035 2037 garnis rombel 02 tugas ke-1

Nama : GARNIS ASTRIYANTI

Rombel : 02 Pendidikan Kimia

Nim : 4301412037

1. What is acid and base by Arrhenius, Bronsted lowry and Lewis?

Answer:

Arhenius:

a. Asam didefinisikan sebagai zat-zat yang dapat memberikan ion hidrogen [H+]

atau ion hidronium [H3O+] bila diarutkan daam air.

b. Basa didefinisikan sebagai zat-zat yang dapat menghasikan ion hidroksida [OH-]

bila dilarutkan dalam air.

Bronsted Lowry:

a. ASAM adalah senyawa yang dapat memberikan proton [H+] kepada senyawa lain

(donor proton).

b. BASA adalah senyawa yang dapat menerima proton [H+] kepada senyawa lain

(akseptor proton).

Lewis:

a.Asam adalah suatu molekul/ion yang dapat menerima pasangan elektron.

b. Basa suatu molekul/ion yang dapat memberikan pasangan elektron.

2. Write the chemical equation for the autoionization of water and the equilibrium law for

Kw?

Answer:

Auto ionisasi Air :

a. Air sebagai asam dan basa

H2O(l) + H2O(l) H3O+(aq) + OH

-(aq)

asam1 basa2 asam2 basa1

b. Ketetapan Kesetimbangan untuk kesetimbangan ionisasi air adalah

𝐾𝑐 = 𝐻3𝑂

+ [𝑂𝐻−]

[𝐻2𝑂][𝐻2𝑂]

Oleh karena konsentras sehingga [𝐻2𝑂]2 harganya juga tetap (konstan), maka

hasil Kc dengan [𝐻2𝑂]2 merupakan konstanta yang disebut Kw ( Tetapan

Kesetimbangan Air )

c. Reaksi ini bertanggung jawab terhadap autoionisasi air dengan persamaan

[H+] [OH

-] = Kw

d. Dimana Kw tetapan hasil ionisasi ion untuk air sebesar 1x10-14

pada suhu 25oC

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

a. in terms of [H+] and [OH

-] and

b. in terms of pH ?

Answer:

a. Asam: Semua zat yang jika dilarutkan dalam air menghasilkan ion H+

basa : Semua zat yang jika dilarutkan dalam air menghasilkan ion OH-

Netral: jika dalam air ditambahkan suatu basa maka konsentrasi [OH-] akan

bertambah, dimana harga Kw tetap. Hal ini menyebabkan konsentrasi [H+]

berkurang.

b. Asam : pH < 7

Basa : pH > 7

Netral : pH = 7

4. At the temperature of the human body, 37oC, 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?

Answer :

𝐻2𝑂 [ 𝐻+] + [𝑂𝐻−]

𝐾𝑐 = 𝐻+ [𝑂𝐻−]

[𝐻2𝑂]

[H+] [OH

-] = Kw

[OH-] = [H

+]

[H+] = 𝐾𝑤

= 2,4 𝑥 10−14

= 1,54 x 10−7

pH = - log [H+]

= - log ( 1,54 x 10−7)

= 7 – log 1,54

= 7 – 0,1875

= 6,8124

pOH = - log [OH-]

= - log (1,55 x 107)

= - log 1,55 + ( - log 10-7

)

= 7 – log 1,55

= 6,8124

KW = [OH-] [H

+]

-log Kw = -log [OH-] + {-log [H

+]}

p = - log

pKw = pOH + pH

pKw = -log [Kw]

= -log (2,4 x 10-14

)

pKw = 14 – log 2,4

pKw = 13,62

pKw = pOH + pH

13,62 = 6,81 + 6,81

Maka air pada suhu tersebut bersifat netral.

5. Deuterium oxide, D2O, ionizes like water. At 20°C 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

20°C. Calculate also the pD and the pDO.

Answer :

D2O D+

+ OD-

𝐾𝑐 = 𝐷+ 𝑂𝐷−

𝐷2𝑂

KW = [OD-] [D

+]

[OD-] = [D

+] = 𝐾𝑤

= 8,9 𝑥 10−16

= 2,98 x 10-8

pD = - log [D+]

= - log (2,98 x 10-8

)

= - log 2,98 + ( - log 10-8

)

= 8 – log 2,98

= 7,53

pOD = - log [OD-]

= - log (2,98 x 10-8

)

= - log 2,98 + ( - log 10-8

)

= 8 – log 2,98

= 7,53

6. 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

Answer :

a) Kw = [H+] [OH

-]

10−14 = [H+] (24 × 10−4)

[H+] =

10−14

24 ×10−4

[H+] = 4,167 × 10−12

b) Kw = [H+] [OH

-]

10−14 = [H+] (1.4 x 10−5)

[H+] =

10−14

1.4 x 10−5

[H+] = 7,14 × 10−10

c) Kw = [H+] [OH

-]

10−14 = [H+] (5.6 x 10−9)

[H+] =

10−14

5.6 x 10−9

[H+] =0,1785 × 10−5

d) Kw = [H+] [OH

-]

10−14 = [H+] (4.2 x 10−13)

[H+] =

10−14

4.2 x 10−13

[H+] =0,0238

7. 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

Answer :

a. 3.5 x 10 -8

M

Kw = [H+] [OH

-]

10-14

= (3,5 x 10-8

) [OH-]

[OH-] = 2,85 x 10

-7 M

b. 0.0065 M

Kw = [H+] . [OH

-]

10-14

= (6,5 x 10-3

) [OH-]

[OH-] = 1,54 x 10

-12 M

c. 2.5 x 10 -13

M

Kw = [H+] [OH

-]

10-14

= (2,5 x 10-13

) [OH-]

[OH-] = 4 x 10

-2 M

d. 7.5 x 10 -5

M

Kw = [H+] [OH

-]

10-14

= (7,5 x 10-5

) [OH-]

[OH-] = 1,3 x 10

-10 M

8. 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?

Answer :

[H+] = 1.9 x 10

-5 mol L

-

pH = - log [H+]

= - log (1.9 x 10-5

)

= - log 1,9 + (- log 10-5

)

= - log 1,9 + 5

= 5 – log 1,9

= 4,7213

9. A soft drink was put on the market with [H+] = 1,4 x 10

-5 mol L-1

. What it’s pH ?

Answer : [H+] = 1,45 x 10

-5 mol L

-

pH = - log [H+]

= - log (1,45 x 10-5

)

= - log 1,45 + (- log 10-5

)

= - log 1,45 + 5

= 5 – log 1,45

= 5 – 0,14612

= 4,85388

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

Answer : Soal no.5

a. POH = - log 2.4 x 10-2

POH = 2-log 2.4

POH = 1.62

PH = 14 – 1.62

PH = 12.38

b. POH = 5 –log 1.4

POH = 4.85

PH = 14- 4.85

= 9.15

c. POH = - log 5.6 x 10-9

POH = 9 – log 5.6

= 8.25

PH = 14 – 8.25

= 5.75

d. POH = 13 – log 4.2

POH = 12.38

PH = 14 – 12.38

= 1.62

Soal no.6

a. (3.5 x 10-8

)

PH = - log 3.5 x 10-8

PH = 8 – log 3.5

= 7.45

b. (0.0065)

PH = 6.5 x 10-2

= 2 – log 6.5

= 1.18

c. 2.5 x 10-13

PH = - log 2.5 x 10 -13

PH = 13 – log 2.5

PH = 12.6

d. 7.5 x 10 -5

M

PH = - log 7.5 x 10 -5

PH = 5 – log 7.5

PH = 4.125

11. 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

Answer : A. pH = 3,14

[H+] = 7,2 . 10

-4

pOH = 10,86

[OH-] = 1,38 . 10

-11

B. pH = 2,78

[H+] = 1,65 . 10

-3

pOH = 11,22

[OH-] = 6,025 . 10

-12

C.

