Asam Basa

0. What do you know about acid by Arrhenius, Brondted Lowry and Lewis

Jawab:

Arrhenius : asam adalah zat

yang mengandung ion H+ jika dilarutkan

dalam air

Bronsted Lowry : asam adalah zat

yang mampu mendonorkan proton

Lewis : asam adalah zat

yang menerima pasangan elektron

1. Write the chemical equation for the autoionization of water and the

equilibrium law for Kw?

Jawab:

Autoionisation of water is H2O

H+ + OH

-

H2O H+ + OH

-

K =

K [H2O]= [H+] [OH

-]

Kw = [H+] [OH

-]

2. How are acidic, basic, and neutral solutions in water defined

a. in terms of [H+] and [OH-] and b. in terms of pH ?

Jawab:

a. Acidic = [H+] > [OH

-]

Basic = [H+] < [OH

-]

Neutral = [H+] = [OH

-]

b. pH 7 neutral ( 298 K )

pH < 7 acid ( 298K )

pH < 7 basic ( 298 K )

3. At the temperature of the human body, 37

oC, the value of Kw is 2.4 x 10

-14.

Calculate the [H+], [OH

-], pH and pOH

of pure water at this temperature. What

is the relation between pH, pOH, and

Kw at this temperature? Is water neutral

at this temperature?

Jawab:

d1 : T = 37oC = 310 K

Kw = 2.4 x 10-14

d2 : [H+], [OH

-], pH, pOH

d3 :

H2O H+ + OH

-

Kw =

Kw = [H+] [OH

-]

2.4 x 10-14

= x2

x =

x = 1.549 x 10-7

[H+] = 1.549 x 10

-7

[OH-] = 1.549 x 10

-7

pH = - log [H+]

= - log (1.549 x 10-7

)

= 7 log 1.549 = 6.8099

pOH = - log [OH-]

= - log (1.549 x 10-7

)

= 7 log 1.549 = 6.8099

Pada suhu ini, pH = pOH yang

merupakan pKw. Air bersifat netral

4. Deuterium oxide, D2O, ionizes like water. At 20C its Kw, or ion product

constant analogous to that of water, is

8.9 x 10-16

. Calculate [D+] and [OD

-] in

deuterium oxide at 20C. Calculate also

the pD and the pDO.

Jawab:

d1 = Kw = 8.9 x 10-16

T = 20C

D2OD+ + OD

-

d2 = [D+], [OD

-], pD, pOD = ...?

d3 = [D+] = [OD

-] =

=

= 2.98 x 10-16

pD = 8 - log 2.98

= 7.53

pOD = 8 - log 2.98

= 7.53

5. Calculate the H+ concentration in each of the following solutions in which the

hydroxide ion concentrations are :

a. 0.0024 M b. 1.4 x 10-5 M c. 5.6 x 10-9 M

d. 4.2 x 10-13 M Jawab:

a. [OH-] = 2.4 x 10

-3 M

pOH = - log [OH-]

= - log 2.4 x 10-3

= 3 - log 2.4

pH = 14- (3- log 2.4 )

= 11 + log 2.4

= 11.38

[H+] = 10

-11.38

= 4.168 x 10-12

M

b. [OH-] = 1.4 x 10

-5 M

pOH = - log [OH-]

= - log 1.4 x 10-5

= 5 - log 1.4

pH = 14- (5- log 1.4 )

= 9 + log 1.4

= 9.146

[H+] = 10

-9.146

= 7.145 x 10-10

M

c. [OH-] = 5.6 x 10

-9 M

pOH = - log [OH-]

= - log 5.6 x 10-9

= 9 - log 5.6

pH = 14- (9- log 5.6 )

= 5 + log 5.6

= 5.748

[H+] = 10

-5.748

= 1.786 x 10-6

M

d. [OH-] = 4.2 x 10

-13 M

pOH = - log [OH-]

= - log 4.2 x 10-13

= 13 - log 4.2

pH = 14- (13- log 4.2 )

= 1 + log 4.2

= 1.62

[H+] = 10

-1.62

= 2.4 x 10-2

M

6. Calculate the OH- concentration in each of following solutions in which the

hydrogen ion concentrations are

a. 3.5 x 10 -8 M b. 0.0065 M c. 2.5 x 10 -13 M

d. 7.5 x 10 -5 M Jawab:

OH- if H+

a. 3.5 x 10 -8 M pH = 8- log 3,5

pOH = 14 - pH

= 14 - (8 - log 3,5)

= 6 + log 3,5

= 6,5

OH- = 10

-6,5

b. 0.0065 M = 6.5 x 10-3 pH = 3- log 6,5

pOH = 14 - pH

= 14 - (3 - log 6,5)

= 11 + log 6,5

= 11,8

OH- = 10

-11,8

c. 2,5 x 10 -13 M pH = 13- log 2,5

pOH = 14 - pH

= 14 - (13 - log 2,5)

= 1 + log 2,5

= 1,39

OH- = 10

-1,39

d. 7,5 x 10-5 M pH = 5- log 7,5

pOH = 14 - pH

= 14 - (5 - log 7,5)

= 9 + log 7,5

= 9,8

OH- = 10

-9,8

7. A certain brand of beer had a hydrogen ion concentration equal to 1.9 x 10

-5 mol

L-1

.What is the pH of the beer?

Jawab:

pH = -log [H+]

= -log [1.9 x10-5

]

= 5-log 1.9

pH = 5 - 0.28

pH = 4.72

8. A soft drink was put on the market with

[ ] = 1,4 x mol . What it's

pH?

Jawab:

pH = - log [H+]

= - log 1,4 x 10-5

pH = 5 - log 1,4

9. Calculate the pH of each of the solutions in Exercises 5 and 6.

Jawab:

Exercise 5

a. [OH-] = x.M = 2.4 x 10

-3

pOH = - log [2.4 x 10-3

]

= 3 - log 2.4

pH = 14 - (3 - log 2.4)

= 11 + log 2.4

b. [OH-] = x.M = 1.4 x 10

-5

pOH = - log [1.4 x 10-5

]

= 5 - log 1.4

pH = 14 - (5 - log 1.4)

= 9 + log 1.4

c. [OH-] = x.M = 5.6 x 10

-9

pOH = - log [5.6 x 10-9

]

= 9 - log 5.6

pH = 14 - (9 - log 5.6)

= 5 + log 5.6

d. [OH-] = x.M = 4.2 x 10

-13

pOH = - log [4.2 x 10-13

]

= 13 - log 4.2

pH = 14 - (13 - log 4.2)

= 1 + log 4.2

Exercise 6

a. [H+] = 3.5 x 10-8 pH = - log [3.5 x 10

-8]

= 8 - log 3.5

b. [H+] = 6.5 x 10-3 pH = - log [6.5 x 10

-3]

= 3 - log 6.5

c. [H+] = 2.5 x 10-13 pH = - log [2.5 x 10

-13]

= 13 - log 2.5

d. [H+] = 7.5 x 10-5 pH = - log [7.5 x 10

-5]

