029 3105 bunga rombel1 tugaske01.docx

Tugas Kapsel ke 01

“pH Asam dan Basa”

Nama : Bunga Mahardhika A

NIM : 4301413105

Rombel : 01

Absen : 29

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

1. Write the chemical equation for the autoionization of water and the equilibrium law for Kw?

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

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

b. in terms of pH ?

3. 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?

4. 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.

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

concentrations are :

a. 0.0024 M

a. 1.4 x 10-5 M

a. 5.6 x 10-9 M

a. 4.2 x 10-13 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

a. 0.0065 M

a. 2.5 x 10 -13 M

a. 7.5 x 10 -5 M

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?

8. A soft drink was put on the market with [ ] = 1,4 x mol . What it's pH ?

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

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

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

12. What is the pH of 0.010 M HCl ?

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

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?

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?

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?

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

solution.

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. HAsO42-

d. (CH3)3NH+

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

20. 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

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

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 !

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.

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

F−?

25. The Ka for HF is 6.8 x 10x. what is the Kb for F-?

26. The barbiturate ion C4HO has Kb = 1,0 x 10 -10 . What is Ka for Barbituric acid ?

27. Hydrogen peroxide, H2O2 is a week acid with Ka = 1.8 x 10 -12. What the value of Kb for the

HO2 ion?

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

Calculate the Ka of its conjugate acid!

29. 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- ?

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?

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!

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.

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.

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?

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

0.15 M HONH2.

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

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

38. What is the pH of a 1.0 M solution of hydrogen peroxide, H2O2? For this solute, Ka = 1.8 x

10-2

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

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)

41. 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?

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

solution ?

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

44. The Kb value of the oxalate ion, C2O42-, is 1.9x10-10. Is a solution of K2C2O4 acidic, basic, or

neutral? Explain.

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.

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

47. Explain why the 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.

49. Calculate the pH of 0.20 M NaCN.

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

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

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?

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 !

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

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+?

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.

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?

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

?

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

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

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.

a. 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.

a. What is the calculate of [H+] in 0.010 M HSO4-, obtained by using the usual

simplifying assumption?

a. How much error is produced by incorrectly using the simplifying assumption?

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?

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?

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

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.

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

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

chemical equation and equilibrium law )

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.

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.

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

a. calculate the pH of the buffer using the Ka for NH4+

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?

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?

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?

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 ?

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

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?

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 ?

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?

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?

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 ?

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

ammonia by dilute hydrochloric acid?

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

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

ought to be true at the equivalence point?

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?

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?

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

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.

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

ANSWER

0. Arrhenius : Acid is molecules that releasing ion [H+] when dissolved in water

Lewis : Acid is molecules which the central atom have empty orbital

Brownsted Lowry : Acid is molecules that can not bind ion [H+]

1. Autoionisation of water is H2O H+ + OH-

H2O H+ + OH-

K =

K [H2O]= [H+] [OH-]

Kw = [H+] [OH-]

2. 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. 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

At this temperature, pH = pOH, and both of them is a half of pKw. The water is neutral.

4. D1 Kw = 8.9 x 10-16

T = 20°C

D2O→D+ + 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. a. Known : [OH-] = 2.4 x 10-3 M

Question : [H+] . . . ?

Answer :

[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. Known : [OH-] = 1.4 x 10-5 M

Question : [H+] . . . ?

Answer :

[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. Known : [OH-] = 5.6 x 10-9 M

Question : [H+] . . . ?

Answer :

[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. Known : [OH-] = 4.2 x 10-13 M

Question : [H+] . . . ?

Answer :

[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 10-2M

6. 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. Info : [H+] = 1.9 x 10-5 mol L-1

Question : pH…….?

Answer : pH = -log [H+]

= -log [1.9 x10-5]

= 5-log 1.9

pH = 5 - 0.28

pH = 4.72

8. Diket :[ ] = 1,4 x mol

Ditanya : pH ....?

Dijawab : pH = - log [ ]

= - log 1,4 x

pH = 5 - log 1,4

9. 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. 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. 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.

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. [H+] = x . Ma pH = - log [ H+]

= 1 x 5 x 10-3 = - log 5 x 10-3

= 5 x 10-3 = 3 - log 5

14. Given : m NaOH = 6.0 gram

V larutan = 1.00 L

Mr NaOH = 40

Asked : pOH and pH = .....?

Solution :

[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. Given :

mass Ba(OH)2 = 0.837 g

Mr Ba(OH)2 = 171

V = 100 mL

Question: pOH and pH of solution = …?

Answer :

15. Known : pH of Ca(OH)2 = 11.60

Asked : [Ca(OH)2]=…..?

Answer : 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

16. Diket: pH HCl=2,5

Volume HCl= 250 Ml

Ditanya: massa HCl pada larutan=?

