2

n = M x V(before

reaction)

0.5x1.0

=0.5

mol

0.5x1.0

=0.5

mol

change -0.01 +0.01

n = M x V(before

reaction)

0.49mol

0.51mol

AcHHAc

co 0.49 0 0.51

change -x x x

ceq 0.49-x x 0.49+x

5108.1)49.0(

)51.0(

x

xxK a

5107.1][ xH

74.4pH 02.0pH

Compare this pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L water.

OHNaNaOH0.01 mol 0.01 mol

ML

mol

V

nOHOH

solution

OHNaOHsolution 01.0

0.1

01.0][][

12121414

2)01.0log(]log[

pOHpH

OHpOH

5712 initialfinal pHpHpH

AHHA

][

][][

HA

AHK a

][

][log]log[log

HA

AHK a

][

][log]log[log

HA

AHK a

][

][log

HA

ApHpK a

][

][log

HA

ApKpH a

][

][log

acid

basepKpH a

Henderson-Hasselbalch Equation

AHHAco [HA]o 0 [A-]o

change -x x x

ceq [HA]o-x x [A-]o+x

][

][log

acid

basepKpH a

xHA

xApKpH o

a

0][

][log

0][

][log

HA

ApKpH o

a

Calculate the pH of a solution containing 0.75 M lactic acid (Ka=1.4x10-4) and 0.25 M sodium lactate.

38.375.0

25.0log)104.1log( 4 pH

A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.

o

oa acid

basepKpH

][

][log

o

oa acid

baseKpH

][

][loglog

b

wawba K

KKKKK

NH3

NH4+

bawba pKpKpKpKpK 14

26.974.41474.4)108.1log( 5 ab pKpK

05.94.0

25.0log26.9

][

][log

o

oa acid

basepKpH

Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution containing 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl.

05.94.0

25.0log26.9

][

][log

o

oaadditionbefore acid

basepKpH

ClHHCl0.1 mol 0.1 mol

43 NHHNHn = M x V(before

reaction)

0.25x1.0=0.25

mol

0.40x1.0

=0.40

mol

change -0.1 +0.1

n = M x V(before

reaction)

0.15mol

0.50mol

73.850.0

15.0log26.9

][

][log

o

oaadditionafter acid

basepKpH

32.005.973.8 initialfinal pHpHpH

Acid addition pH ↓

Base addition pH ↑

Buffering Capacity It represents the amount of H+ or OH the buffer can

absorb without a significant change in pH.

High buffering capacity: solution absorbs large amounts of acid or base without significant change in pH.

Low buffering capacity: small amounts of acid or base can produce a significant change in pH.

Factors affecting the buffering capacity:

1) The concentrations of buffer components• The more concentrated the components of a buffer , the

greater the buffer capacity.2) The ratio [A-]/[HA] • The closer this ratio is to 1, the higher the buffering

capacity.

The relation between buffer capacity and pH change.

AcHHAc 1][

][

0

HA

A o

0.01 mol solid

NaOH added to

1 L buffer.

Factors affecting the buffering capacity:

1) The concentrations of buffer components2) The ratio [A-]/[HA] • The closer this ratio is to 1, the higher the buffering

capacity.

0][

][log

HA

ApKpH o

a

1][

][

0

HA

A o

apKpH maximum buffer capacity

ProblemA chemist needs a solution buffered at pH= 4.30 and can choose from the following acids and their sodium salts:

1. chloroacetic acid, Ka= 1.35 x 10-3 2. propanoic acid, Ka= 1.3 x 10-5

3. benzoic acid, Ka= 6.4 x 10-5 4. hypochlorous acid, Ka= 3.5 x 10-8

Calculate the ratio [A-]/[HA] required for the best system to yield a pH of 4.30.

pKa

2.874.894.197.46

0][

][log

HA

ApKpH o

a

0][

][log19.430.4

HA

A o

29.1][

][

0

HA

A o

Strong acid – Strong base titration

HCl(aq) + NaOH(aq) H2O + NaCl(aq)

H+(aq) + OH-

(aq) H2Onet ionic equation:

http://www.google.jo/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwia-dC50KTJAhXGLhoKHaB9ASsQjRwIBw&url=http://wps.prenhall.com/wps/media/objects/3083/3157555/blb1703.html&psig=AFQjCNErbuxo3YPSz4lr7W1esxMRyqD72Q&ust=1448302237069614