= 5 - log 7.5

10. Calculate the molar concentrations of H

+ and OH

- in solution that have the

following pH values.

a. 3.14 b. 2.78 c. 9.25 d. 13.24 e. 5.70

Jawab:

We have,

a. pH = 3.14 so, pOH = 14 - 3.14 = 10.86

[H+] = 10

- 3.14

[OH-] = 10

- 10.86

b. pH = 2.78 so, pOH = 14 - 2.78 = 11.22

[H+] = 10

- 2.78

[OH-] = 10

- 11,22

c. pH = 9.25 so, pOH = 14 - 9.25 = 4.75

[H+] = 10 - 9.25

[OH-] = 10

- 4.75

d. pH = 13.24 so, pOH = 14 - 13.24 = 0.76

[H+] = 10

- 13.24

[OH-] = 10

- 0.76

e. pH = 5.70 so, pOH = 14 - 5.70 = 8.30

[H+] = 10

- 5.70

[OH-] = 10

- 8.30

11. Calculate the molar concentration of H+

and OH-

in solution that have the

following pOH values .

a. 8.26 b. 10.25 c. 4.65 d. 6.18 e. 9.70

Jawab:

a. pOH = 8.26

[ OH- ] = antilog 8.26

= 10 -8.26

= 5x10-9

M

pH = 14 - 8.26 = 5.74

[ H+ ] = antilog 5.74

= 10 -5.74

= 1.819x10-6

M

b. pOH = 10.25

[ OH- ] = antilog 10.25

= 10 -10.25

= 5.6x10-11

M

pH = 14 - 10.25 = 3.75

[ H+ ] = antilog 3.75

= 10 -3.75

= 1.77827x10-4

M

c. pOH = 4.65


= 10 -4.65

= 2.2387x10-5

M

pH = 14 - 4.65 = 9.35

[ H+ ] = antilog 9.35

= 10 -9.35

= 4.467x10-10

M

d. pOH = 6.18


= 10 -6.18

= 6.6x10-7

M

pH = 14 - 6.18 = 7.82

[ H+ ] = antilog 7.82

= 10 -7.82

= 1.5x10-8

M

e. pOH = 9.70


= 10 -9.70

= 1.995x10-10

M

pH = 14 - 9.70 = 4.3

[ H+ ] = antilog 4.3

= 10 -4.3

= 5.0118x10-5

M

12. What is the pH of 0.010 M HCl ? Jawab:

Ma = 1 x 10

-2 mol/liter

[H+] = x . Ma

= 1 . 1 x 10-2

mol/liter

= 1 x 10-2

mol/liter

pH = - log [1 x 10-2

]

= 2

13. What is the pH of 0.0050 M solution of HNO3 ?

Jawab:

[H+] = x . Ma

= 1 x 5 x 10-3

= 5 x 10-3

pH = - log [ H+]

= - log 5 x 10-3

= 3 - log 5

14. A sodium hydroxide solution is prepared by dissolving 6.0 g NaOH in

1.00 L of solution. What is the pOH and

the pH of the solution?

Jawab:

D1 : m NaOH = 6.0 gram

V larutan = 1.00 L

Mr NaOH = 40

D2 : pOH and pH = .....?

D3:

[NaOH] = x

= x

= 0.15 M

[OH-] = 0.15 M

= 1.5 x 10-1

M

pOH = - log [OH-]

= - log 1.5 x 10-1

= 1 - log 1.5

= 0.824

pH = 14 - 0.824

= 13.176

15. A solution was made by dissolving 0.837 g Ba(OH)2 in 100 mL final

volume. What is the pOH and the pH of

the solution?

Jawab:

D1: mass Ba(OH)2 = 0.837 g

Mr Ba(OH)2 = 171

V = 100 mL

D2: pOH and pH of solution = ? D3 :

16. A solution of Ca(OH)2 has a measured

pH of 11.60. What is the molar

concentration of Ca(OH)2 in the

solution?

Jawab:

pOH = pKw - pH

=14 - 11.60

=2.4

pOH = - log [OH-]

[OH-] = 10

-2.4

= 3.98 x 10-3

Ca(OH)2 (aq) Ca2+

(aq)+ 2OH-

(aq)

[Ca(OH)2]=

= 1.99 x 10-3

So, [Ca(OH)2] = 1.99 x 10-3

17. A solution of HCl has a pH of 2.50. How many grams of HCl are there in

250 mL of this solution.

Jawab:

HCl H+ +Cl

-

pH HCl =2,5

[HCl] = [H+]= 3,16. 10

-3M

Mol HCl = M HCl. V HCl

=3,16. 10-3

. 2,5. 10-1

= 7,9. 10

-4 mol

Massa HCl = mol HCl. Mr HCl

= 7,9. 10-4

. 36,5

= 288,35. 10-4

gram

= 2,8835. 10-4

gram

18. Write the chemical equation for the ionization of each of the following weak

acids in water (For any polyprotic acids

, write only the equation for the first

step in the ionization).

a. HNO2 b. H3PO4 c. HAsO4

2-

d. (CH3)3NH+ Jawab:

a. HNO2 H+ + NO2

-

b. H3PO4 3H+

+ PO43-

c. HAsO42-

AsO43-

+ H+

d. (CH3)3NH+ (CH3)3N + H

+

19. For each of the acids in exercise 18, write the appropriate Ka expression

Jawab:

a. HNO2 H+

+ NO2-

b. H3PO4 3 H

+ + PO4

3-

c. HAsO4

2- H

+ + AsO4

3-

d. (CH3)3NH

+ (CH3)3N + H

+

20. Write the chemical equation for the ionization of each of following weak

bases in water.

a. (CH3)3N b. AsO4

3-

c. NO2-

d. (CH3)2N2H2 Jawab:

a. (CH3)3N + H2O CH3 N H

+ + OH

-

CH3

b. AsO43-

+ H2O HAsO42-

+ OH-

c. NO2-

+ H2O HNO2 + OH-

d. (CH3)2N2H2 + H2O H N NH2 + OH

-

CH3 CH3

21. For each of the bases in Exercise 20, write the appopriate Kb expression.

Jawab:

a. (CH3)3N + H2O (CH3)3NH+

+ OH-

b. AsO4

3- +H2O H3AsO4+ 3OH

-

H3AsO4 + H2O As(OH)5

c. NO2

- + H2O HNO2 + OH

-

d. (CH3)2N2H2 + H2O

(CH3)3NH3+ + OH

-

22. Benzoic acid, C6H5CO2H, is an organic acid whose sodium salt, C6H5CO2Na,

has long been used as a safe foods

additive to protect beverages and many

foods againts harmful yeasts and

bacteria. The acid is monoprotic. Write

the equation for it's Ka !

Jawab:

C6H5COOH+NaOH C6H5COONa

+ H2O

C6H5COOH C6H5COOH + H+

Ka

23. Write the equation for the equilibrium that the benzoate ion, C6H5CO2

- (review

exercise 22), would produce in water as

functions as a Bronsted base. Then write

the expression for the Kb of the

conjugate base of benzoic acid.