pH = 9,25

[H+] = 5,62 . 10

-10

pOH = 4,75

[OH-] = 1,77. 10

-5

D. pH = 13,24

[H+] = 5,75 . 10

-14

pOH = 0,76

[OH-] = 1,73 . 10

-1

E.

pH = 5,70

[H+] = 1,99 . 10

-6

pOH = 8,3

[OH-] = 5,01 . 10

-9

12. 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

Answer: A. pOH = 8,26

[OH-] = 5,49. 10

-9

pH = 5,74

[H+] = 1,81 . 10

-16

B. pOH = 10,25

[OH-] = 5,62. 10

-11

pH = 3,75

[H+] = 1,77 . 10

-4

C..

pOH = 4,65

[OH-] = 2,23. 10

-5

pH = 9,35

[H+] = 4,46 . 10

-10

D. pOH = 6,18

[OH-] = 6,60. 10

-7

pH = 7,82

[H+] = 1,51 . 10

-8

E pOH = 9,70

[OH-] = 1,99. 10

-10

pH = 4,3

[H+] = 5,01. 10

-5

13. What is the pH of 0.010 M HCl ?

Answer:

HCl = H+

+ Cl-

[H+] = valensi . M

= 1 . 0,010 M

= 0,010 M

pH = - log 0,010

= 2 – log 1

= 2 - 0

= 2

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

Answer:

HNO3 = H+

+ NO3 -

[H+] = valensi . M

= 1 . 0,0050

= 0,0050

pH = - log 0,0050

= 3 – log 5

= 3 – 0,698

= 2,303

15. 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?

Answer:

M = 𝑔𝑟

𝑀𝑟x

1000

𝑝

=6,0

40x

1000

𝑝

= 0,15

NaOH = Na+

+ OH-

OH- = 1 . 0,15

= 0,15

pOH = 1 – log 1,5

pH = 14 –pOH

= 14 – ( 1 – log 1,5 )

= 13 + log 1,5

= 13 + 0,176

= 13,176

16. 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?

Answer:

M = 𝑔𝑟

𝑀𝑟 x

1000

𝑉

= 0,837

171 x

1000

100

= 0.0489

= 4,89 x 10-2

[OH-] = b x M

= 2 x 4,89 x 10-2

= 9,78 x 10-2

pOH = - log [OH-]

= - log 9,78 x 10-2

= 2 - log 9,78

= 1, 009

pH = 14 – pOH

= 14 – 1,009

= 12,991

17. 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?

Answer:

pH = 11,60

pOH = 2,4

-log [OH-] = 2,4

= 3 – log 3,98

[OH-]

= 3,98 x 10

-3

Ca(OH)2 Ca

2+ + 2OH

-

1,99 x 10-3

1,99 x 10-3

3,98 x 10-3

M Ca(OH)2 = 1,99 x 10-3

18. A solution of HCl has a pH of 2.50. How many grams of HCl are there in 250 mL of this

solution.

Answer :

pH HCl = 2,50

Vlar = 250 mL

Massa HCl ?

pH = 2,5

[H+] = 3,16 . 10

-3

M = n

v

3,16 . 10 -3

= n

0,25

n = 7,9 . 10 -4

mol

massa HCl = 7,9 . 10 -4

mol x 36,5

= 28,8 . 10 -3

mg

19. 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. HAsO42-

d. (CH3)3NH+

Answer: 𝑎) HNO2 aq + H2O l NO2−

aq + H3O+ aq

𝑏) H3PO4 aq + H2O l H2PO4−

aq + H3O+ aq

𝑐) HAsO4(aq )2− + H2O l AsO4

3− aq + H3O+

aq

𝑑) CH3 3NH(aq )+ + H2O l (CH3)3N(aq ) + H3O+

aq

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

Answer :

a. HNO2 H+

+ NO2-

𝐾𝑎 = 𝐻+ [𝑁𝑂2

−]

[𝐻𝑁𝑂2]

b. H3PO4 H+ + H2PO4

2-

𝐾𝑎 = 𝐻+ [𝐻2𝑃𝑂4

2−]

[𝐻3𝑃𝑂4]

c. HAsO42-

H+ + AsO4

3+

𝐾𝑎 = 𝐻+ [𝐴𝑠𝑂4

3+]

[𝐻𝐴𝑠𝑂42−]

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

𝐾𝑎 = 𝐻+ [ 𝐶𝐻3 3𝑁]

[ 𝐶𝐻3 3]𝑁𝐻+

21. Write the chemical equation for the ionization of each of following weak bases in water.

a. (CH3)3N

b. AsO43-

c. NO2-

d. (CH3)2N2H2

Answer :

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

+

b. AsO43-

+ H2O OH- + HAsO4

2-

c. NO2- + H2O OH

- + HNO2

d. (CH3)2N2H2 + H2O OH- + (CH3)3N2H3

+

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

Answer :

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

+

𝐾𝑏 = 𝑂𝐻− [ 𝐶𝐻3 3𝑁𝐻+]

[ 𝐶𝐻3 3]𝑁

b. AsO43-

+ H2O OH- + HAsO4

2-

𝐾𝑏 = 𝑂𝐻− [𝐻𝐴𝑠𝑂4

2−]

[𝐴𝑠𝑂43−]

c. NO2- + H2O OH

- + HNO2

𝐾𝑏 = 𝑂𝐻− [𝐻𝑁𝑂2]

[𝑁𝑂2−]

d. (CH3)2N2H2 + H2O OH- + (CH3)3N2H3

+

𝐾𝑏 = 𝑂𝐻− [ 𝐶𝐻3 3𝑁2𝐻3

+]

[ 𝐶𝐻3 3]𝑁2𝐻2

23. 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 !

Answer :

C6H5CO2H(aq ) + H2O(l) → C6H5CO2(aq )− + H3O(aq )

+

𝐾𝑎 = [ C6H5CO2

− ][ H3O+ ]

[ C6H5CO2H ]

24. 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.

Answer :

1. C6H5CO2- + H2O C6H5CO2H + OH

-

𝐾𝑏 = 𝑂𝐻− [𝐶6𝐻5𝐶𝑂2𝐻 𝐶6𝐻5𝐶𝑂2

−]

[𝐶6𝐻5𝐶𝑂2−]

25. The pKa of HCN is 9.21 and that of HF is 3.17. Which is the strong Bronsted base CN−

or F−?

Answer :

pKa HCN = 9,21

-log Ka = 9,21

= 10 – 0,79

= 10 – log 6,16

Ka = 6,16 x 10-10

pKa HF = 3,17

-log Ka = 3,17

= 4 – 0,83

= 4 – log 6,76

Ka = 6,76x 10-4

Semakin kuat asam, maka semakin kecil basa konjugasinya, maka yang lebih kuat basa

konjugasinya adalah HCN. Asam HCN memiliki Ka lebih rendah dari pada HF sehingga

asam HCN lebih lemah dan basa konjugasinya yaitu CN- memiliki sifat kebasaan yang

lebih kuat dari pada F-.

26. The Ka for HF is 6.8 x 10x. what is the Kb for F

-?

Answer :

Kb = Kw

Ka

Kb = 10−14

6,8x10x

= 1,47 x 10-15-x

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

. What is Ka for Barbituric acid ?

Answer :

Ion C4HO memiliki Kb = 1,0 x 10-10

. Nilai Ka untuk asam barbituratnya adalah

Kw = Ka x Kb

10-14

= Ka x 1,0 x 10-10

Ka = 10−14

1,0 𝑥 10−10

Ka = 1,0 x 10-4

28. Hydrogen peroxide, H2O2 is a week acid with Ka = 1.8 x 10-12

. What the value of Kb for

the HO2 ion?