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

17. The ionization of :

a. HNO2 H+ + NO2-

b. H3PO4 3H+ + PO43-

c. HAsO42- AsO4

3- + H+

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

18. a. HNO2 H+ + NO2-

b. H3PO4 3 H+ + PO43-

c. HAsO4 2- H+ + AsO43-

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

CH3

19. a. (CH3)3N + H2O ↔ CH3 — N — H+ + OH-

CH3

b. AsO43- + H2O ↔HAsO4

2- + OH-

c. NO2- + H2O ↔ HNO2 + OH-

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

CH3 CH3

20. Diket : a. (CH3)3N

b. AsO43-

c. NO2-

d. (CH3)2N2H2

Dit : write for the appropriate Kb expression from the exercise 20 ?

Jawab :

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

b. AsO43- +H2O H3AsO4 + 3OH-

H3AsO4 + H2O As(OH)5

c. NO2- + H2O HNO2 + OH-

d. (CH3)2N2H2 + H2O (CH3)3NH3+ + OH-

21. C6H5COOH + NaOH C6H5COONa + H2O

C6H5COOH C6H5COOH + H+

Ka

22. C6H5CO2- + H2O C6H5OH + OH-

Kb =

23. pKa HCN = 9.21 → Ka HCN = 6.17×10-10

pKa HF = 3.17 → Ka HF = 6.76×10-4

Kb CN− = = 0.16×10-4

Kb F− = = 0.15×10-10

So, the strong Bronsted base is CN−

24. Ka for HF = 6.8x10-4

What is Kb for F-?

HF H+ + F-

Kw = Ka x Kb

10-14 = 6.8x10-4 x Kb

Kb = 1.47x10-11

25. Kw = Kb x Ka

10-14 = 1,0 x 10 -10 x Ka

Ka =

Ka = 10 -4

26. D1 : Ka of H2O2 = 1.8 x 10-12

D2 : Kb of HO2 ion

D3 :

Kb =

=

= 5.56 x 10-3

27. D1 Kb of CH3NH2 = 4.4 x 10-4

D2 Ka = ...?

D3 Ka =

=

= 2.27 x 10-11

28. Known : Ka = 1.4 x 10-4

T 25oC

Question : Kb . . .?

Answer :

Kw = Ka xKb

10-14 = 1.4 x 10-4 x Kb

Kb =

Kb = 7.14 x 10-11

29. 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.

30. Info : 0.1 M HIO4

[H+] = 3.8 x 10-2 mol L-1

Question : Ka and pKa…….?

Answer :

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

31. Diket : [HC2H2ClO2] = 0,10 M dan α =11 %

Ditanya : dan pKa....?

Dijawab : α =

0,11 = pKa = - log Ka

= (0,11)2 x 0,1 pKa = - log 1,21x

= 1,21 x

= 3 - log 1,21

32. 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

33. Given : M HONH2 = 0.15 Molar

pH = 10.12, so pOH = 3.88 → [OH-] = 10-3.88

Ask : Kb ? and pKb ?

Solution : [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

34. D1 : HONH2

M = 0.15 mol/lt

pH = 10.12

Kb = 1.15x10-7 (from exercise 34)

D2 : α ?

D3 :

α: = = 0.000878

35.

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

36. [H+] = pH = - log [ H+]

= = -log 1.64 x 10-3

= = 3 - log 1.64

= 1.64 x 10-3

37. Given : [H2O2] = 1.0 M

Ka = 1.8 x 10-2

Asked : pH = ....?

Solution : H2O2 O2 + 2H+ + 2e-

[H+] =

=

= 0.134 M

pH = - log [H+]

= - log 0.134

= 0.87

38. Given :

M HC6H5O = 0.050 mol/L

Ka = 1.3 x 10-10

Question : [H+] = …?

% HC6H5O is ioniozed= …?

Answer:

HC6H5O C6H50 + H+

B 0.05 - -

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

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

39. Known : pKb of Cod = 5.79

[Cod] = 0.020 M

Asked : pH of Cod = ….?

Answer : 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

40. Diket pKb ND3= 4,96

Kb = 1,096. 10-5

T = 250C

Ditanya: pH 0.2 M ND3= ?

Jawab:

[OH] =

=

=

= 1,48. 10-3

pOH = 3-log 1,48

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

= 11+ log 1,48

= 11,17

41. Given : pH = 2.54

Question : [CH3COOH] ….?

Answer :

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

42. 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-

43. 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

Basa

44. Diket : there are 0,5 M solution are prepared of each : NaI, KF, (NH4)2SO4,

KCN, KC2H3O2, CsNO3, and KBr.

Dit : Write formula of those

a. Acidic

b. Basic

c. Neutral

Jawab :

a. Acidic is (NH4)2SO4 and CsNO3

b. Basic is KF, KCN, KC2H3O2

c. Neutral is NaI and KBr

45. 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 .