Jawab:

C6H5CO2- + H2O C6H5OH +

OH-

Kb =

24. The pKa of HCN is 9.21 and that of HF is 3.17. Which is the strong Bronsted

base CN or F

?

Jawab:

pKa HCN=9.21 Ka HCN= 6.1710-10 pKa HF = 3.17 Ka HF = 6.7610-4

Kb CN = = 0.1610

-4

Kb F = = 0.1510

-10

So, the strong Bronsted base is CN

25. The Ka for HF is 6.8 x 10x. what is the

Kb for F-?

Jawab:

HF H+ + F

-

Kw = Ka x Kb

10-14

= 6.8x10-4

x Kb

Kb = 1.47x10-11

26. The barbiturate ion C4HO has Kb = 1,0 x 10

-10 . What is Ka for Barbituric acid ?

Jawab: Kw = Kb x Ka

10-14

= 1,0 x 10 -10

x Ka

Ka =

Ka = 10 -4

27. Hydrogen peroxide, H2O2 is a week acid with Ka = 1.8 x 10

-12. What the value of

Kb for the HO2 ion?

Jawab:

Kb =

=

= 5.56 x 10-3

28. Methylamine, CH3NH2 resambles ammonia in odor and basicity. Its Kb is

4.4 x 10-4

. Calculate the Ka of its

conjugate acid!

Jawab:

Ka =

=

= 2.27 x 10-11

29. Lactic acid, HC3H5O3, is responsible for the sour taste of sour milk. At 25

oC its

Ka = 1.4 x 10-4

. What is the Kb of its

conjugate base, tha lactate ion, C3H5O3-?

Jawab:

Kw = Ka xKb

10-14

= 1.4 x 10-4

x Kb

Kb =

Kb = 7.14 x 10-11

30. Iodic acid, HIO3 has a pKa of 0.77 a. What is the formula an the Kb of

its conjugate base?

b. Its is conjugate base a stronger or

a weaker base than the acetate ion?

Jawab:

a. the formula

HIO3 H+ = IO3-

pKa = 0.77

Ka = 5,88 x 10 -1

Ka x Kb = 10-14

Kb =

= 1,7 x 10-14

b. Ka CH3COO- = 1,8 x 10

-5

Kb CH3COO- = 5,5 x 10

-10

SO, HIO3= is stronger conjugate base

then asetate ion.

31. Periodic acid,HIO4,is an important oxidizing agent and a moderately strong

acid. In a 0.10 M solution , [H+] = 3.8 x

10-2

mol L-1

. Calculate the Ka and pKa

for periodic acid!

Jawab:

3.8 x 10-2

=

( 3.8 x 10-2

)2 = Ka x 0.1

1.44 x 10-3

= Ka x 0.1

Ka =

Ka = 1.44 x 10-2

pKa = -log Ka

= -log (1.44 x 10-2

)

= 2 - log 1.44

pKa = 1.84

32. Choloacetic acid, HC2H2ClO2, is a stronger monoprotic acid than acetic

acid. In a 0,10 M solution, this acid is 11

% ionized. Calculate the Ka and pKa for

Choloacetic acid.

Jawab:

=

0,11 =

= (0,11)2 x 0,1

= 1,21 x

= 3 - log 1,21

pKa = - log Ka

pKa = - log 1,21x

33. Ethylamine, CH3CH2NH2, has a strong, pungent odor similar to that ammonia.

Like ammonia, it is a Bronsted base. A

0.10 M solution has a pH of 11.86.

Calculate the Kb and pKb for

ethylamine.

Jawab:

M = 0.1 mol/L

pH = 11.86

pOH = 14 - 11.86 = 2.14

[OH-] = 7.2 x 10-3

7.2 x 10-3

5.184 x 10-5

: 0.1 = Kb

Kb = 5.184 x 10-4

34. Hidroxylamine, HONH2, like ammonia, is a Bronsted base. A 0.15 M solution

has a pH of 10.12. What are Kb and

pKb for Hidroxylamine?

Jawab:

[OH-] =

10-3.88

=

10-7.76

=

Kb =

Kb = 1.158 x 10 -7

pKb = - Log Kb

= - Log 1.158 x 10 -7

= 7 Log 1.158 = 6.936

35. Refer to data in the preceding question to calculate the percentage ionization of

the base in 0.15 M HONH2.

Jawab:

: = = 0.000878

36. What is the pH of 0.125 M pyruvic acid ? It's Ka is 3.2 x 10

-3

Jawab:

Ma = 0.125 mol/liter

Ka = 3,2 x 10-3

[H+] =

=

=

= 2 x 10-2

pH = - log [2 x 10-2

]

= 2 - log 2

= 2 - 0.301

= 1.699

37. What is pH of 0.15 M HN3 ? for HN3, Ka = 1.8 x 10

-5

Jawab:

[H+] =

=

=

= 1.64 x 10-3

pH = - log [ H+]

= -log 1.64 x 10-3

= 3 - log 1.64

38. What is the pH of a 1.0 M solution of hydrogen peroxide, H2O2? For this

solute, Ka = 1.8 x 10-2

Jawab:

H2O2 O2 + 2H+ + 2e

-

[H+] =

=

= 0.134 M

pH = - log [H+]

= - log 0.134

= 0.87

39. Phenol, also known as carbolic acid, is sometimes used as a disinfectant. What

are the concentrations of all of the

substance in a 0.050 M solution of

phenol, HC6H50? What percentage of

the phenol is ionized? For this acid, Ka=

1.3 x 10-10

Jawab:

HC6H5O C6H50 + H

+

B 0.05 - -

R 2.55 x 10-6 2.55 x 10-6 2.55x10-6

A 0.05 - 2.55 x 10-6 2.55 x 10-6

2.55 x 10-6

40. Codeine, a cough suppressant extracted from crude opium, is a weak base with a

pKb of 5.79. What will be the pH of a

0.020 M solution of codeine? (Use Cod

as a symbol for codeine)

Jawab:

pKb = - log Kb

Kb = 10 -5.79

= 10 -6

[OH-]=

=

=

= 1.4 x 10 -4

pOH = - log [OH-]

= - log 1.4 x 10 -4

= 4 - log 1.4

= 4 - 0.146

= 3.854

pH = pKw - pOH

= 14 - 3.854

= 10.146

So, pH of Cod = 10.146

41. Deuteroammonia, ND3, is a weak base with a pKb of 4.96 at 25

oC. What is the

pH of a 0.20 M solution of this

compound?

Jawab:

[OH] =

=

=

= 1,48. 10-3

pOH = 3-log 1,48

pH = 14 - (3-log 1,48)

= 11+ log 1,48

= 11,17

42. A solution of acetic acid has a pH of 2.54. What is the concentration of acetic

acid in this solution ?