Answer :

Kb = Kw

Ka

Kb = 10−14

1,8x10−12

= 5,55 x 10-3

29. Methylamine, CH3NH2 resambles ammonia in odor and basicity. Its Kb is 4.4 x 10-4

.

Calculate the Ka of its conjugate acid!

Answer :

Ka = Kw

Kb

Ka = 10−14

4,4 x10−4

= 2,27 x 10-11

30. Lactic acid, HC3H5O3, is responsible for the sour taste of sour milk. At 25oC its Ka = 1.4

x 10-4

. What is the Kb of its conjugate base, tha lactate ion, C3H5O3- ?

Answer :

Kw = Ka x Kb

10-14

= 1,4 x 10-4

x Kb

Kb = 10−14

1,4 𝑥 10−4

Kb = 0,714 x 10-10

31. 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?

Answer :

a. HIO3 IO3- + H

+

HIO3 + H2O IO3- + H3O

+

Ka = [IO 3

−]+ [H3O+]

[HIO 3]

IO3- + H2O HIO3+ OH

-

Kb = [HIO 3]+ [OH −]

[IO 3−]

b. pKa = -log Ka

0,77 = -log Ka

Ka = 1,698 . 10-1

untuk HIO3

Ka ion asetat = 1,75 x 10-5

Ka HIO3 > Ka ion asetat maka HIO3 basa konjugasinya merupakan basa lemah.

Jika harga Ka semakin besar maka semakin kuat asam dan semakin lemah basa

konjugasinya. Karena HIO3 yang bersifat asam kuat di dalam air. Di dalam air

praktis tidak terdapat lagi molekul HIO3, semuanya terion membentuk IO3- dimana

ion ini merupakan basa lemah.

32. 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!

Answer :

[H+] = 𝐾𝑎 𝑥 𝑀

3,8 x 10-2

= 𝐾𝑎 𝑥 0,1

(3,8 x 10-2

)2 = Ka x 0,1

14,44 x 10-4

= Ka x 0,1

Ka = 14,44 x 10-3

pKa = - log 14,44 x 10-3

pKa = 3 – log 14,44

pKa = 3 – 1,16

pKa = 1,84

33. 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.

Answer :

∝ = 𝐾𝑎

𝑀

(0, 11)2 =

𝐾𝑎

0,1

0,0121 x 0,1 = Ka

Ka = 1,21 x 10-3

pKa = -log Ka

= -log 1,21 x 10-3

= 2,9172

34. 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.

Answer :

𝑝𝑂𝐻 = 14 − 11,86

= 2,14

−𝑙𝑜𝑔 𝑂𝐻− = 3 − 0,86

= 3 − log 7,24

𝑂𝐻− = 7,24 . 10−3

𝑂𝐻− = 𝐾𝑏 . 𝑀

𝑂𝐻− 2 = 𝐾𝑏 . 𝑀

7,24 . 10−3 2 = 𝐾𝑏 . 0.1

5,24. 10−4 = 𝐾𝑏

𝑝𝐾𝑏 = −𝑙𝑜𝑔𝐾𝑏

= 3,28

35. 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?

Answer :

𝑝𝑂𝐻 = 14 − 10,12

= 3,88

−𝑙𝑜𝑔 𝑂𝐻− = 4 − 0,12

= 4 − log 1,138

𝑂𝐻− = 1,138 . 10−4


𝑂𝐻− 2 = 𝐾𝑏 . 𝑀

1,138 . 10−4 2 = 𝐾𝑏 . 0,15

1,158 . 10−7 = 𝐾𝑏

𝑝𝐾𝑏 = − log 𝐾𝑏

= 6,936

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

in 0.15 M HONH2.

Answer : 𝛼 = 𝐾𝑏

𝑀 . 100%

= 1,158 . 10−7

0,15. 100%

= 0,0878%

37. What is the pH of 0.125 M pyruvic acid ? It’s Ka is 3.2 x 10-3

Answer :

𝐻+ = 𝐾𝑎 . 𝑀

= 3,2 . 10−3 . 0,125

= 0,02

𝑝𝐻 = − log 2 . 10−2

= 1,69

38. What is pH of 0.15 M HN3 ? for HN3, Ka = 1.8 x 10-5

Answer :

𝐻+ = 𝐾𝑎 . 𝑀

= 1,8 . 10−5 . 0,15

= 1,643 . 10−3

𝑝𝐻 = − log 1,643 . 10−3

= 2,784

39. What is the pH of a 1.0 M solution of hydrogen peroxide, H2O2? For this solute, Ka =

1.8 x 10-2

Answer :

𝐻+ = 𝐾𝑎 . 𝑀

= 1,8. 10−2 . 1

= 1,3416 . 10−1

𝑝𝐻 = − log 1,3416 . 10−1

= 0,8723

40. 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

Answer :

𝛼 = 𝐾𝑏

𝑀. 100%

= 1,3. 10−10

0,050. 100%

= 5, 099 . 10−3%

41. 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)

Answer :

𝑝𝐾𝑏 = 5,79

= 6 − 0,21

𝐾𝑏 = 6 − log 1,62

𝐾𝑏 = 1,62 . 10−6


𝑂𝐻− = 1,8 . 10−4

𝑝𝑂𝐻 = 4 − 𝑙𝑜𝑔1,8

= 3,745

𝑝𝐻 = 10,255

42. Deuteroammonia, ND3, is a weak base with a pKb of 4.96 at 25oC. What is the pH of a

0.20 M solution of this compound?

Answer :

pKb = 4,96

-log Kb = 5 – 0,04

-log Kb = 5 – log 1,09

Kb = 1,09 x 10-5

[OH-] = 𝐾𝑏 𝑥 𝑀𝑏

[OH-] = 1,09 𝑥 10−5𝑥 0,2

[OH-] = 2,18 𝑥 10−6

[OH-] = 1,47 x 10

-3

pOH = - log [OH-]

pOH = - log (1,47 x 10-3

)

pOH = 3 – log 1,47

pH = 11 + log 1,47

43. A solution of acetic acid has a pH of 2.54. What is the concentration of acetic acid in this

solution ?

Answer :

pH = 2,54

=3 - 0,46

= 3 – log 2,88

[H+] = 2,88 x 10

-3

[H+] = 𝐾𝑎 × 𝑀𝑎

[H+]

2 = Ka x Ma

8,29 x 10-6

= 1,8 x 10-5

x Ma

Ma = 𝟖,𝟐𝟗 𝐱 𝟏𝟎−𝟔

𝟏,𝟖 𝐱 𝟏𝟎−𝟓

Ma = 0,46 M

44. 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

Answer :

Aspirin merupakan asam asetilalisiklik

Asam monoprotik merupakan asam yang hanya dapat melepaskan 1 ion H+ setiap

molekul dalam setiap larutan.

Kb = 𝑲𝒘

𝑲𝒂 Ka > Kb , maka larutan ini bersifat ASAM

= 𝟏𝟎−𝟏𝟒

𝟑,𝟐𝟕 𝒙 𝟏𝟎−𝟒

= 3,058 x 10-10

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

, is 1.9x10-10

. Is a solution of K2C2O4 acidic,

basic, or neutral? Explain.

Answer :

KbC2O42-

=1,9x10-10

Kw = Kb . Ka

10−14 = 1,9 . 10−10 . Ka

Ka = 5,26 . 10−5

Ka > Kb

5,26 . 10−5 > 1,9 . 10−10 (ASAM)

46. 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.

