Acid Ba See Quil

7/28/2019 Acid Ba See Quil

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Acid Base Equilibria

Chapter 17


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6/4/2013 CHEM 106, Acid-Base Equilibria 2

Topics of Discussion

Solutions of a weak acid and base Acid-Ionization Equilibria Polyprotic Acids

Base-Ionization Equilibria Acid-base Properties of Salt Solutions

Solutions of Weak Acid/Base with Another Solute

Common-Ion Effect Buffers Acid-Base Titration Curves


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Weak acid

HA + H 2O H3O+ + A -

K a is acid dissociation constant

HA H+ + A -

aC23

C2

3

K 0]K [H[HA]

]][A0[H

K 0][HA][H

]][A0[H

aK [HA]]][A[H


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Some weak acids

N a m e F o r m u l a K a p K aI o d i c a c id H IO3 0 .1 7 0 .2 3

N i t r o u s a c id H N O2 7 .1 x 1 0- 4

3 .1 5H y d r o f lu o r i c a c id H F 6 .8 x 1 0- 4 3 .1 7F o r m i c a c id H C O O H 1 .8 x 1 0- 4 3 .7 4A c e t i c a c id C H3C O O H 1 .8 x 1 0

- 5 4 .7 4

H y p o c h l o r o u s a c i d H O C l 3 . 0 x 1 0- 8

7 .5 2


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Weak Base

B + H 20 BH + + OH -

NH 3 + H 2O NH 4+ + OH -

bK

[B]

]][OH[BH

b3

4 K ][NH

]][OH[NH


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Some weak bases

N a m e F o r m u la K b p K b

M e t h y l a m i n e C H3N H2 4 .4 x 1 0- 4

3 .3 6H y d r a z i n e N2H4 1 .7 x 1 0- 6 5 .7 7

A m m o n i a N H3 1 .8 x 1 0- 5 4 .7 4

H y d r o x i l a m in e H O N H2 6 .6 x 1 0- 9 8 .1 8A n i l in e C6H5N H2 4 .4 x 1 0

- 1 0 9 .3 6


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Equilibrium calculations

K a and K b from percentage ionization %ionization = (amount ionized)/(amount available)x100

ExampleIn a 0.01 M solution of butyric acid the acid is 4 %ionized at 20 0C. Calculate K a and pK a for butyric acid at

this temperature.


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Solution

HBu + H 2O H3O + Bu - H 2O + H B u H 3O

+ + B u-

In i t i a l 0 .0 1 0 0C h a n g e - 0 . 0 1 x 0 . 0 4 + 0 . 0 0 0 4 + 0 . 0 0 0 4E q u i li b r i u m 0 .0 0 9 9 6 0 .0 0 0 4 0 .0 0 0 4

K a

= (0.0004) 2/(0.00996)

= 1.6x10 -5

pK a = 4.8


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Equilibrium calculations

K a and K b from initial concentrations and pHExampleIn a 0.1 M solution of formic acid, the pH is2.38 at 25 0C. Calculate the K a and pK a for formic acid at this temperature.

HCOOH + H 2O HCOO - + H 3O+


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Solution

[H 3O+] = 10 -2.38 = 0.0042 M

H 2O + H C O O H H 3O+ + H C O O-

I n i t i a l 0 .1 0 0C h a n g e - 0 .0 0 4 2 + 0 .0 0 4 2 + 0 .0 0 4 2e q u i l i b r i u m 0 .0 9 6 0 .0 0 4 2 0 .0 0 4 2

[HCOOH]]O][H[HCOOK 3a

K a = 1.84x10 -4


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Simplifying assumption

HA H+ + A - Accuracy of calculations may be set within 5%

[H+

]/[HA] initial < 0.05[H+]2/[HA] initial = K a [HA] initial > K a 400

[HA] equilibrium = [HA] initial - x (x is negligible)[HA] equilibrium ~ [HA] initial


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Example

What is the pH of 0.010 M solution of dimethyl amine, (CH 3)2 NH? K b = 9.6x10 -4

(CH 3)2 NH + H 2O (CH 3)2 NH 2+ + OH - Is simplification possible?[(CH 3)2 NH] initial > 400 K b0.01 > (400)(9.6x10 -4) = 0.384

No simplification. Quadratic solution .