Jawab:

pH = - log [H+]

2.54 = - log [H+]

[H+] = 10

-2.54

[H+] = 2.88 x 10

-3

[H+] =

2.88 x 10-3

=

(2.88 x 10-3

)2= 1.8 x 10

-3 x Ma

= 1.8 x 10-5

x Ma

3.2 x 10-1

= Ma

0.32 = Ma

43. Aspirin is acetylsalicyclic acid, a monoprotic acid whose Ka value is 3,27

x 10-4

. does a solution of the sodium salt

of aspirin in water test acidic, basic, or

neutral ? Explain

Jawab:

The sodium salt of aspirin is basic, because

it from acetylsalyclic acid and sodium

hydroxide. Weaker acid with stronger base

want to produce basic salt or if sodium

hydroxide dissolved in water to produce

acetylsalicyclic acid and OH-

NaAcetylsalicyclic + H2O

HAcetylsalicyclic + OH-

44. The Kb value of the oxalate ion, C2O42-

,

is 1.9x10-10

. Is a solution of K2C2O4

acidic, basic, or neutral? Explain.

Jawab:

K2C2O4 merupakan basa karena K2C2O4

merupakan garam yang terbentuk dari basa

kuat dan asam lemah sehingga garamnya

bersifat basa. K2C2O4 dapat diperoleh dari

mereaksikan basa kuat yaitu KOH dan asam

lemah H2C2O4.

Reaksi:

H2C2O4 + 2 KOH K2C2O4 + 2 H2O

45. Consider the following compounds and suppose that 0.5M solutions are

prepared of each : NaI, KF, (NH4)2SO4,

KCN, KC2H3O2, CsNO3, and KBr.

Write the formulas of those that have

solutions that are

a. Acidic, b. Basic, and c. Neutral.

Jawab :

a. Acidic is (NH4)2SO4 and CsNO3 b. Basic is KF, KCN, KC2H3O2 c. Neutral is NaI and KBr

46. Will an aqueous solution of ALCl3 turn litmus red or blue ? explain?

Jawab:

yes it will. based on Lewis' acid-base

theory , AlCl3 is an acid . because its

configuration have a vacant orbital . it

means that the central atom of AlCl3 is

not octet yet . so we can assume that

AlCl3 is an acidic acid based on Lewis'

acid-base theory.

Berdasrkan teori asam basa Lewis ,

senyawa AlCl3 merupakan senyawa

yang bersifat asam karena atom

pusatnya belum mencapai konfigurasi

oktet atau bisa dikatakan masih

mempunyai orbital kosong. Jadi, larutan

AlCl3 dapat merubah warna kertas

lakmus biru menjadi merah karena sifat

keasamannya

47. Explain why the beryllium ion is a more acidic cation than the calcium ion.

Jawab:

in the periodic system of elements, Be is

located above Ca so it is more likely to

be acidic because of the periodic system

of elements in one group the greater the

atomic number or from top to the down

it will be more likely to be basic. So,

beryllium ion is a more acidic cation

than the calcium ion.

48. Ammonium nitrate is commonly used in fertilizer mixtures as a source of

nitrogen for plant growth. What effect,

if any, will this compound have on the

acidity of the moisture in the ground?

Explain.

Jawab:

Ammonium nitrate (NH4NO3)

NH4NO3 NH4+

+ NO3

NH4+ + H2O NH3 + H3O

+

NO3 + H2O

If any this compound in the ground, the

acidity of the moisture in the ground

will increase. There is H3O+ as a

product from the chemical equations

above.

49. Calculate the pH of 0.20 M NaCN. Jawab:

[NaCN] = 0.20 M

[OH-] =

=

=

= 1.82x10-5

pOH = 5 - Log 1.82

pH = 9 + Log 1.82

50. Calculate the pH of 0,04 M KNO2 ? Jawab:

KNO2 K+ + NO2

-

K+

+ H2O NO2

- + H2O HNO2 + [OH

-]

[OH-] = x [NO2]

= x

= 10 -12

pOH = 12

pH = 2

51. Calculate the pH of 0.15 M CH3NH3Cl. For CH3NH2, Kb = 4.4 x 10

-4

Jawab:

[H+] =

=

= 1.846 x 10-6

pH = - log [H+]

= - log (1.846 x 10-6

)

= 6 - log 1.846

= 5.734

52. A weak base B forms the salt BHCl, composed of the ions BH

+ and Cl

-. A

0.15 M solution of the salt has a pH of

4.28. What is the value of Kb for the

base B?

Jawab:

pOH = 14 - 4.28

= 9.72

[OH-] = 10

-9.27

= 1.9 x 10-10

[OH-] =

1.9 x 10-10

=

3.61 x 10-20

= Kb x 0.15

Kb = 24.067 x 10-20

= 2.4067 x 10-19

53. Calculate the number of grams of NH4Br that have to be dissolved in 1.00

L of water at 25oC to have a solution

with a pH of 5.16 !

Jawab:

pH = 5.16

[H+] = 10-5.16

= 6.9 x 10-6

M NH4Br =

=

=

[H+] =

6.9 x10-6

=

(6.9 x 10-6

)2 = 5.5 x 10

-10 x

5.5 x 10

-10 x gram NH4Br =

4.761 x 10-11

x 76

Massa NH4Br = 6.5 gram

54. The conjugate acid of a molecular base has a hypohetical formula. BH

+, and has

pKa of 5.00. A solution of salt of this

cation, BHY, tests slightly basic. Will

the conjugate acid of Y-, HY, have a

pKa greater than 5.00 or less than 5.00?

explain

Jawab:

Conjugate acid is BH+ that pKa = 5

BHOH BH+ + OH- pKa=5, Ka = 10-5

BHOH + HY BHY + H2O BHY BH+ + Y-

[ OH-] =

=

=

=

= 10-5

x [BHY] x 10

pOH = 5 - log [BHY]

pOH = pKa HY

So, pKa less than 5

55. Many drugs that are natural Bronsted bases are put into aqueous solution as

their much more soluble salt with strong

acids. The powerful painkiller

morphine, for example, is very slightly

soluble in water, but morphine nitrate is

quite soluble. We may represent

morphine by the symbol Mor and its

conjugate acid as H-Mor+. The pKb of

morphine is 6.13. What is the calculated

pH of a 0.20 M solution of H-Mor+?

Jawab:

pKb = 6.13

Kb = 10-pKb

= 10-6.13

Kb = 7.41 x 10-7

Mor + H2O H-

Mor+ + OH

-

0.2 M 0.2 M

[OH-]=

=

= 7.41 x 10-7

pOH= 7 - log 7.41

=6.13

So, pH = 14 - 6.13

= 7.87

56. Quinine, an important drug in treating malaria, is a weak Bronsted base that we

may represent as Qu. To make it more

soluble in water, it is put into a solution

as its conjugate acid, which we may

represent as H- . What is the

calculate pHof a 0,15 M solution of H-

? Its pKa is 8,52 at 25 0C.

Jawab:

D1 : [H- ] = 0,15 M, pKa = 8,52

H- + H2O Qu+H3

D2 : pH....?

