004 2024 Nurullah Aulia Sugiarti Rombel 01 Tugas Ke 01

QUESTIONS

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 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.

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. According to Arrhenius opinion, acid is a species that releases H+ when dissolved in the

water.

According to Bronsted Lowry opinion, acid is a species that gives proton (proton donor)

According to Lewis opinion, acid a substance that accepts a pair of electrons.

1. Chemical equation for the autoionization of water

H2O (l) H+

(aq) + OH-(aq)

K[H2O] = [H+][ OH

-]

Kw = [H+][ OH

-]

2. In terms of [H+] and [OH-], acidic is the substance where the [H+] > [OH-],

basic is the substance that [H+]

3. Given : T = 37oC

Kw = 2,4x10-14

Question : - [H+], [OH

-], pH and pOH

- Relation between pH, pOH, and Kw at this temperature

- Is water neutral at this temperature?

Answer :

Kw = [H+][ OH

-]

At this temperature, pH=pOH, both are half of pKw

At this temperature, the water is almost neutral because pH is

4. Given : Kw = 8.9 x 10-16

T = 20C

D2O D+ + OD

-

Questions :[D+], [OD

-], pD, pOD = ...?

Answer :

[D+] = [OD

-] =

=

= 2.98 x 10-16

pD = 8 - log 2.98

= 7.53

pOD = 8 - log 2.98

= 7.53

5. a. Given : [OH-] = 2.4 x 10-3 M

Questions : [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. Given : [OH-] = 1.4 x 10

-5 M


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. Given : [OH-] = 5.6 x 10

-9 M


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. Given : [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 10-2

M

6. a. Given : [ H+ ] = 3.5 x 10 -8 M

Questions : [OH- ]..?

Answer :

pH = 8- log 3,5

pOH = 14 - pH

= 14 - (8 - log 3,5)

= 6 + log 3,5

= 6,5

[OH- ]= 10

-6,5

b. Given : [H+]=0.0065 M = 6.5 x 10-3

Questions : [OH-] ?

Answer :

pH = 3- log 6,5

pOH = 14 - pH

= 14 - (3 - log 6,5)

= 11 + log 6,5

= 11,8

[OH- ]

= 10 -11,8

c. Given : [H+]=2,5 x 10 -13 M

Questions :[OH-] ?

Answer :

pH = 13- log 2,5

pOH = 14 - pH

= 14 - (13 - log 2,5)

= 1 + log 2,5

= 1,39

[OH- ]

= 10 -1,39

d. Given : [H+]=7,5 x 10-5 M

Questions : [OH- ] ?

Answer :

pH = 5- log 7,5

pOH = 14 - pH

= 14 - (5 - log 7,5)

= 9 + log 7,5

= 9,8

[OH- ]

= 10 -9,8

7. Given : [H+] = 1.9 x 10-5mol L-1

Questions : pH.?

Answer : pH = -log [H+]

= -log [1.9 x10-5

]

= 5-log 1.9

pH = 5 - 0.28

pH = 4.72

8. Given : [H+] = 1,4 x 10-5mol L-1

Questions : pH ... ?

Answer :

[H+] = 1,4 x 10

-5mol L

-1

pH = - log [H+]

= - log 1,4 x 10-5

mol L-1

= 5 log 1,4

= 5 0,146

pH = 4,854

9. Exercise 5

a. Given : [OH-] = 0,0024 M = 2,4 x 10

-3 M

Asked : pH ... ?

Answer :

[OH-] = 2,4 x 10

-3 M

pOH = - log [OH-]

= - log 2,4 x 10-3

M

= 3 log 2,4

= 3 0.38

pOH = 2,62

pH = 14 pOH

= 14 2,62

pH = 11,38

b. Given : [OH-] = 1,4 x 10

-5 M

Asked : pH ... ?

Answer :

[OH-] = 1,4 x 10

-5 M

pOH = - log [OH-]

= - log 1,4 x 10-5

M

= 5 log 1,4

= 5 0.146

pOH = 4,854

pH = 14 pOH

= 14 4,854

pH = 9,146

c. Given : [OH-] = 5,6 x 10

-9 M

Asked : pH ... ?

Answer :

[OH-] = 5,6 x 10

-9 M

pOH = - log [OH-]

= - log 5,6 x 10-9

M

= 9 log 5,6

= 9 0.748

pOH = 8,252

pH = 14 pOH

= 14 8,252

pH = 5,748

d. Given : [OH-] = 4,2 x 10

-13 M

Asked : pH ... ?