Answer :

Asam : (NH 4 ) 2 SO 4

Basa : KF, KCN, KC2 H3O2

Netral : NaI, KBr, CsNO3

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

Answer :

Larutan AlCl3 dapat memerahkan lakmus biru, karena Larutan AlCl3 bersifat asam. Di

dalam air garam ini akan terhidrolisis sebagian (kation dari basa lemah terhidrolisis,

sedangkan anion dari asam kuat tidak). Dalam air, AlCl3 terionisasi sempurna

membentuk ion Cl- dan Al

3+

AlCl3 Al3+

+ 3Cl-

Reaksi Hidrolisisnya adalah

Al3+

(aq) + 3H2O(aq) Al(OH)3(aq) + 3H+(aq)

Cl-(aq) + H2O(l) (tidak ada reaksi)

Adanya sisa H+

pada reaksi hidrolisis diatas menyebabkan larutan bersifat asam.

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

Answer :

Pada sistem periodik unsur, semakin kebawah tataletak suatu unsur, maka tingkat ke-

basaan suatun unsur semakin kuat. Sementara semakin keatas letak unsur maka tingkat

keasaman semakin kuat. Be terletak diatas Ca sehingga Be bersifat lebih asam daripada

Ca.

49. 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.

Answer :

NH4NO3 NH4+ + NO3

-

NH4+ tergolong basa lemah, sedangkan NO3

- tergolong asam lemah, maka keduanya

mengalami hidrolisis. Hidrolisis NH4+ menghasilkan ion H

+, sedangkan hidrolisis

NO3- menghasilkan OH

-, maka akan netral. Sehingga tidak berpengaruh terhadap

keasaman tanah.

50. Calculate the pH of 0.20 M NaCN.

Answer :

M = 0,2 M

[OH-]

= 𝑘𝑤

𝑘𝑎 . 𝑀

= 10−14

1,8 𝑥 10−5 . 2.10−1

= 2. 10−10

[OH-] = 1,4. 10

-5

pOH = - log [OH-]

pOH = - log ( 1,4 x 10-5

)

pOH = 5 - log 1,4

pH = 9 + log 1,4

51. Calculate the pH of 0,04 M KNO2 ?

Answer :

M = 0,04 M

[OH-]

= 𝑘𝑤

𝑘𝑏 . 𝑀

= 10−14

1,8 𝑥 10−5 . 4 𝑥 10−2

= 2,22. 10−11

[OH-] = 4,71 x 10

-6

pOH = - log [OH-]

pOH = - log ( 4,71 x 10-6

)

pOH = 6 - log 4,71

pH = 8 + log 4,71

52. Calculate the pH of 0.15 M CH3NH3Cl. For CH3NH2, Kb = 4.4 x 10-4

Answer :

[H+]

=

𝑘𝑤

𝑘𝑏 . 𝑀

= 10−14

4,4.10−4 . 1,5.10−1

= 3,4 10 −10

= 1,8 x 10-5

pH = - log [H+]

pH = - log ( 1.8 x 10-5

)

pH = 5 – log 1,8

53. 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?

Answer :

pH = 4,28

M = 0,15 M

pH = 4,28

-log [H+] = 5 – 0,72

-log [H+] = 5 – log 5,25

[H+]

= 5,25 x 10-5

[H+]

= 𝑘𝑤

𝑘𝑏 . 𝐺

5,25 x 10-5

= 10−14

𝑘𝑏 𝑥 1,5 𝑥 10−1

(5,25 x 10-5

)2 =

10−14

𝐾𝑏 x 1,5 x 10

-1

2,75 x 10-9

x Kb = 1,5.10−15

Kb = 5,45 10 -7

54. 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 !

Answer :

NH4Br NH3 + HBr

pH = -log [H+]

5,16 = - log [H+]

(6-0,84) = - log [H+]

0,84 . 10-6

= [H+]

Kb NH3 = 1,8 . 10 -5

[H+]

= 𝑘𝑤

𝑘𝑏 . [𝐺]

0,84 . 10-6

= 10−14

1,8 .10 −5 𝑥 [𝐺]

Masing-masing ruas dikuadratkan

0,7056 . 10-12

= 𝟏𝟎−𝟏𝟒

𝟏,𝟖 .𝟏𝟎−𝟓 x [G]

[G] = 1,27008 .10-3

n = M.V

= 1,27008 .10-3

mol/L . 1 L

= 1,27008 .10-3

mol

n = 𝒎

𝑴𝑴

m = MM . n

= 98 . 1,27008 .10-3

= 0,1244678 gram

55. 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

Answer :

Pka BH+ = 5 maka bereaksi dengan HY membentuk BHY (basa). Karena hasil yang dibentuk

bersifat basa maka HY mempunya pKa lebih besar daripada pKa BH+.

[BH+] [Y

-] = 10−14

10−5 [Y-] = 10−14

[Y-] = 10−9

pKa = 9

Jadi pKa lebih dari 5.

56. 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+?

Answer :

pKb = 6.13

Kb = 7 – log 7.41

Kb = 7.41 x 10-7

Kw = Ka x Kb

10-14

= Ka x 7.41 x 10-7

Ka = 𝟏𝟎−𝟏𝟒

𝟕.𝟒𝟏 𝒙 𝟏𝟎−𝟕

Ka = 0.135 x 10-7

[H+] = 𝐾𝑎 𝑥 𝑀

[H+] = 0.135 x 10−7 𝑥 0.2

[H+] = 5.2 x 10-5

pH = - log [H+]

= - log 5.2 x 10-5

pH = 5 - log 5.2

pH = 4.2

57. 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-𝑄𝑢+. What is the calculate pHof a 0,15 M

solution of H-𝑄𝑢+ ? Its pKa is 8,52 at 25 0C.

Answer :

pKa = 8.52

Ka = 9 – log 3.01

Ka = 3.01 x 10-9

[H+] = 𝐾𝑎 𝑥 𝑀

[H+] = 3.01 x 10−9 𝑥 0.15

[H+] = 4.5 x 10-5

pH = - log [H+]

= - log 4.5 x 10-5

pH = 5 - log 4.5

pH = 4.3

58. 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?

Answer :

Konsentrasi asalm basa berkaitan dengan fase zat tersebut. Pada Kp hanya berlaku fase

gas sedangkan untuk Kc berlaku fase gas dan aquous.

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

solution ?

Answer :

HF(aq) H+

(aq) + F-(aq)

α = 𝐾𝑎

𝑀

α = 7.2 𝑥 10−4

0.15

α = 0.069

α = 6.9 %

Ka = 7.2 x 10-4

[H+] = 𝐾𝑎 𝑥 𝑀

[H+] = 7.2 x 10−4 𝑥 0.15

[H+] = 0.01

pH = - log [H+]

pH = - log 10-2

pH = 2

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

solution?

Answer :

α = 𝐾𝑎

𝑀

α = 1.8 𝑥 10−5

0.001

α = 0.018

α = 1.8 %

Ka = 1.8 𝑥 10−5

[H+] = 𝐾𝑎 𝑥 𝑀

[H+] = 1.8 𝑥 10−5 𝑥 0.001

[H+] = 1.3 x 10-4

pH = - log [H+]

= - log 1.3 x 10-4

pH = 4 - log 1.3

61. What is the pH of a 1.0 x 10-7

M solution of HCl ?

Answer :

M HCl = 1,0 x 10-7

HCl -> H+ + Cl

-

Dalam hal ini berlaku ketentuan :

(H+) (OH

-) = Kw

(Cl-) = HCl

(H+) =(OH

-) + (Cl

-), prinsip penetralan muatan sehingga berlaku ;

(H+) = (OH

-) + (HCl)

(H+) = Kw / (H

+) + (HCl)

(H+)2 = Kw + (HCl)(H+)

(H+)2 – (HCl) (H

+) – Kw = 0

(H+)2 – 10

-7 (H

+) – 10

-14 = 0

(H+) = 1.62 x 10

-7 M

pH =- log 1.62 x10-7

= 6.79

62. 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?