D3 : pKa = 8,52

pKa = - log Ka

8,52 = - log Ka

Ka = 3,02 x

H- + H2O Qu + H3

M 0,15 - - -

R -

S 0,15 - -

Ka =

3,02 x

3,02x

= 2,13 x

pH = - log

= - log 2,13 x

pH = 5 - log 2,13

57. Generally, under what conditions are we unable to use the initial concentration of

an acid or base as though it were the

equilibrium concentration in the mass

action expression?

Jawab:

Initial concentration is unable to

calculate equilibrium concentration

when mole of both of components

which react is same.

58. What is the percentage ionization in a 0.15 M solution of HF ? What is the pH

of the solution ?

Jawab:

D1 : M HF = 0.15 Molar

Ka HF = 6.5 X 10-4

D2 : pH ?

D3 : [H+] =

=

= 9.87 x 10-3

pH = - Log [H+]

= - Log 9.87 x 10-3

= 3 - Log 9.87

= 2.00568

59. What is the percentage ionization in 0.0010 M acetic acid ? What is the pH

of the solution?

Jawab:

: = = 0.134

[ H+ ] = =

= 1.34x10-4

pH = -log 1.34x10-4

= 4 log 1.34 = 3.8729

60. What is the pH of a 1.0 x 10-7 M solution of HCl ?

Jawab:

Ma = 1.0 x 10

-7 mol/liter

Dalam hal ini berlaku ketentuan :

[H+] [OH

-] = Kw

[Cl-] = [HCl]

[H+] = [OH

-] + [Cl

-] ;

prinsip penetralan muatan

pH = - log 1,62 x 10

-7 = 6.79

61. The hydrogen sulfate ion HSO4-, is a

moderately strong Bronsted acid with a

Ka of 1.0x10-2

.

a. Write the chemical equation for the ionization of the acid and give the

appropriate Ka expression.

b. What is the value of [ H+] in 0.010 M HSO4

- (furnished by the salt,

NaHSO4) ? Do NOT make

simplifying assumptions; solve the

quadratic equation.

c. What is the calculate of [H+] in 0.010 M HSO4

-, obtained by using the

usual simplifying assumption?

d. How much error is produced by incorrectly using the simplifying

assumption?

Jawab:

a. HSO4- H

+ + SO4

2-

b. Ka = because [H+] =

[SO42-

]

So, . Ka =

[H+]

2= Ka x [HSO4

-]

[H+]

2 = 10

-2 x 10

-2 [H

+] =

10-2

c. [H+] =

=

= 1 x 10-2

d. The error has happened in using the simplifying assumption is 0 %

because product of point c is equal to

with point b.

62. Para-Aminobenzoic acid (PABA) is a powerful sunscreening agent whose salt

were once used widely in suntanning......

The parent acid, which we may

symbolize as H-Paba, is a weak acid

with a pKa of 4.92 (.....oC). What is the

[H+] and pH of 0.030 M solution of this

acid?

Jawab:

15. Given : pKa H- Paba = 4.92 [H-Paba] =

0.030 M

Asked : [H+] and pH = ...?

Solution :

NH2 COOH NH2

COO- + H

+

pKa = - log Ka

4.92 = - log Ka

Ka = 1.2 x 10-5

Ka =

1.2 x 10-5

=

[H+] =

[H+] = 6 x 10

-4

pH = - log [H+]

= 4 - log 6

= 3.22

63. Barbituric acid, HC4H3N2O3 (which we will abbreviate H-Bar), was discovered

by the Nobel Prize-winning organic

chemist Adolph von Baeyer and named

after his friend, Barbara. It is the parent

compound of widely sleeping drugs, the

barbituretes. Its pKa is 4.01. what is the

[H+] and pH of a 0.050 M solution of H-

Bar?

Jawab:

D1: pKa = 4.01

M = 0.050 mol/L

D2 : [H+] and pH = ?

D3 :

64. Write ionic equation that illustrate how each pair of compounds can serve as a

buffer pair.

a. H2CO3 and NaHCO3 (the "carbonate" buffer in blood)

b. NaH2PO4 and Na2HPO4 (the "phosphate" buffer in side body

cells)

c. NH4Cl and NH3 Jawab:

a. H2CO3(aq) + NaOH(aq) NaHCO3(aq) + H2O(l)

Ionic equation:

2H+ (aq) + CO3

2-(aq) + Na

+ (aq) + OH

-(aq)

Na+

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

Weak acid : H2CO3 Conjugation base : HCO3

-

b. H3PO4 (aq)+ NaOH (aq) NaH2PO4(aq) + H2O

Ionic equation:

3H+

(aq) + PO43-

(aq) + Na+

(aq) + OH-(aq)

Na+

(aq) + H2PO4-(aq) + H2O(l)

H2PO4(aq) + NaOH (aq) HPO4 (aq)

+ H2O (l)

Ionic equation:

Na+ (aq) + H2PO4

- (aq) + Na

+ (aq) +

OH(aq) 2Na+

(aq) + HPO42-

(aq) +

H2O(l) Weak acid : H2PO4

-

Conjugation base : HPO42-

c. NH3(aq) + HCl (aq) NH4Cl (aq) Ionic equation:

NH3(aq) + H+

(aq) + Cl-(aq)

NH4+

(aq) + Cl-(aq)

Weak base : NH3 Conjugation acid : NH4

+

65. Which buffer would be better able to hold a steady pH on the addition of

strong acid, buffer 1 or buffer 2?

Explain.

Buffer 1 is a solution containing 0.10 M

NH4Cl and 1 M NH3.

Buffer 2 is a solution containing 1 M

NH4Cl and 0.10 M NH3.

Jawab:

D1: a. buffer 0.10 M NH4Cl and 1 M

NH3

b. buffer1 M NH4Cl and 0.10 M NH3

d2: buffer would be better able to hold a

steady pH=? D3: pH larutan a sebelum ditambah asam

kuat

[OH-] = Kb NH3.

= 1,8 .10-5

= 1,8. 10-4

= 4- log 1,8

pOH = 3,745

pH = 10,255

pH larutan b sebelum ditambah asam

kuat

[OH-] = Kb NH3.

= 1,8 .10-5

= 1,8. 10-6

= 6- log 1,8

pOH = 5,74

pH = 8,26

pH larutan a setelah ditambah asam kuat

Misalnya asam kuat yang ditambahkan

HCl 0.05 M

[OH-] = Kb NH3.

= 1,8 .10-5

=1,8 .10-5

= 6. 10-6

= 6- log 6

pOH = 5,22

pH = 8,78

pH larutan b setelah ditambah asam

kuat

[OH-] = Kb NH3.