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Solution

x2 + 9.6x10 -4x - 9.6x10 -6 = 0x = 2.65x10 -3

pOH = -log(2.65x10 -3) = 2.58 pH = 11.42

H 2 O + ( C H 3)2 N H ( C H 3)2 N H 2+ + O H -

I n i t i a l 0 .0 1 0 0C h a n g e - x + x + xE q u i l i b r i u m 0 .0 1 - x x x

42

106.901.0 x K x x

b


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Polyprotic AcidsAcids with two or more acidic protonsH2SO 3(aq) HSO 3-(aq) + H +(aq) K a1= 1.3x10 -2

HSO 3-(aq) SO 32-(aq) + H +(aq) K a2= 6.3x10 -8

For polyprotic acids: K a1

> K a2

> K a3

Number of protons to be dissociated are larger. First proton separates from singly negative charged ion.

For 0.25 M sulfuroz acid; pH=?, SO3

2- =?H 2 O + H 2 SO 3 H 3 O

+ + HSO 3-

Initial 0.25 0 0Change -x +x +xEquilibrium 0.25-x x x


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Polyprotic Acids-II

xapp= 0.06, quadratic equation is needed to obtain x.SO 32- =?

1.29 pH ,051.0103.125.0 32

2

1 O H x x x

x

K a

H 2O + HSO 3- H 3O

+ + SO 32-

Initial 0.051 0.051 0Change -x +x +xEquilibrium 0.051-x 0.051+x x

8-2382 6.3x10SOx103.6051.0)051.0(

x x x x

K a


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c - ase roper es o aSolutions

Ionization of formic acidHCOOH + H 2O H3O+ + HCOO -

Hydrolysis of formate ionHCOO - + H 2O HCOOH + OH -

a3 K [HCOOH]

]][HCOOO[H

W ba

b

K OH O H HCOO

OH HCOOH x

HCOOH HCOOO H

K K

K HCOO

OH HCOOH

]][[][

]][[][

]][[.

][]][[

33


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For any acid-base conjugate

pairsK a x K b = K W

pK a + pK b = pK W = 14The value of K a = 1.8x10 -5 for CH 3COOH.What is the value of K b for CH 3COO -?Solution

105

14

a

W b 5.6x101.8x10

1x10K K

K


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Solutions of salts of weak acidsand bases

HCOOH + H 2O HCOO - + H 3O+

Strong acid weak conjugate baseHCl H+ + Cl -

H2O H+ + OH -

acid conjugate base

Very strong acid Very weak base

Very weak acidVery strong base


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ExampleWhat is the pH of 0.1 M solution of NaOCl?

K a = 3.0x10 -8SolutionOCl - + H 2O HOCl + OH -

78

14

b 3.3x103x101x10

][OCl][HOCl][OH

K

(x2/0.1) = 3.3x10 -7

[HOCl] = [OH -] = x = 1.8x10 -4

pOH = -log(1.8x10-4) = 3.74

pH = 14.00 - 3.74 = 10.26


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Ammonia and its conjugateacid ammonium ion

NH 3 + H 2O NH 4+ + OH -

NH 4+ + H 20 NH 3 + H 3O+

105

14

b

w

4

33a 5.6x101.8x10

1.0x10K K

][NH]O][H[NH

K

5

3

4 b 1.8x10][NH]][OH[NH

K


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Examples of some weak acidsand bases

Metal ions with high charge densities are weak acids

[Al(H 2O) 6]3+ + H 2O [Al(H 2O) 5OH] 2+ + H 3O+

Metal ions with small charges are nonacids Aquous ions of Gr IA : Li +, Na +, K +, Rb +, or Cs + (except Be 2+) and Gr IIA : Mg 2+, Ca 2+, Sr 2+, andBa 2+ do not affect pH of solution.


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Acid-base properties of a salt

If only the cation of salt is acidic, thesolution will be acidic.

If only the anion of the salt is basic, thesolution will be basic.If a salt has a cation that is acidic and ananion that is basic, the pH of the solution isdetermined by the relative strength of theacid or base.


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ExamplePredict whether the following salts are acidic,

basic or neutral. NaOCl NaNO 2 KCl NH 4Br

NaOCl Na + + OCl -

OCl - + H 2O HOCl + OH - (basic)

NaNO 2 (basic)KCl (neutral) NH 4Br (acidic)


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Common-Ion EffectThe common-ion effect is the shift in an ionic

equilibrium cause by the addition of a solute that provides an ion that takes part in the equilibrium.What is the degree of ionization of 0.10 M formic

acid, HCOOH, solution? K a= 1.84x10 -4 a. When it is together with 0.20 M HCl. b.When it is pure.