Titration of50 mL 0.200 M HNO3 with 0.100 M NaOH

A. No NaOH has been added.

699.0)200.0log(

]log[]log[ 3

pH

HNOHpH

B. 10 mL NaOH has been added.

mmolmLVMn Lmol

NaOHNaOHNaOH 110100.0

mmolmLVMn Lmol

HNOHNOpresentinitiallyHNO 1050200.0333

HNO3(aq) + NaOH(aq) H2O + NaCl(aq)

reactedHNOpresentinitiallyHNOremainingHNO nnn

333

mmoln

nnn

remainingHNO

addedNaOHpresentinitiallyHNOremainingHNO

91103

33

MmL

mmol

V

n

V

nH HNOH 15.0

1050

9][ 3

82.0)15.0log(]log[ HpH

C. 20 mL NaOH (total) has been added.

mmolmLVMn Lmol



mmolnpresentinitiallyHNO 10

3

mmoln

nnn

remainingHNO


82103

33

MmL

mmol

V

n

V

nH HNOH 11.0

2050

8][ 3

94.0)11.0log(]log[ HpH

D. 50 mL NaOH (total) has been added.

31.1pH

So on until:

E. 100 mL NaOH (total) has been added.

mmolmLVMn Lmol




3

mmoln

nnn

remainingHNO


010103

33

)(10][

0.05050

0][

7

3

ionizationwaterfromMH

MmL

mmol

V

n

V

nH HNOH

7)10log(]log[ 7 HpH

Stoichiometric point or point of equivalence

F. 150 mL NaOH (total) has been added.

mmolmLVMn Lmol




3

mmoln

nnn

nnn

excessinNaOH

presentinitiallyHNOaddedNaOHexcessinNaOH

reactedNaOHaddedNaOHexcessinNaOH

51015

3

MmL

mmol

V

n

V

nOH excessNaOHOH 025.0

15050

5][

4.1214

6.1)025.0log(]log[

pOHpH

OHpOH

mmolnnexcessinNaOHOH

5


mmolmLVMn Lmol




3

mmoln

nnn

nnn

excessinNaOH

presentinitiallyHNOaddedNaOHexcessinNaOH


101020

3

MmL

mmol

V

n

V


20050

10][

6.1214

4.1)04.0log(]log[

pOHpH

OHpOH


10

Weak acid – Strong base titration

Titration of50 mL 0.100 M HAc with 0.100 M NaOH

A. No NaOH has been added.

pH of 0.10 M HAc?

87.2

108.11.0][ 5

pH

KcxH ao

B. 10 mL NaOH has been added.

mmolmLVMn Lmol


HAc(aq) + OH-(aq) H2O + Ac-

(aq)

reactedHAcpresentinitiallyHAcremainingHAc nnn

mmoln

nnn

remainingHAc

addedNaOHpresentinitiallyHAcremainingHAc

415

mmolmLVMn Lmol

HAcHAcpresentinitiallyHAc 550100.0

mmolnnaddedNaOHformedAc

1

MmL

mmol

V

nHAc HAc 067.0

1050

4][

M

mL

mmol

V

nAc Ac 0167.0

1050

1][

14.4067.0

0167.0log)108.1log(

][

][log 5

0

HA

ApKpH o

a

HAc

Aa

Vn

V

n

ao

a n

npKpK

HA

ApKpH

HAc

A

loglog][

][log

0

C. 25 mL NaOH has been added.

mmolmLVMn Lmol

NaOHNaOHNaOH 5.225100.0


(aq)

mmoln

nnn

remainingHAc


5.25.25

mmolnn

addedNaOHformedAc5.2

74.45.2

5.2log74.4log

HAc

Aa n

npKpH

D. 40 mL NaOH has been added.

mmolmLVMn Lmol



(aq)

mmoln

nnn

remainingHAc


145

mmolnn

addedNaOHformedAc4

34.51

4log74.4log

HAc

Aca n

npKpH

E. 50 mL NaOH has been added.

mmolmLVMn Lmol



(aq)

mmoln

nnn

remainingHAc


055

mmolnn

addedNaOHformedAc5

HAc

Aca n

npKpH

logmorenoexistsHAc

becauseapplicablenot

OHHAcOHAc 2 ][

][][

Ac

OHHAcKb

co

change

ceq

MmL

mmol

V

nAc Ac

o 05.05050

5][

a

wobob K

KcKcx

x

xK

05.0

2

MK

KcxOH

a

wo

6

5

14

1027.5108.1

100.105.0][

72.814

28.5)1027.5log(]log[ 6

pOHpH

OHpOH


mmolmLVMn Lmol


HAc(aq) + NaOH(aq) H2O + NaAc(aq)

mmolnpresentinitiallyHAc 5

mmoln

nnn

nnn

excessinNaOH

presentinitiallyHAcaddedNaOHexcessinNaOH


156

MmL

mmol

V

n

V


6050

1][

96.1114

04.2)091.0log(]log[

pOHpH

OHpOH


1

G. 75 mL NaOH (total) has been added. 30.12pH

Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka =6.2x10-10) when dissolved in water. If a 50.0-mL sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solutiona. after 8.00 mL of 0.100 M NaOH has been added.