= 1,8 .10-5

=1,8 .10-5

= 8,6. 10-7

= 7- log 8,6

pOH = 6,065

pH = 7,935

Perubahan pH larutan a sebelum

ditambah asam kuat dan pH larutan a

setelah ditambah asam kuat = (10,255 -

8,78)

= 1,475

Perubahan pH larutan b sebelum

ditambah asam kuat dan pH larutan b

setelah ditambah asam kuat = (8,26 -

7,935)

= 0,325

Kesimpulannya larutan penyangga yang

pHnya cenderung hanya berubah sedikit

adalah larutan b karena perubahan

pHnya hanya 0,325

66. What is the pH of a solution that contains 0.15 M HC2H3O2 and 0.25 M

C2H3O2-? Use Ka = 1.8 x 10

-5 for

HC2H3O2 Jawab: D1 : C2H2H3O2 H+ +C2H3O2

-

M HC2H3O2 : 0.15 M

M C2H3O2- : 0.25 M

D2 : pH .? D3 :

1.8x10-5

=

1.8x10-5

x 0.15 = 0.25 x [H+]

= [H+]

1.08 x 10-5

= [H+]

pH = -log 1.08 x 10-5

= 5-log 1.08

= 5-0.033

= 4.967

67. Rework the preceding problem using the Kb for the acetate ion. ( be sure to write

the poper chemical equation and

equilibrium law )

Jawab:

D1 : M HC2H3O2 = 0.15 M

M C2H3O2- = 0.25 M

Ka = 1.8 x 10 -5

D2: pH dengan menggunakan Kb dari

ion asetat

D3:

68. By how much will the pH change if 0.050 mol of HCl is added to 1.00 L off

the buffer in Exercise 66.

Jawab:

[H+] = Ka x

= 1.8 x 10-5

x

=1.081 x 10-5

pH= -log [H+]

= - log 1.081 x 10-5

= 5 - log 1.081

= 4.9664

Perubahan pH= 4.9666 - 4.9664

= 0,0002

Perubahan pH sangat kecil karena

jumlah HCL yang di tambahkan sangat

sedikit sedangkan volume buffernya

besar.

69. By how much will the pH change if 50.0 mL of 0.10 M NaOH is added to 500mL

of the buffer in Exercise 66.

Jawab:

D1 : 500 mL of Buffer that contain 0,15

M HC2H3O2 and 0,25 M C2H3O2- ?

Ka = 1,8 x 10-5 for HC2H3O2

D2 : How much will pH Change ?

D3: HC2H3O2 + NaOH C2H3O2Na + H2O

m = 75 mmol 5 mmol 125 mmol

r = 5 mmol 5 mmol _ 5 mmol + s = 70 mmol 0 130 mmol

Atau dengan cara lain

70. A buffer is prepared containg 0.25 M NH3 and 0.14 M NH4+

a. calculate the pH of the buffer using the Kb for NH3

b. calculate the pH of the buffer using the Ka for NH4

+

Jawab:

NH3 (aq) + H2O(l) NH4+ (aq) + OH

-

a. misal Kb = 1.8 x 10-5 dan volume larutan dianggap sama , maka

[OH-]=Kb .

= Kb .

= Kb .

= 1.8 x 10-5

. 1.7857143

= 3.21428574 x 10-5

pOH = 5 - log 2.21428574

pOH = 4.492915518

pH = 9.507084482

b. misal Ka = 10-5

[H+] = Ka

= 10-5

= 10

-5

= 10-5

x 0.56

= 5.6 x 10-6

pH = 6 - log 5.6

71. By how much will the pH change if 0.020 mL of HCl is added to 1.00 L of

the buffer in Exercise 70?

Jawab:

D1 : Buffer

0.25 NH3 and 0.14 NH4+ 1.00 L

D2 : the change of pH if 0.020 mol HCl

is added to 1.00 L.?

D3 : the mol of base = 0.25 M x 1.00 L

= 0.25 mol

The mol of conjugate acid = 0.14 M

x 1.00 L = 0.14 mol

* If using Kb of NH3+

[OH-] = Kb

= 1.8x10-5

x

= 3.21x10-5

pOH = -log 3.21x10-5

= 4.5

pH = 14 - 4.5 =9.5

If added 0.020 mol HCl

NH3 (aq) + H+

(aq) NH4+

m 0.25mol 0.020mol 0.14 mol

r - 0.020mol 0.020mol +0.02mol

s 0.23mol - 0.16mol

[OH-] = 1.8x10

-5 x

= 2.6x10-5

pOH = 5-log 2.6

= 5-0.41

= 4.59

pH = 14 - 4.59 = 9.41

the change of pH = 9.5 - 9.41 = 0.09

*If using Ka of NH4+

[H+] = Ka

= 10-5

x

= 0.56x10-5

= 5.6x10-6

pH = 6 - log 5.6

pH = 5.25

If added 0.020mol HCl

[H+] = Ka

= 10-5

x

= 1.43x10-5

pH = 5 - log 1.43

pH = 4.84

Change of pH = 5.25 4.84 = 0.41

72. By how much will the pH change if 75 ml of 0.10 M KOH is added to 200 ml

of the buffer in exercize 70?

Jawab:

pH buffer

[OH] = Kb

= 110-5

= 1.7910-5

pOH = log 1.7910-5 = 5 - log 1.79

= 5 0.25 = 4.75

pH = 14 4.75 = 9.25

Added by 75 ml 0.10 KOH

NH4+

(aq)+OH

(aq)NH3(aq)+H2O(l) m 28 mmol 7.5 mmol 50 mmol r 7.5 mmol 7.5 mmol 7.5 mmol

s 20.5 mmol 57.5 mmol

[OH] = Kb

= 110-5

= 2.810-5

pOH = log 2.810-5 = 5 log 2.8 = 5 - 0.45

= 4.55

pH = 14 - 4.55

= 9.45

So, pH is change from 9.25 to 9.45

73. How many grams of sodium acetat, NaC2H3O2, would have to be added to

1.0 L of 0.15 M acetic acid (pKa 4.74)

to make the solution a buffer for pH

5.00?

Jawab:

Jawab:

[H+] =

10-5

=

X =

= 0.0833

Mass of CH3COOH = mol x Mr

= 0.0833 x 60

= 5 gram

74. How many grams of sodium formate, NaCHO2, would have to be added to 1.0

L of 0.12 M formic acid (Pka 3.74) to

make the solution a buffer for pH 3.80 ?

Jawab:

CHO2H 1.0 L 0.12 M = 0.12 mol

Mr NaC2HO2 = 68

pKa = 3.74 Ka = 1.8 x 10 -4

pH = 3.80 [H+] = 1.58 x 10

-4

kasus buffer asam

[H+] = Ka

1.58 x 10 -4

= 1.8 x 10 -4

1.58 x 10 -4

=

X = 0,14 mol

n =

0.14 =

massa NaC2HO2 = 9.3 gram

75. What mole ratio of NH4Cl to NH3 would buffer a solution at pH 9.25?

Jawab:

pOH = 4.75

[OH-] = Kb

1.77 x 10-5

= 1.8 x 10-5

=

=

76. How many grams of ammonium choride would have to be dissolved in 500 mL

of 0.20 M NH3 to prepare a solution

buffered at pH 10.00?

Jawab:

D1 :500 mL of 0.20 M NH3

pH buffer = 10.00

D2: m NH4Cl = ...?

D3: pH = 10, pOH = 4, [OH-] = 10

-4

[OH-] = Kb

10-14

= 1.8 x 10-14

.