46. 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.

47. 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.

48. [NaCN]= 0.20 M

[OH-] =

=

=

= 1.82x10-5

pOH = 5 - Log 1.82

pH = 9 + Log 1.82

49. KNO2 K+ + NO2-

K+ + H2O

NO2- + H2O HNO2 + [OH-]

[OH-] = x [NO2]

= x

= 10 -12

pOH = 12

pH = 2

50. D1 : M CH3NH3Cl = 0.15

Kb = 4.4 x 10-4

D2 : pH of CH3NH2

D3 :

[H+] =

=

= 1.846 x 10-6

pH = - log [H+]

= - log (1.846 x 10-6)

= 6 - log 1.846

= 5.734

51. D1 0.15 M BHCl, pH = 4.28

D2 Kb for the base B = ...?

D3 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

52. Known : V H2O = 1.00 L

pH = 5.16

Mr NH4Br = 76

T 25oC

Kb = 1.8 x 10-5

Question : massa of NH4Br . . .?

Answer :

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

53. 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

54. Info : 0.2 M H-Mor+

pKb = 6.13

Question : pH…….?

Answer : 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

55. Diket : [H- ] = 0,15 M, pKa = 8,52

Reaksi : H- + H2O Qu + H3

Ditanya : pH....?

Dijawab : 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 = pH = - log

3,02 x = - log 2,13 x

3,02 x pH = 5 - log 2,13

= 2,13 x

56. Initial concentration is unable to calculate equilibrium concentration when mole of both of

components which react is same.

57. Given : M HF = 0.15 Molar

Ka HF = 6.5 X 10-4

Ask : pH ?

Solution : [H+] =

=

= 9.87 x 10-3

pH = - Log [H+]

= - Log 9.87 x 10-3

= 3 - Log 9.87

= 2.00568

58. D1 : CH3COOH

M = 10-3 mol/lt

Ka = 1.8x10-5

D2 : α and pH ?

D3 :

α: = = 0.134

[ H+ ] = = = 1.34x10-4

pH = -log 1.34x10-4 = 4 – log 1.34 = 3.8729

59.

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. Answer:

60. 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.

61. 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

62. Given :

pKa = 4.01

M = 0.050 mol/L

Question : [H+] and pH = …?

Answer :

63. Answer:

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) + PO4

3-(aq) + Na+

(aq) + OH-(aq) Na+

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

NaH2PO4(aq) + NaOH (aq) Na2HPO4 (aq) + H2O (l)

Ionic equation:

Na+ (aq) + H2PO4- (aq) + Na+ (aq) + OH-

(aq) 2Na+(aq) + HPO4

2-(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+

64. Diket: a. buffer 0.10 M NH4Cl and 1 M NH3

b. buffer1 M NH4Cl and 0.10 M NH3

ditanya: buffer would be better able to hold a steady pH=?

Jawab:

pH of A solution before added strong acid

[OH-] = Kb NH3.

= 1,8 .10-5

= 1,8. 10-4

= 4- log 1,8

pOH = 3,745

pH = 10,255

pH of B solution before added strong acid

[OH-] = Kb NH3.

= 1,8 .10-5

= 1,8. 10-6

= 6- log 1,8

pOH = 5,74

pH = 8,26

pH of A solution after added strong acid

For example, strong acid which added 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 solution B after added strong acid

[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

Changes in pH of the solution A before plus a strong acid and a pH of the solution after the

addition of a strong acid = (10,255 - 8,78)

= 1,475

Changes in pH of the solution B before plus a strong acid and a pH of the solution after the

addition of a strong acid = (8,26 - 7,935)

= 0,325

In conclusion buffer solution pH is likely to have changed little b solution for pH changes

only 0.325

65. Given : C2H2H3O2 H+ +C2H3O2-

M HC2H3O2 : 0.15 M

M C2H3O2- : 0.25 M

Question : pH ….?

Answer :

1.8 x 10-5 =

1.8 x 10-5 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

= 4.97

66. Diket : M HC2H3O2 = 0.15 M

M C2H3O2- = 0.25 M

Ka = 1.8 x 10 -5

Dit : pH dengan menggunakan Kb dari ion asetat

Jawab :

67. [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.

68. Diket : 500 mL of Buffer that contain 0,15 M HC2H3O2 and 0,25 M C2H3O2- ?

Ka = 1,8 x 10-5 for HC2H3O2

Dit : How much will pH Change ?

Jawab :

HC2H3O2 + NaOH C2H3O2Na + H2O

m = 75 mmol 5 mmol 125 mmol

r = 5 mmol 5 mmol _ 5 mmol +

s = 70 mmol o 130 mmol

Atau dengan cara lain

69. 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

= 5.25

70. Given : Buffer

0.25 NH3 and 0.14 NH4+ 1.00 L

Asked : the change of pH if 0.020 mol HCl is added to 1.00 L.?