Answer :

a. HSO4- H

+ + SO4

- Ka = 1.0x10

-2

𝐾𝑎 = 𝐻+ (𝑆𝑂4

−)

(𝐻𝑆𝑂4−)

b. b. Dengan menggunakan persamaan kuadrat, dapat memecahkan persamaan

berikut:

x2+10

-2x-10

-4 = 0

Dengan menggunakan rumus kuadrat

x = −𝑏 ± 𝑏2−4𝑎𝑐

2𝑎

= −10−2 ± 10−22−4.10−4

2

=−10−2±10−2

2

x1 = 0 M atau x2 = - 0,01 M

jawaban secara fisik tidak mungkin, sebab konsentrasi ion yang dihasilkan

sebagai akibat ionisasi tidak mungkin negative. Jadi yang mungkin adalah 0

M.

c. HSO4- H

+ + SO4

-

M 0,01 - -

R –x +x +x

S (0,001-x) x x

Ka = H+ [SO 4−]

[HSO 4−] = 1.0x10

-2

𝑥2

0,001−x= 1.0x10

-2

Dengan menggunakan pendekatan 0,001-x = 0,001

𝑥2

0,001−x≈

𝑥2

0,001= 1.0x10

-2

x2

= 1,0 x 10-5

x = 3,162 x 10

-3 M

d. Pemeriksaan terhadap pendekatan : 3,162 x 10−3

1,0 𝑥 10−2 x100% = 31,622%

ini menunjukkan bahwa x lebih besar dari 5 persen dari konsentrasi awal

sehingga pendekatan ini tidak sah.

63. 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?

Answer :

pKa = 4,92

Ka = 1,202 x 10-5

[H+] = Ka x M

= 1,202 x 10-5

x 3 x 10-2

= 6,004 x 10-4

pH = - log [H+]

= - log 6,004 x 10-4

= 4 – log 6,004 = 3,222

64. 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?

Answer :

pKa = 4,01

Ka = 9,77 x 10-5

[H+] = Ka x M

= 9,77 x 10-5

x 5 x 10-2

= 2,21 x 10-3

pH = - log [H+]

= - log 2,21 x 10-3

= 3 – log 2,21 = 2,656

65. 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

Answer :

a) H2CO3 + OH- + Na

+ Na

+ + HCO3

- + H2O

HCO3- + H

+ H2CO3

b) HPO42-

+ H+ + 2Na

+ 2Na

+ + H2PO4

-

H2PO4- + OH

- + Na

+ HPO4

2- + Na

+ + H2O

c) NH3 + H+ + Cl

- NH4

+ + Cl

-

NH4

+ + OH

- + Cl

- NH3 + Cl

- + H2O

66. 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.

Answer :

buffer 1

Basa : NH3 1M

Asam konjugasi : NH4Cl

𝑂𝐻− = 𝐾𝑏 𝑏𝑎𝑠𝑎

𝑔𝑎𝑟𝑎𝑚

= 1,8 × 10−5 1𝑀

0,1𝑀

= 1,8 × 10−4

𝑝𝑂𝐻 = 4 − 𝑙𝑜𝑔1,8

= 4 − 0,2553

= 3,7447

𝑝𝐻 = 14 − 3,7447

= 10,2553

𝑏𝑢𝑓𝑓𝑒𝑟 2

Basa : NH3 0,1 M

Asam konjugasi : NH4Cl 1 M

𝑂𝐻− = 𝐾𝑏 𝑎𝑠𝑎𝑚


= 1,8 × 10−50,1 𝑀

1 𝑀

= 1,8 × 10−6

𝑝𝑂𝐻 = 6 − 𝑙𝑜𝑔1,8

= 5,7447

𝑝𝐻 = 14 − 5,7447

= 8,2553

Buffer yang dapat lebih mempertahankan pH jika ditambah dengan asam kuat adalah, sesuai

yaitu buffer 1.

67. 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

Answer :

dimisalkan volume larutan 1 L.

HC2H3O2 ↔ H+

+ C2H3O2−

(buffer asam)

[H+] = Ka ×

[CH 3COOH ]

[CH 3COO −]

= 1,8 × 10−5

× 0,15 𝑀

0,25 𝑀

= 1,8 × 10−5

× 0,6

= 1,08 × 10−5

pH = −log [H+]

= −log 1,08 × 10−5

= 5 − log 1,08

= 5 – 0,0334

= 4,9666

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

poper chemical equation and equilibrium law )

Answer :

69. 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.

Answer :

pH buffer ke I : 10.2553

Mol basa mula-mula = V x M

= 1 L x 1 M

= 1 mol

Reaksi :

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

m: 1 mol 0.05 mol 0 mol

r : -0.05mol -0.05 mol +0.05 mol

s : 0.95mol 0 mol 0.05mol

Garam = 0.05 mol + 0.1 mol

= 0.15 mol

Basa = 0.95 mol

[OH-] = 1.8 x 10

-5 x

𝟎.𝟗𝟓

𝟎.𝟏𝟓 mol

= 1.14 x 10-4

pOH = 4-log 1.14

= 3.9431

pH = 14 3.9431

= 10.0569

70. 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.

Answer :

pOH = –log Kb x 𝒃𝒂𝒔𝒂

𝒈𝒂𝒓𝒂𝒎

= - log 1.8 x 10-5

x 𝟏 𝒎𝒐𝒍𝒂𝒓

𝟎.𝟎𝟏 𝒎𝒐𝒍𝒂𝒓

= 3.7

pH = 14 – pOH

= 14 – 3.7

= 10.25

NH4Cl + NaOH NH4OH + NaCl

m : 50 mmol 5 mmol

r : 5 mmol 5 mmol 5 mmol

s : 45 mmol - 5 mmol

pOH = –log Kb x 𝑏𝑎𝑠𝑎


= - log 1.8 x 10-5

x 0.005 𝑚𝑜𝑙

0.045 𝑚𝑜𝑙

= 5.69

pH = 14 – pOH

= 14 – 5.69

= 8.31

Perubahan pH = 10.25 – 8.31

= 1.94

Buffer 2



= - log 1.8 x 10-5

x 0.1 𝑚𝑜𝑙

1 𝑚𝑜𝑙

= 5.74

pH = 14 – pOH

= 14 – 5.74

= 8.25

NH4Cl + NaOH NH4OH + NaCl

m : 500 mmol 5 mmol

r : 5 mmol 5 mmol 5 mmol

s : 495 mmol - 5 mmol



= - log 1.8 x 10-5

x 0.005 𝑚𝑜𝑙

0.495 𝑚𝑜𝑙

= 6.74

pH = 14 – pOH

= 14 – 8.74

= 7.26

Perubahan pH = 8.25 – 7.26

= 0.99

71. 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+

Answer :

𝑎. 𝑁𝐻3 + 𝐻2𝑂 ⇌ 𝑁𝐻4+ + 𝑂𝐻−

𝐾𝑏 = 𝑁𝐻4

+ 𝑂𝐻−

𝑁𝐻3

1.8𝑥10−5 =0.14 𝑂𝐻−

0.25

𝑂𝐻− = 3.214𝑥10−5

𝑝𝑂𝐻 = − log 𝑂𝐻−

= −𝑙𝑜𝑔3.214 ∗ 10−5

= 5 − log 3.214

𝑝𝐻 = 9 + log 3.214

= 9.507

b. 𝐾𝑎 =𝐾𝑤

𝐾𝑏

=1𝑥10−14

1.8𝑥10−5

= 5.5𝑥10−10

𝐻+ = 𝐾𝑎𝑥𝑎

𝑔

= 5.5𝑥10−10𝑥0.14

0.25

= 3.08𝑥10−10

𝑝𝐻 = − log 𝐻+

= − log 3.08 ∗ 10−10

= 10 − log 3.08

= 9.51

72. 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?