H 2 O + HCOOH H 3 O+ + HCOO -

Initial 0.10 0.20 0Change -x +x +xequilibrium 0.10-x 0.20+x x

53a 102.910.0

)20.0([HCOOH]

]O][H[HCOOK x x

x x x

P x Ion Deg

x Ion Deg 2

4

102.4..

105.9..


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Buffers

Buffer solutions are used to keep the pH constant.Dilution of solution, addition of acid or base to buffer solution affects the pH very little.

Buffer consists of a weak Bronsted acid and itsconjugate base or vice versa.Examples:

NH 3 and NH 4Cl (NH 3 / NH 4+)

CH 3COOH and CH 3COONa (CH 3COOH / CH 3COO -)


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How a buffer work?

Buffer: HA H+ + A -

If extra acid is added:

H+ + A - HA (extra acid is neutralized byconjugate base A -)If extra base is added:OH - + HA H2O + A - (extra base isneutralized by acid HA)


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Acetic acid / acetate buffer

What is the pH of a buffered solution madeup 0.015 M sodium acetate and 0.10 Macetic acid? K a = 1.8x10 -5

H C2H 3O 2 H + + C2H 3O 2-I n i t i a l 0 .1 0 0 .0 0 .0 1 5C h a n g e - x + x + xe q u i l i b r i u m 0 .1 0 - x x 0 .0 1 5 + x

)1.0()015.0(108.1 5

x x x x [H+] = x = 1.2x10 -4

pH = 3.92


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Ammonia / ammonium ion buffer

Calculate the pH of a buffer solution prepared by dissolving 0.10 mol NH 3 and 0.2 mol NH 4Cl in 1 L water. K b = 1.8x10 -5

H2O + NH 3 NH 4+ + OH -

Initial 0.1 0.2 0.0Change -x +x +xequilibrium 0.1-x 0.2+x x


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Henderson-Hasselbalch

EquationHA H+ + A -

-logK a= log[HA] - log[H +] - log[A -][HA]

]][H[AK

a

[HA]][Alog pK pH a


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Preparation of a buffer solutionwith a given pH

How do you prepare an acetate/acetic acid buffer solution with a pH = 5.0? (K a = 1.8x10 -5)

Solution pH = pKa + log([A -]/[HA])5.0 = -log(1.8x10 -5) + log([A -]/[HA])10 (5.00 - 4.74) = [A -]/[HA][A -]/[HA] = 1.82


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Selecting a weak acid for thepreparation of a buffer solution

A buffer solution is effective if 0.1 < [HA] / [A -] < 10 pH = pK a + log([A -]/[HA])

pH = pK a 1


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Example

How a buffer solution with pH = 8.00 be prepared?Solution:

pH = pK a + log([A -]/[HA])Hypochlorous acid, HOCl: Ka = 3.0x10 -8

pKa = 7.528.00 = 7.52 + log([OCl -]/[HOCl])[OCl -]/[HOCl] = 3.02


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Buffer capacity

The ratio [A -]/[HA] defines the pH of buffer solution.

The magnitude of concentrations [A -] and[HA] defines the buffer capacity.It is the ability of a buffer solution to

compensate extra added acid or base.


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Example (Buffer capacity)

A 1 L buffer solution is prepared using 1.00M acetic acid, HA, and 1.00 M NaAc. In an

experiment 0.11 mol of hydroxide ion isgenerated without any volume change. Canthe buffer solution handle this without a pH

change of 0.1 unit?K a = 1.8x10 -5


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Solution

pH =4.84 (buffer solution just tolerates excess base)

5a 1.8x10[HA]

]][A[HK

O H- + H A A- + H2OIn i t i a l 0 .1 1 1 .0 1 .0

E q u i l i b r i u m 0 .0 0 .8 9 1 .1 1

51.8x100.89

](1.11)[H

4.74 pH ,1.8x10H 5

HA + H 2O H 3O+

+ A-


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Acid-base titrations

The pH of an unknown acid may be found bytitrating it with a standard base solution until the

end point is reached.The equivalence point occurs whenstoichiometrically equivalent acid and basereacts.Titration curves are drawn, pH vs addedreagent, and equivalence point is determined.