Answered : 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

+

B 0.25mol 0.020mol 0.14 mol

C - 0.020mol 0.020mol +0.02mol

A 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

= 10-5x

= 0.56x10-5

= 5.6x10-6

pH = 6 - log 5.6

pH = 5.25

If added 0.020mol HCl

[H+] = Ka

= 10-5x

= 1.43x10-5

pH = 5 - log 1.43

pH = 4.84

Change of pH = 5.25 – 4.84 = 0.41

71. 75 ml 0.10 M KOH added to 200 ml buffer of 0.25 M NH3 and 0.14 M NH4+

pH buffer

[OH−] = Kb ×

= 1×10-5 ×

= 1.79×10-5

pOH = −log 1.79×10-5

= 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)

Before 28 mmol 7.5 mmol 50 mmol

Change 7.5 mmol 7.5 mmol 7.5 mmol

After 20.5 mmol − 57.5 mmol

[OH−] = Kb ×

= 1×10-5 ×

= 2.8×10-5

pOH = −log 2.8×10-5

= 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

72. 1.0 L CH3COOH 0.15

pKa = 4.74, Ka = 1.8x10-5

pH Buffer= 5.00, [H+] = 1x10-5

How Many CH3COONa would have be added?

[H+] =

10-5 =

X = =0.0833

Mass of CH3COOH = mol x Mr

= 0.0833 x 60

= 5 gram

73. 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

74. D1 : pH of buffer solution = 9.25

D2 : mole ratio of NH4Cl to NH3

D3 :

pOH = 4.75

[OH-] = Kb

1.77 x 10-5 = 1.8 x 10-5

=

=

75. 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

76. Known : V NH3 = 125 mL

M NH3= 0.10 mol/L

pH = 9.15

Kb = 1.8 x 10-4.85

Question : massa of NH4Cl . . .?

Answer :

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

77. 25 mL 0,1 M HCl Ka CH3COOH = 1,8 x 10 -5

0,1 mol CH3COOH, 0,11 mol CH3COONa

a. [ 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

b. Sama. Pengenceran dengan penambahan air pada buffer, pH akan tetap karena penentu

pH buffer adalah jumlah mol bukan konsentrasi buffer

78. Info : 0.15 M HCl pH from No.78 = 4.79

100 mL of buffer pH after = 4.79 + 0.05 = 4.84

pH decrease by 0.05 pH unit [H+] = 4.45 x 10-5

Question : a) Volume of HCl would have to added….?

b) 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 :

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 10-5a

4.89 x 10-6 - 1.8 x 10-6 = 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) The amount of HCl is added is the same, namely 330 ml, Buffer coupled with

water (dilution) pH fixed for determining the pH of the buffer is not the number

of moles of buffer concentration.

ame as the mean one acidic or alkaline solution is weak. Choosing indicators affect titration

process as to determine the end point of the titration with marked changes in color,

according to the expected pH range so that the selection of the indicators used are not

wrong.

79. 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

80. 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.

81. 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.

82. 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.

83. Given:

V NH3 = 25 mL

M NH3 = 0.10 mol/L

Kb NH3 = 1.8 x 10-5

M HBr = 0.10 mol/L

Question : what is pH at equivalence point and a good indicator = …?

Answer:

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

NH3 + HBR NH4Br

B 2.5 mmol 2.5 mmol -

R 2.5 mmol 2.5 mmol 2.5 mmol

A - - 2.5 mmol

84. Given : [HCl] = 0.1000 M

[NaOH] = 0.1000 M

V HCl = 25 mL

Asked : 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


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


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


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


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


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


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


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


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

85. Given :

V = 25 mL

M = 0.1000 mol/L = 10-1 mol/L

Ka = 1.8 x 10-5

Question :

Calculate 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 :

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.


B 2.5 mmol 1.25 mmol - -

R 1.25 mmol 1.25 mmol 1.25 mmol 1.25 mmol

A 1.25 mmol - 1.25 mmol 1.25 mmol

log1.8 = 4.744

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


B 2.5 mmol 2.5 mmol - -

R 2.5 mmol 2.5 mmol 2.5 mmol 2.5 mmol

A - - 2.5 mmol 2.5 mmol

86. Answer:

a. Known : 25.00 mL of 0.1000 M NH3

Asked : pH….?

Answer : [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. Known : 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

Asked : pH after mixed….?

Answer: 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. Asked : pH after half the NH3 has been neutralized….?

Answer : NH3(aq) + HCl(aq) NH4Cl(aq)

Before: 2,5 1.25 -

React: 1.25 1.25 1.25

After : 1.25 mmol - 1.25 mmol

[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. Asked : pH at the equivalence point….?

Answer : 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

029 3105 bunga rombel1 tugaske01.docx

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