Answer :

𝑁𝐻3 + 𝐻+ → 𝑁𝐻4+

𝑀 ∶ 0.25 0.02 0.14

𝑅 ∶ 0.02 0.02 0.02

𝑆 ∶ 0.23 − 0.16

𝑂𝐻− = 𝐾𝑏

𝑏

𝑔

= 1.8𝑥10−50.23

0.16

= 2.5875𝑥10−5

𝑝𝑂𝐻 = 5 − log 2.5875

= 4.58

𝑝𝐻 = 14 − 4.58

= 9.42

73. 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?

Answer :

𝑁𝐻4+ + 𝑂𝐻− → 𝑁𝐻3 + 𝐻2𝑂

𝑀 ∶ 0.028 0.0075 0.05 −

𝑅 ∶ 0.0075 0.0075 0.0075 −

𝑆 ∶ 0.0205 − 0.0575 −

𝑂𝐻− = 𝐾𝑏

𝑎

𝑔

= 1.8𝑥10−50.0575

0.0205

= 2.804𝑥10−5

𝑝𝑂𝐻 = 5 − log 2.804

= 4.55

𝑝𝐻 = 14 − 4.55

= 9.45

74. 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?

Answer :

Pka = - log Ka

4,74 = - log Ka

5 – 0,26 = - log Ka

Ka = 5 – log 1,82

Ka = 1,82 x 10-5

Larutan Buffer memiliki pH = 5, maka [H+] = 10

-5

[H+] = Ka x

𝑛 .𝑎𝑠𝑎𝑚

𝑛 .𝑏𝑎𝑠𝑎 𝑘𝑜𝑛𝑗𝑢𝑔𝑎𝑠𝑖𝑛𝑦𝑎

10-5

= 1,82 x 10-5

x 0,15

𝑛 .𝑏𝑎𝑠𝑎 𝑘𝑜𝑛𝑗𝑢𝑔𝑎𝑠𝑖

n. basa konjugasi = 0,273 mol

n.basa konjugasi = n. Natrium asetat = 0,273 mol.

Massa natrium asetat = 0,273 mol x 82 gram/mol = 22,386 gram

75. 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 ?

Answer :

Larutan asam formiat (Pka = 3,74)

Pka = - log Ka

3,74 = - log Ka

4 – 0,26 = - log Ka

Ka = 4 – log 1,82

Ka = 1,82 x 10-4

Larutan Buffer memiliki pH = 3,8, maka [H+] = 1,6 x 10

-4

[H+] = Ka x

𝑛 .𝑎𝑠𝑎𝑚

𝑛 .𝑏𝑎𝑠𝑎 𝑘𝑜𝑛𝑗𝑢𝑔𝑎𝑠𝑖𝑛𝑦𝑎

1,6 x 10-4

= 1,82 x 10-4

x 0,12

𝑛 .𝑏𝑎𝑠𝑎 𝑘𝑜𝑛𝑗𝑢𝑔𝑎𝑠𝑖

n. basa konjugasi = 1,365 mol

n.basa konjugasi = n. Natrium formiat = 1,365 mol.

Massa natrium formiat = 1,365 mol x 68 gram/mol = 92,82 gram

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

Answer :

Perbandingan mol NH4Cl dan NH3 dari larutan penyangga dengan pH =9,25

Maka

pOH =4,75

[OH-] = 1,78 x 10

-5

[OH-] = Kb x

𝑛 .𝑏𝑎𝑠𝑎

𝑛 .𝑎𝑠𝑎𝑚 𝑘𝑜𝑛𝑗𝑢𝑔𝑎𝑠𝑖𝑛𝑦𝑎

1,78 x 10-5

= 10-5

x 𝑛 .𝑁𝐻3

𝑛 .𝑁𝐻4+

10−5

1,78 𝑥 10 −5 = 𝑛 . 𝑁𝐻4+

𝑛 .𝑁𝐻3

Perbandingan NH4+ dan NH3 adalah 1 : 1,78

77. 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?

Answer :

NH3 = 0,2 M x 0,5 liter = 0,1 mol

Buffer basa dengan pH = 10, maka pOH = 4

[OH-] = 10

-4

[OH-] = Kb x



10-4

= 1,8 x 10-5

x 0,1

𝑛 .𝑎𝑠𝑎𝑚 𝑘𝑜𝑛𝑗𝑢𝑔𝑎𝑠𝑖

n.asam konjugasi = 1,8 x 10-2

mol

Massa ammonium klorida = 1,8 x 10-2

mol x 53,5 gram/mol = 0,963 gram

Maka ammonium klorida yang dibutuhkan sebanyak 0,963 gram.

78. 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 ?

Answer :

n. NH3 = 0,1 M x 0,125 lter = 0,0125 mol

Buffer basa dengan pH = 9,15, maka pOH = 4,85

[OH-] = 1,4 x 10

-5

[OH-] = Kb x



1,4 x 10-5

= 1,8 x 10-5

x 0,0125

𝑛 .𝑎𝑠𝑎𝑚 𝑘𝑜𝑛𝑗𝑢𝑔𝑎𝑠𝑖

n.asam konjugasi = 0,016 mol

Massa ammonium klorida = 0,016 mol x 53,5 gram/mol = 0,86 gram

Maka ammonium klorida yang dibutuhkan sebanyak 68,48 gram.

79. 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?

Answer :

a. [H+] = Ka .a

g

[H+] = 1,8x10−5 x 0,1 mol

0,11 mol

= 1,636 x 10-5

pH awal = -log [H+]

= -log 1,636 x 10-5

= 4,786

Penambahan HCl (0,0025 mol) menyebabkan perubahan komposisi penyangga :

CH3COO-(aq) + H

+(aq) CH3COOH(aq)

Mula-mula : 0,11 mol 0,0025 mol 0,1 mol

Reaksi : -0,0025 mol -0,0025 mol +0,0025 mol

Sisa : 0,1075 mol - 0,1025 mol

[H+] = Ka .a

g

[H+] = 1,8x10−5 x 0,1025 mol

0,1075 mol

= 1,716 x 10-5

pH akhir = -log [H+]

= -log 1,716 x 10-5

= 4,765

Penambahan HCl (0,0025 mol) menyebabkan perubahan komposisi penyangga :

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH

-(aq)

Mula-mula : 0,11 mol 0,15 mol 0,1 mol -

Reaksi : -0,0125 mol -0,0125 mol +0,0125 mol +0,0125

mol

Sisa : 0,0975 mol 0,1375 mol 0,1025 mol 0,0125

mol

[H+] = Ka .a

g

[H+] = 1,8x10−5 x 0,1025 mol

0,0975 mol

= 1,05 x 10-5

pH = -log [H+]

= -log 1,05 x 10-5

= 4,978

80. 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?

Answer :

[H+] = Ka .a

g

[H+] = 1,8X10−5 x 0,02 mol

0,022 mol

= 1,636 x 10-5

pHawal = -log [H+]

= -log 1,636 x 10-5

= 4,786

Penambahan HCl (0,15X mol) menyebabkan perubahan komposisi penyangga :

pH = pHawal- penurunan pH

= 4,786-0,05

` = 4,736

pH = 4,736

- log [H+] = 5 - 0,264

= 5 – log 1,836

[H+] = 1,836 x 10

-5

CH3COO-(aq) + H

+(aq) CH3COOH(aq)

Mula-mula : 0,022 mol 0,15X mol 0,02 mol

Reaksi : -0,15X mol -0,15X mol +0,15X mol

Sisa : 0,022-0,15X mol - 0,02 +0,15X mol

[H+] = Ka .a

g

1,836 x 10−5 = 1,8X10−5 x 0,02 + 0,15X mol

0,022 − 0,15X mol

X = 8,05 x 10-3

L = 8,05 mL

Jadi volume HCl adalah 8,05 x 10-3

mL.