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Titration of 25 mL of 0.2 MHCl with 0.2 M NaOH

VolumeNaOH (mL)

Total volume(mL)

Excess ion(mol/L)

pH

0 25.00 0.2 (H +) 0.7010.00 35.00 8.571x10 -2 1.0720.00 45.00 2.222x10 -2 1.6524.00 49.00 4.082x10 -3 2.3924.90 49.90 4.000x10 -4 3.4024.99 49.99 4.000x10 -5 4.4025.00 50.00 0 7.0025.01 50.01 3.999x10 -5 (OH -) 9.6025.10 50.10 3.992x10 -4 10.6026.00 51.00 3.922x10 -5 11.5950.00 75.00 6.666x10 -2 12.82

24.99


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Titration curveTitration of 25 mL 0.2 M HCl with 0.2M NaOH

0

2

4

6

8

10

12

14

0 10 20 30 40 50

mL NaOH added

p H

Equivalence point


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Titration of weak acid by strongbase

CH 3COOH + OH - CH 3COO - + H 2OBefore the titration begins Dissociation of weak acid

During the titration but before theequivalence point Formation of a buffer solution

At the equivalence point Hydrolysis of conjugate base

After the equivalence point Concentration of OH -

Titration of 25 mL 0.2 M HCl with 0.2M NaOH

0

2

4

6

8

10

12

14

15 20 25 30

mL NaOH added

p H


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Before the titration begins

What is the pH of 0.2 M CH 3COOH?Ka = 1.8x10 -5

CH 3COOH + H 2O CH 3COO - + H 3O+

Solution:5

2

3

33 1.8X100.2

x

COOH][CH

]COO][CHO[H

[H3O+] = 1.9x10 -3 M pH = 2.72


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Before the equivalence point...

Adding 10 mL 0.2 M NaOH to 25 mL 0.2 M CH 3COOH.

O H- + C H3C O O H C H3C O O-+ H2O

In i t i a l 2 m m o l 5 m m o l 0 m m o lE q u i l ib r iu m - 3 m m o l 2 m m o l

51.8x10(0.0857)

](0.0571)[H

[CH 3COOH] = (3 mmol)/(35 mL) = 0.0857 M

[CH 3COO -] = (2 mmol)/(35 mL) = 0.0571 M

[H+] = 2.7x10 -5 M pH = 4.57


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At the equivalence point

25 mL 0.2 M NaOH is added to 25 mL 0.2 MCH 3COOH. What is the pH of solution?Acetic acid is converted to acetate ion completely.CH

3COO - + H

2O CH

3COOH + OH -

10

a

W

3

3

3

5.6x10K

K

]COO[CH

COOH]][CH[OH

0.1M50mL5mmol

]COO[CH

[OH -] = 7.5x10 -6 M

pOH = 5.12 pH = 14.00 - 5.12 = 8.88


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After the equivalence point

35 mL 0.2 M NaOH is added to 25 mL 0.2 M

CH 3COOH. What is the pH of solution?OH - neutralizes CH 3COOH. Excess OH - definesthe pH.

[OH-

] = (35x0.2-25x0.2 mmol)/(60 mL) = 0.033M pOH = 1.48 pH = 12.5


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Titration of 25 mL 0.2 M HA by 0.2 M

NaOH

2

4

6

8

10

12

14

0 10 20 30 40

mL NaOH added

p H

pH = 8.88 atequivalence pt


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Titration of a weak base bystrong acid

NH 3 + H 2O NH 4+ + OH - Before the titration begins Dissociation of weak base

During the titration but before theequivalence point Formation of a buffer solution

At the equivalence point Hydrolysis of conjugate acid

After the equivalence point Concentration of H +

Titration of 25 mL 0.2 M NH3 and 0.2 M

HCl

0

2

4

6

8

10

12

14

15 20 25 30

mL HCl

p H


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Titration of 25 mL 0.2 M NH3 and 0.2M HCl

0

2

4

68

10

12

0 10 20 30 40

mL HCl

p H


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Titration of 25 mL 0.2 M NH3 and 0.2 MHCl

0

2

4

6

8

10

12

14

15 20 25 30

mL HCl

p H

Titration of 25 mL 0.2 M HA by 0.2 MNaOH

2

4

6

8

10

12

14

15 20 25 30

mL NaOH added

p H

Titration of 25 mL 0.2 M HCl with 0.2M NaOH

0

2

4

6

8

10

12

14

15 20 25 30

mL NaOH added

p H


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Acid-base indicators

HIn(aq) H+(aq) + In -(aq)

pH = pK a + log([In -]/[HIn])

pH > pK a + 1 ([In -]/[HIn]= 10, base color) pH < pK a - 1 ([In -]/[HIn]= 0.1, acid color)

Acid formcolor1

Base formcolor2

[HIn]]][In[H

K in

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