10-14

= 0.018 x n AK

n AK = 5.556 x 10-3

n AK =

5.556 x 10-3

=

m = 0.297 gram

77. How many grams of ammonium chloride have to be dissolved into 125

mL of 0.10 M NH3 to make it a buffer

with a pH of 9.15 ?

Jawab:

D1 : V NH3 = 125 mL

M NH3= 0.10 mol/L

pH = 9.15

Kb = 1.8 x 10-4.85

D2: massa of NH4Cl . . .?

D3:

Mol NH3 = 12.5 mmol

pH = 9.15

pOH = 4.85

[OH-] = Kb

1.4 x 10-5

= 1.8 x 10-5

2.25 x 10-4

= n asam konj. x 1.4 x 10-5

n as.konj = 16.07 mmol

n as.konj =

massa NH4Cl = 16.07 x 53.5

= 859.75 mg

= 0.86 gram

78. Suppose 25.00 mL of 0.100 M HCl is added to an acetate buffer prepared by

dissolving 0.100 mol of acetic acidand

0.110 of sodium acetate in 500 mL of

solution. What are the initial and final

pH value? what would be the pH if the

same amount of HCl solution were

added to 125 mL of pure water?

Jawab:

Awal :

[ H+] = Ka

= 1,8 x 10 -5

x

= 1,64 x 10-5

pH = 5 - log 1,64

= 4, 79

[H+] = Ka x

= 1,8 x 10-5

x

= 1,8x 10 -5

x

= 1,72 x 10-5

pH = 5 - log 1,72

= 4,76

Final:

Pengenceran dengan penambahan

air pada buffer, pH akan tetap

karena penentu pH buffer adalah

jumlah mol bukan konsentrasi

buffer

79. How many milliliters of 0.15 M HCl would have to be added to 100 mL of

the buffer described in exercise 78 to

make the pH decrease by 0.05 pH unit?

How many milliliters of the same HCl

solution would, if added to 100 mL of

pure water, make the pH decrease by

0.05 pH unit?

Jawab:

a) [H+] =

4.45 x 10-5

=

4.89 x 10-6

- 4.45 x 10-5

a = 1.8 x 10-6

+ 1.8

x 10-5

a

4.89 x 10-6

- 1.8 x 10- = 1.8 x 10

-5a + 4.45

x 10-5

a

3.09 x 10-6

= 6.25 x 10-5

a

a =

a = 0.049

mol HCl = 0.049

M . V = 0.049

0.15 . V = 0.049

V=

V= 0.33 L

V= 330 mL

b) Jumlah HCl yang ditambahkan adalah sama yaitu 330 mL,Buffer ditambah

dengan air (pengenceran) pH larutan

tetap karena penentu pH buffer adalah

jumlah mol bukan konsentrasi buffer.

80. What can make the titrated solution at the equivalence point in an acid-base

titration have a pH not equal to 7,00 ?

Ho w does this possibility affect the

choice of an indicator ?

Jawab:

Apabila dalam titrasi asam-basa, titk

ekuivalen dapat tercapai pada pH tidak

sama dengan 7 berarti salah satu larutan

asam atau basanya bersifat lemah.

Pemilihan indikator sangat

mempengaruhi proses titrasi karena

untuk mengetahui titik akhir titrasi

dengan ditandai perubahan warna,

sesuai dengan rentang pH yang

diperkirakan sehingga pemilihan

indikator yang digunakan tidak salah.

81. Explain why ethyl red is a better indicator than phenolphtalein in the

titration of dilute ammonia by dilute

hydrochloric acid?

Jawab:

Methyl red is a better indicator than

phenolphtalein in the titration of dilute

ammonia by dilute hydrochloric acid

because the result of ammonia and

hydrochloric acid is a solution that has

pH < 7 ( influenced by hydrochloric

acid as a strong acid and ammonia is a

weak base ). Which is pH range of

methyl red is 4.4 -6.2 and pH range of

phenolphtalein is 8.3 -10.0

82. What is a good indicator for titrating potassium hydroxide with hydrobromic

acid? Explain.

Jawab: For titrating potassium

hydroxide with hydrobromic acid we

use metilred indicator as a good

indicator. Titrating potassium hydroxide

with hydrobromic acid is example of

titration strong base and weak acid. The

equivalen point is occur in value of pH

smaller than 7, so we must use indicator

that have trayek of pH under 7, for

example indicator metilred that have

trayek of pH from 4.8 until 6.

83. In the titration of an acid with base,what condition concerning the quantities of

reactans ought to be true at the

equivalence point?

Jawab:

the quantities of reactans ought to be

true at the equivalence point when the

mols equivalence of acis as same as the

mols equivalence of base.

84. When 50 mL of 0.10 M formic acid is titrated with 0.10 M sodium hydroxide,

what is the pH at the equivalence point?

(Be sure to take into account the change

in volume during the titration). What is

a good indicator for this titration?

Jawab:

Va = 50 ml = 0.05 L ; Ma = 0.1 mol/liter

; na = 0.05 x 0.1 = 0.005 mol

Mb = 0.1 mol/liter

Va x Ma = Vb x Mb 0.05 x 0.1 = Vb x 0.1

Vb = 0.05 L

nb = 0.05 x 0.01 = 0.005 mol

Vtotal = 0.05 + 0.05 = 0.1

Ka = 1.8 x 10-4

M : 0.005 mol 0.005 mol

R : 0.005 mol 0.005 mol 0.005

mol -

S : - - 0.005

mol

[OH-] =

=

=

= 1.67 x 10-6

pOH = 6 - log 1.67

pH = 14 - 6 + log 1.67

= 8 + log 1.67

= 8 + 0.223

= 8.223

Because it is on route pH 5.2 to 6.8, the

indicator used is bromine cresol purple.

85. When 25 mL of 0.10 M aqueous ammonia is titrated with 0.10 M

hydrobromic acid, what is the pH at the

equivalence point? What is a good

indicator?

Jawab: D1: V NH3 = 25 mL

M NH3 = 0.10 mol/L

Kb NH3 = 1.8 x 10-5

M HBr = 0.10 mol/L

D2 : what is pH at equivalence point

and a good indicator = ?

D3:

At equivalence point means that the

number of acid moles equal to the

moles of base.

NH3 + HBR NH4Br m 2.5 mmol 2.5 mmol -

r 2.5 mmol 2.5 mmol 2.5 mmol

s - - 2.5 mmol

86. For the titratin of 25.00 mL of 0.1000 M HCl with 0.1000 M NaOH, calculate the

pH of the resulting solution after each of

the following quantities of base has been

added to the original solution (you must

take into account the change in total

volume). Construct a graph showing the

titration curve for this experiment.

a. 0 mL b. 10.00 mL c. 24.90 mL d. 24.99 mL e. 25.00 mL f. 25.01 mL g. 25.10 mL h. 26.00 mL i. 50.00 mL

Jawab:

D1 : [HCl] = 0.1000 M

[NaOH] = 0.1000 M

V HCl = 25 mL

D2 : pH and the titration curve = ...?