Penambahan HCl (0,15X mol) menyebabkan perubahan komposisi penyangga :

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH

-(aq)

Mula-mula :0,022 mol 0,015 + 0,15 X mol 0,02 mol -

Reaksi : -0,015 + 0,15 X mol -0,015 + 0,15 X mol +0,015+0,15Xmol

+0,015 + 0,15 X mol

Sisa : 0,007+ 0,15 X mol - 0,035+0,15Xmol

+0,015 + 0,15 X mol

pH = pHawal- penurunan pH

= 4,786-0,05

` = 4,736

pH = 4,736

- log [H+] = 5 - 0,264

= 5 – log 1,836

[H+] = 1,836 x 10

-5

[H+] = Ka .a

g

[H+] = 1,8x10−5 x 0,035 + 0,15X mol

0,007 + 0,15X mol

1,836 x 10-5

= 1,8x10−5 x 0,035+0,15X mol

0,007+0,15X mol

X = 8,49 L = 8499,66 mL

81. 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 ?

Answer :

pH titik ekivalen dari titrasi ini adalah kurang dari 7, sehingga indikator yang digunakan

adalah etil red dengan trayek pH 4,2 – 6,3; karena indikator ini mengalami perubahan

warna yang tajam di sekitar titik ekivalen.

pH titik ekivalen dari titrasi ini adalah 7, sehingga indikator yang digunakan adalah PP

dengan trayek pH 8,3 – 10; karena indikator ini mengalami perubahan warna yang tajam

di sekitar titik ekivalen.

82. Explain why ethyl red is a better indicator than phenolphtalein in the titration of dilute

ammonia by dilute hydrochloric acid?

Answer :

Etil merah merupakan indikator yang lebih baik digunakan daripada indikator fenolftalein

dalam titrasi antara larutan ammonia encer dengan larutan asam klorida encer karena titik

ekivalen lebih kecil daripada 7. Sebab, garam yang terbentuk menghasilkan basa lemah

dan ion H+. Nilai pH pada kesetimbangan agak lebih kecil daripada di kasus titrasi asam

kuat dengan basa kuat. Kurvanya curam namun perubahannya cepat di dekat titik

kesetimbangan. Akibatnya titrasi masih mungkin asalkan indikator yang tepat dipilih,

yakni indikator dengan rentang indikator yang sempit yaitu indikator etil merah yang

memiliki trayek pH 4,4 – 6,2.

83. What is a good indicator for titrating potassium hydroxide with hydrobromic acid?

Explain.

Answer :

Indikator yang paling tepat digunakan untuk mentitrasi Kalium hidroksida dengan Asam

hidrobromat adalah indikator Bromthymol biru yang memiliki trayek pH 6,0-7,6 karena

pada titrasi ini titik ekivalen dicapai pada pH sekitar 7. Apabila trayek pH terlalu jauh dari

7, titik akhir titrasi akan sangat menyimpang titik ekivalen.

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

ought to be true at the equivalence point?

Answer :

Kondisi jumlah reaktan yang benar pada titik ekuivalen adalah secara stoikhiometri

dimana konsentrasi asam sama dengan konsentrasi basa atau titik dimana jumlah basa

yang ditambahkan sama dengan jumlah asam yang dinetralkan : [H+] = [OH

-] yang

biasanya ditandai dengan berubahnya warna indikator.

85. 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?

Answer :

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)

Mula-mula : 0,005 mol 0,1X mol - -

Reaksi : 0,1X mol 0,1X mol 0,1X mol 0,1X mol

Sisa : 0,005-0,1X mol - 0,1X mol 0,1X mol

0,005 = 0,1X

X = 0,05 L = 50 mL

Jadi volume NaOH adalah 50 mL.

[OH−] = Kw

Ka [M]

[OH−] = Kw

Ka

mol

V campuran

[OH−] = 10−14

1,8 x 10−5 5 x 10−3

10−1

= 2,77 𝑥 10−11

= 5,26 x 10-6

[OH-] = -log [OH

-]

= -log 5,26 x 10-6

= 5,279

pH = 14 – pOH

= 14 - 5,279

= 8,721

Indikator yang paling tepat digunakan adalah indikator fenolftalein yang memiliki trayek

pH 8,3-10. Karena pH larutan di titik ekuivalen lebih besar dari 7. Hal ini disebabkan oleh

garam yang terhidrolisis menghasilkan asam lemah dan ion OH-.

86. 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?

Answer :

V NH3 = 25 mL = 0,025 L

M NH3 = 0,1 M

M HBr = 0,1 M

Kb NH3 = 1,8 x 10-5

Kw = 10-14

a. Mencari PH pada titik ekivalen

M1 V1 = M2 V2

0.1 M x 25 ml = 0.1M x V2

V2 = 25 ml

n NH4OH = M V n HBr = M V

= 0.1 M x 25 ml = 0.1 M x 25 ml

= 2.5 mmol = 2.5 mmol

NH4OH+ HBr → NH4Br + H2O

m 2.5 mmol 2.5 mmol - -

r 2.5 mmol 2.5 mmol 2.5 mmol 2.5 mmol

s - - 2.5 mmol 2.5 mmol

PH saat ekivalen = 𝟏

𝟐 ( pKw - pKa - log G )

= 𝟏

𝟐( 14 - 3.745 - log

𝟓 𝒎𝒎𝒐𝒍

𝟏𝟎𝟎 𝒎𝒍 )

= 𝟏

𝟐 ( 14 - 3.745 – (- 1.301 )

=𝟏

𝟐 ( 14 – 3.745 + 1.301 )

= 5.278

Indikator yang paling tepat digunakan adalah indikator Metil merah yang memiliki trayek

pH 4,4 – 6,2. Karena titik ekivalen lebih kecil daripada 7. Sebab, garam yang terbentuk

menghasilkan basa lemah dan ion H+. Nilai pH pada kesetimbangan agak lebih kecil

daripada di kasus titrasi asam kuat dengan basa kuat. Kurvanya curam namun

perubahannya cepat di dekat titik kesetimbangan.

87. 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

Answer :

V HCl = 25 mL = 2,5 x 10-2

L

M HCl = 0,1 M

M NaOH = 0,1 M

a. 0 mL

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

Mula-mula : 2,5 x 10-3

mol 0 mol - -

Bereaksi : 0 mol 0 mol 0 mol 0 mol

Sisa : 2,5 x 10-3

mol - 0 mol 0 mol

M HCl = 𝐦𝐨𝐥

𝐕 𝐜𝐚𝐦𝐩𝐮𝐫𝐚𝐧

= 𝟐,𝟓 . 𝟏𝟎−𝟑

𝟐,𝟓 .𝟏𝟎−𝟐

= 0,1 M

[H+] = M x valensi [H

+]

= 0,1 M x 1

= 0,1

pH = -log [H+]

= -log 0,1

= 1

b. 10 mL = 0,01 L



mol 10-3

mol - -

Bereaksi : 10-3

mol 10-3

mol 10-3

mol 10-3

mol

Sisa : 1,5 x 10-3

mol - 10-3

mol 10-3

mol

M HCl = mol

V campuran

= 1,5 . 10−3

3,5 .10−2

= 4,2857 x 10-2

M


+]

= 4,2857 x 10-2

M x 1

= 4,2857 x 10-2

pH = -log [H+]

= -log 4,2857 x 10-2

= 1,3679

c. 24,9 mL = 0,0249 L



mol 2,49 x 10-3

mol - -

Bereaksi : 2,49 x 10-3

mol 2,49 x 10-3

mol 2,49x 10-3

mol 2,49 x 10-3

mol

Sisa : 10-5

mol - 2,49x 10-3

mol 2,49 x 10-3

mol

M HCl = mol

V campuran

= 10−5

4,99 .10−2

= 2 x 10-4

M


+]