mmolmLVMn Lmol

NaOHNaOHNaOH 8.08100.0

HCN(aq) + OH-(aq) H2O + CN-

(aq)

reactedHCNpresentinitiallyHCNremainingHCN nnn

mmoln

nnn

remainingHAc

addedNaOHpresentinitiallyHCNremainingHCN

2.48.05

mmolmLVMn Lmol

HAcHAcpresentinitiallyHCN 550100.0

mmolnnaddedNaOHformedCN

8.0

49.82.4

8.0log21.9log

HCN

CNa n

npKpH

b. at the halfway point of the titration.

mmoln

nn

addedNaOH

presentinitiallyHCNaddedNaOH

5.2521

21

mLmmol

M

nV

VMn

Lmol

NaOH

NaOHNaOH

NaOHNaOHaddedNaOH

251.0

5.2


(aq)

mmoln

nnn

remainingHCN


5.25.25

mmolnn

addedNaOHformedCN5.2

21.95.2

5.2log21.9log

HCN

CNa n

npKpH

halfway point of the titration:

apKpH

c. at the equivalence point of the titration.

mmoln

nn

addedNaOH

presentinitiallyHCNaddedNaOH

5

mLmmol

M

nV

VMn

Lmol

NaOH

NaOHNaOH

NaOHNaOHaddedNaOH

501.0

5


(aq)

mmoln

nnn

remainingHCN


055

mmolnn

addedNaOHformedCN5

OHHCNOHCN 2 ][

][][

CN

OHHCNKb

co

change

ceq

MmL

mmol

V

nCN Ac

o 05.05050

5][

a

wobo K

KcKcxOH ][ pH=10.96

Titration Curves of Polyprotic Acids

V/2 3V/2V 2V

pH = pKa1

pH = pKa22

pKpKpH 21 aa

pOH from Kb equil’m of A2–

39

A chemist dissolves 2.00 mmol of a solid acid in 100.0 mL water and titrates the resulting solution with 0.0500 M NaOH. After 20.0 mL NaOH has been added,

the pH is 6.00. What is the Ka value for the acid?

mmoln presentinitiallyHA 2

mmoln

mmolVMn

remainingHAc

NaOHNaOHaddedNaOH

2.48.05

120050.0

mmolnnn

nnn

addedNaOHpresentinitiallyHAremainingHA

reactedHApresentinitiallyHAremainingHA

112

HAc

Aa n

npKpH

log

6100.1

6

1

1log

a

a

a

K

pK

pKpH

Weak base – Strong acid titration

The pH curve for the titration of 100.0 mL of 0.050 M NH3 with 0.10 M HCl. Note the pH at the equivalence point is less than 7, since the solution contains the weak acid NH4

+..

Acid –Base Indicators

• Organic Dyes• weak acid-weak base conjugate pair

HIn (aq) H+ (aq) + In– (aq)

Acid form base form

(one color) (another color)

– If you put a small amount of HIn in solution• Will be one color if acidic (phenolphthalein = colorless)• Another color if basic (phenolphthalein = bright magenta)• Color will change when pH rises or falls

42][

]][[HIn

InHK In

Color Change

• Bromothymol BlueHIn (aq) H+ (aq) + In– (aq)

Acid form base form

(one color) (another color)

10][

][

HIn

In– If , the color of In- will be observed.

10

1

][

][

HIn

In– If , the color of HIn will be observed.

][

][log

HIn

InpKpH In

aIn pKpK

10][

][

HIn

In– If , the color of In- will be observed.

][

][log

HIn

InpKpH In

][

][log

HIn

InpKpH In

1

1 InpKpH

10

1

][

][

HIn

In– If , the color of HIn will be observed.

][

][log

HIn

InpKpH In

1

1 InpKpH

Color Change occurs in the range pH=pKIn1

Bromothymol Blue

KIn=1.0x10-7

Color Change occurs in the range pH=pKa1pH=71=6-8

pH<6 pH>8pH=6-8

pH

Colors and approximate pH range of some common acid-base indicators.

Is this indicator suitable for your titration?

100.0 mL of 0.10 M HCl with 0.10 M

NaOH.

titration of 50 mL of 0.1 M HC2H3O2 with 0.1 M NaOH.

Best indicator when pH at the point of equiv. is nearest to pKIn

To find if a given indicator is suitable, calculate pH at the point of equiv.

Part II Applications of Acid-Base Equilibria. Common Ion Effect Calculate [H + ] and the percent...

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Transcript of Part II Applications of Acid-Base Equilibria. Common Ion Effect Calculate [H + ] and the percent...