Solution :

a. V NaOH = 0 mL HCl + NaOH NaCl +H2O

M 2.5 0

R 0 0 0 0

S 2.5 - 0 0

[HCl] =

= 0.1

[H+] = 1 x 0.1

= 0.1

pH = - log [H+]

= - log 0.1

= 1

b. V NaOH = 10.00 mL HCl + NaOH NaCl + H2O

M 2.5 1

R 1 1 1 1

S 1.5 - 1 1

[HCl] =

= 0.043

[H+] = 1 x 0.043

= 0.043

pH = - log [H+]

= - log 0.043

= 1.37

c. V NaOH = 24.90 mL HCl + NaOH NaCl + H2O

M 2.5 2.49

R 2.49 2.49 2.49 2.49

S 0.01 - 2.49 2.49

[HCl] =

= 2.0 x 10-4

[H+] = 1 x (2.0 x 10

-4)

= 2.0 x 10-4

pH = - log [H+]

= - log 2.0 x 10-4

= 3.698

d. V NaOH = 24.99 mL HCl + NaOH NaCl + H2O

M 2.5 2.499

R 2.499 2.499 2.499 2.499

S 1x10-3

- 2.499 2.499

[HCl] =

= 2.0 x 10-5

[H+] = 1 x (2.0x10

-5)

= 2.0 x 10-5

pH = - log [H+]

= - log 2.0 x 10-5

= 4.7

e. V NaOH = 25.00 mL HCl + NaOH NaCl + H2O

M 2.5 2.5

R 2.5 2.5 2.5 2.5

S - - 2.5 2.5

Titration in equivalent point

pH = 7 (neutral)

f. V NaOH = 25.01 mL HCl + NaOH NaCl + H2O

M 2.5 2.501

R 2.5 2.5 2.5 2.5

S - 1x10-3

2.5 2.5

[NaOH] =

= 1.9996 x 10-5

[OH-] = 1 x (1.9996 x 10

-5)

= 1.9996 x 10-5

pOH = - log [OH-]

= - log 1.9996 x 10-5

= 4.699

pH = 14 - 4.699

= 9.3

g. V NaOH = 25.10 mL HCl + NaOH NaCl + H2O

M 2.5 2.51

R 2.5 2.5 2.5 2.5

S - 0.01 2.5 2.5

[NaOH] =

= 1.996 x 10-4

[OH-] = 1 x (1.996 x 10

-4)

= 1.996 x 10-4

pOH = - log [OH-]

= - log 1.996 x 10-4

= 3.7

pH = 14 - 3.7

= 10.3

h. V NaOH = 26.00 mL HCl + NaOH NaCl + H2O

M 2.5 2.6

R 2.5 2.5 2.5 2.5

S - 0.1 2.5 2.5

[NaOH] =

= 1.96 x 10-3

[OH-] = 1 x (1.96 x 10

-3)

= 1.96 x 10-3

pOH = - log [OH-]

= - log 1.96 x 10-3

= 2.7

pH = 14 - 2.7

= 11.3

i. V NaOH = 50.00 mL HCl + NaOH NaCl + H2O

M 2.5 5

R 2.5 2.5 2.5 2.5

S - 2.5 2.5 2.5

[NaOH] =

= 0.033

[OH-] = 1 x (0.033)

= 0.033

pOH = - log [OH-]

= - log 0.033

= 1.477

pH = 14 - 1.477

= 12.523

87. For the titration of 25.00 mL of 0.1000 M acetic acid with 0.1000 M NaOH,

calculate the pH:

a. Before the addition of any NaOH solution,

b. After 10.00 mL of the base has been added,

c. After half of the HC2H302 has been neutralized, and

d. At the equivalence point. Jawab:

a. Before the addition of any NaOH solution, it means calculate pH of a

weak acid.

b. After 10.00 mL of the base has been added means that we calculate pH of

acid buffer.

CH3COOH+ NaOH CH3COONa + H2O

B 2.5 mmol 1 mmol - -

R 1 mmol 1 mmol 1 mmol 1 mmol

A 1.5 mmol - 1 mmol 1 mmol

c. At a half of the HC2H3O2 means that

the number of acid moles equal to a

half the moles of base. CH3COOH + NaOH CH3COONa+H2O

B 2.5 mmol 1.25 mmol - -

R 1.25 mmol 1.25 mmol 1.25mmol 1.25mmol

A 1.25 mmol - 1.25 mmol 1.25mmol

log1.8 = 4.744

d. At equivalence point means that the number of acid moles equal to the

moles of base. CH3COOH + NaOH CH3COONa +H2O

B 2.5 mmol 2.5 mmol - -

R 2.5 mmol 2.5 mmol 2.5 mmol 2.5mmol

A - - 2.5 mmol 2.5mmol

88. For the titration of 25.00 mL of 0.1000 M ammonia with 0.1000 M HCl,

calculate the pH

a. before the addition of any HCl solution,

b. after 10.00 mL of the acid has been added,

c. after half of the NH3 has been neutralized, and

d. at the equivalence point Jawab:

a. [OH-] =

=

=

= 10 -3

pOH = - log [OH-]

= - log 10-3

= 3

pH = pKw - pOH

= 14 - 3

= 11

So, pH of NH3 before the

addition of any HCl solution

are 11

b. Moles of NH3 = n x M = 25 x 0.1 = 2.5 mmol

Moles of HCl = n x M = 10 x 0.1 = 1

mmol

NH3(aq) + HCl(aq) NH4Cl(aq)

Before: 2,5 1 -

React: 1 1 1

After : 1.5mmol - 1 mmol

[OH-] = Kb x

= 10-5

x

= 1.5 x 10-5

pOH = - log [OH-]

= - log 1.5 x 10-5

= 5 - log 1.5

= 5 - 0.176

= 4.824

pH = pKw - pOH

= 14 - 4.824

= 9.176

So, pH after 10.00 mL of

HCl has been added were

9.176

c. pH after half the NH3 has been neutralized.?

NH3(aq) + HCl(aq) NH4Cl(aq)

Before: 2,5 1.25 -

React: 1.25 1.25 1.25

After : 1.25 mmol - 1.25mmol

[OH-] = Kb x

= 10-5

x

= 10-5

pOH = - log [OH-]

= - log 10-5

= 5

pH = pKw - pOH

= 14 - 5

= 9

So, pH after half of the NH3 has been neutralized were 9

d. pH at the equivalence point.? NH3(aq) + HCl(aq) NH4Cl(aq)

Before: 2,5 2.5 -

React: 2.5 2.5 2.5

After : - - 2.5 mmol

Looking for volume total:

Moles of NH3 = moles of HCl

25 x 0.1 = V x 0.1

V = 25 mL

[H+] =

=

=

= 0,7 x 10 -5

pH = - log [H+]

= - log 0,7 x 10 -5

= 5 - log 0.7

= 5 - (-0.15)

= 5.15

So, pH at the equivalence point are 5.15

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