= 2 x 10-4

M x 1

= 2 x 10-4

pH = -log [H+]

= -log 2 x 10-4

= 3,69

d. 24,99 mL = 0,02499 L



mol 2,499 x 10-3

mol - -

Bereaksi : 2,499 x 10-3

mol 2,499 x 10-3

mol 2,499x10-3

mol 2,499 x 10-

3 mol

Sisa : 10-6

mol - 2,499x10-3

mol 2,499 x 10-

3 mol

M HCl = mol

V campuran

= 10−6

4,99 .10−2

= 2 x 10-5

M


+]

= 2 x 10-5

M x 1

= 2 x 10-5

pH = -log [H+]

= -log 2 x 10-5

= 4,69

e. 25 mL = 0,025 L



mol 2,5 x 10-3

mol - -


mol 2,5 x 10-3

mol 2,5 x 10-3

mol 2,5 x 10-3

mol

Sisa : - - 2,5 x 10-3

mol 2,5 x 10-3

mol

pH = 7 karena larutan HCl dan larutan NaOH habis bereaksi.

f. 25,01 mL = 0,02501 L



mol 2,501 x 10-3

mol - -


mol 2,5 x 10-3

mol 2,5 x 10-3

mol 2,5 x 10-3

mol

Sisa : - 10-6

mol 2,5 x 10-3

mol 2,5 x 10-3

mol

M NaOH = mol

V campuran

= 10−6

5,001.10−2

= 1,99 x 10-5

M

[OH-] = M x valensi [OH

-]

= 1,99 x 10-5

M x 1

=1,99 x 10-5

pOH = -log [OH-]

= -log 1,99 x 10-5

= 4,701

pH = 14 – pOH

= 14 – 4,701

= 9,299

g. 25,10 mL = 0,0251 L



mol 2,51 x 10-3

mol - -


mol 2,5 x 10-3

mol 2,5 x 10-3

mol 2,5 x 10-3

mol

Sisa : - 10-5

mol 2,5 x 10-3

mol 2,5 x 10-3

mol

M NaOH = mol

V campuran

= 10−5

5,01.10−2

= 1,99 x 10-4

M


-]

= 1,99 x 10-4

M x 1

=1,99 x 10-4

pOH = -log [OH-]

= -log 1,99 x 10-4

= 3,701

pH = 14 – pOH

= 14 – 3,701

= 10,299

h. 26 mL = 0,026 L



mol 2,6 x 10-3

mol - -


mol 2,5 x 10-3

mol 2,5 x 10-3

mol 2,5 x 10-3

mol

Sisa : - 10-4

mol 2,5 x 10-3

mol 2,5 x 10-3

mol

M NaOH = mol

V campuran

= 10−4

5,1.10−2

= 1,96 x 10-3

M


-]

= 1,96 x 10-3

M x 1

= 1,96 x 10-3

pOH = -log [OH-]

= -log 1,96 x 10-3

= 2,707

pH = 14 – pOH

= 14 – 2,707

= 11,293

i. 50 mL = 0,05 L



mol 5 x 10-3

mol - -


mol 2,5 x 10-3

mol 2,5 x 10-3

mol 2,5 x 10-3

mol

Sisa : - 2,5 x 10-3

mol 2,5 x 10-3

mol

2,5 x 10-3

mol

M NaOH = mol

V campuran

= 2,5 . 10−3

7,5.10−2

= 3,33 x 10-2

M


-]

= 3,33 x 10-2

M x 1

= 3,33 x 10-2

pOH = -log [OH-]

= -log 3,33 x 10-2

= 1,477

pH = 14 – pOH

= 14 – 1,477

= 12,523

88. 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.

Answer :

V CH3COOH = 25 mL = 0,025 L

M CH3COOH = 0,1 M

M NaOH = 0,1 M

Ka CH3COOH = 1,8 x 10-5

Kw = 10-14

a. [H+] = Ka .M

[H+] = 1,8 . 10−5 . 10−1

= 1,8 . 10−6

= 1,34 . 10-3

pH = -log [H+]

= -log 1,34 . 10-3

= 2,87

b.



mol 10-3

mol - -

Reaksi : 10-3

mol 10-3

mol 10-3

mol 10-3

mol

Sisa : 1,5 x 10-3

mol - 10-3

mol 10-3

mol

[H+] = Ka .

a

g

= 1,8 x 10-5

. 1,5 . 10−3

10−3

= 2,7 x 10

-5

pH = -log [H+]

= -log 2,7 x 10-5

= 4,568

c. 1

2 . M CH3COOH . V CH3COOH = M NaOH . V NaOH

1

2 . 0,1 M . 2,5 x 10

-2 L = 0,1 M . V NaOH

V NaOH = 1,25 x 10-2

L

[H+] = Ka .

a

g

= 1,8 x 10-5

. 1,25 . 10−3

1,25 . 10−3

= 1,8 x 10

-5

pH = -log [H+]

= -log 1,8 x 10-5

= 4,74

d.

M CH3COOH . V CH3COOH = M NaOH . V NaOH

0,1 M . 2,5 x 10-2

L = 0,1 M . V NaOH

V NaOH = 2,5 x 10-2

L



mol 2,5 x 10-3

mol - -

Reaksi : 2,5 x 10-3

mol 2,5 x 10-3

mol 2,5 x 10-3

mol 2,5x10-3

mol

Sisa : - - 2,5 x10-3

mol 2,5 x10-

3 mol

pH = 7 karena [H+] = [OH

-] , dimana larutan CH3COOH dan larutan NaOH habis

bereaksi.

89. 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

Answer :

a. 𝑂𝐻− = 𝐾𝑏 . 𝑀

= 1,8 . 10−5 . 0,1

= 1,3 . 10−3

𝑝𝑂𝐻 = − log 1,3 . 10−3

= 3 − log 1,3

= 2,8860

𝑝𝐻 = 14 − 2,8860

= 11,1139

b. NH3+ HCl NH4Cl

m 2,5mmol 1 mmol -

r 1mmol 1mmol 1mmol

s 1,5 mmol - 1mmol

[OH-]= Kb.

nb

nak

= 1,8 . 10−5 . 1,5

1

=2,7 . 10-5

𝑝𝑂𝐻 = − log 2,7 . 10−5

= 5 − log 2,7

= 4,5686

𝑝𝐻 = 14 − 4,5686

= 9,431

a. NH3+ HCl NH4Cl

m 2,5 mmol 1 ,25mmol -

r 1,25mmol 1,25mmol 1,25mmol

s 1,25mmol - 1,25mmol

[OH-]= Kb.

nb

nak

= 1,8 . 10−5 . 1,25

1,25

=1,8 . 10-5

𝑝𝑂𝐻 = − log 1,8 . 10−5

= 5 − log 1,8

= 4,74

𝑝𝐻 = 14 − 4,74

= 9,26

b. NH3+ HCl NH4Cl

m 2,5 mmol 2,5mmol -

r 2,5mmol 2,5mmol 2,5mmol

s - - 2,5mmol

[OH-] =

𝐾𝑤

𝐾𝑏 . [𝑔𝑎𝑟𝑎𝑚]

= 10−14

10−5. [

2,5

50]

= 10−9

1,8. [

1

20]

= 1

36. 10−9

= 0,0277 . 10 −9

= 27,7 . 10 −12

= 5,19 . 10−6

𝑝𝑂𝐻 = − log 5,19. 10−6

= 6 − log 5,19

= 5,284

𝑝𝐻 = 14 − 5,284

= 10,715

035 2037 garnis rombel 02 tugas ke-1

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