Answer :

[OH-] = 4,2 x 10

-13 M

pOH = - log [OH-]

= - log 4,2 x 10-13

M

= 13 log 4,2

= 13 0.623

pOH = 12,377

pH = 14 pOH

= 14 12,377

pH = 1,623

Exercise 6

a. Given : [H+] = 3,5 x 10-8 M

Asked : pH ... ?

Answer :

[H+] = 3,5 x 10

-8 M

pH = - log [H+]

= - log 3,5 x 10-8

M

= 8 log 3,5

= 8 0,544

pH = 7,456

b. Given : [H+] = 0,0065 M = 6,5 x 10-3 M

Asked : pH ... ?

Answer :

[H+] = 6,5 x 10

-3 M

pH = - log [H+]

= - log 6,5 x 10-3

M

= 3 log 6,5

= 3 0,813

pH = 2,187

c. Given : [H+] = 2,5 x 10-13 M

Asked : pH ... ?

Answer :

[H+] = 2,5 x 10

-13 M

pH = - log [H+]

= - log 2,5 x 10-13

M

= 13 log 2,5

= 13 0,398

pH = 12,602

d. Given : [H+] = 7,5 x 10-5 M

Asked : pH ... ?

Answer :

[H+] = 7,5 x 10

-5 M

pH = - log [H+]

= - log 7,5 x 10-5

M

= 5 log 7,5

= 5 0,875

pH = 4,125

10. a. Given : pH = 3,14

Asked : [H+] and [OH

-] ... ?

Answer :

pH = 3,14

pH = 4 0,86

[H+] = 4 log 7,24

[H+] = 7,24 x 10

-4 M

pOH = 14 pH

= 14 3,14

pOH = 10,86

pOH = 11 0,14

[OH-] = 11 log 1,38

[OH-] = 1,38 x 10

-11 M

b. Given : pH = 2,78


-] ... ?

Answer :

pH = 2,78

pH = 3 0,22

[H+] = 3 log 1,66

[H+] = 1,66 x 10

-3 M

pOH = 14 pH

= 14 2,78

pOH = 11,22

pOH = 12 0,78

[OH-] = 12 log 6,026

[OH-] = 6,026 x 10

-12 M

c. Given : pH = 9,25


-] ... ?

Answer :

pH = 9,25

pH = 10 0,75

[H+] = 10 log 5,62

[H+] = 5,62 x 10

-10 M

pOH = 14 pH

= 14 9,25

pOH = 4,75

pOH = 5 0,25

[OH-] = 5 log 1,78

[OH-] = 1,78 x 10

-5 M

d. Given : pH = 13,24


-] ... ?

Answer :

pH = 13,24

pH = 14 0,76

[H+] = 14 log 5,75

[H+] = 5,75 x 10

-14 M

pOH = 14 pH

= 14 13,24

pOH = 0,76

pOH = 1 0,24

[OH-] = 1 log 1,74

[OH-] = 1,74 x 10

-1 M = 0,174 M

e. Given : pH = 5,70


-] ... ?

Answer :

pH = 5,70

pH = 6 0,30

[H+] = 6 log 1,995

[H+] = 1,995 x 10

-6 M

pOH = 14 pH

= 14 5,70

pOH = 8,30

pOH = 9 0,70

[OH-] = 9 log 5,01

[OH-] = 5,01 x 10

-9 M

11. a. Given : pOH = 8,26


-] ... ?

Answer :

pOH = 8,26

pOH = 9 0,74

[OH-] = 9 log 5,495

[OH-] = 5,495 x 10

-9 M

pH = 14 pOH

= 14 8,26

pH = 5,74

pH = 6 0,26

[H+] = 6 log 1,82

[H+] = 1,82 x 10

-6 M

b. Given : pOH = 10,25


-] ... ?

Answer :

pOH = 10,25

pOH = 11 0,75

[OH-] = 11 log 5,62

[OH-] = 5,62 x 10

-11 M

pH = 14 pOH

= 14 10,25

pH = 3,75

pH = 4 0,25

[H+] = 4 log 1,78

[H+] = 1,78 x 10

-4 M

c. Given : pOH = 4,65


-] ... ?

Answer :

pOH = 4,65

pOH = 5 0,35

[OH-] = 5 log 2,24

[OH-] = 2,24 x 10

-5 M

pH = 14 pOH

= 14 4,65

pH = 9,35

pH = 10 0,65

[H+] = 10 log 4,47

[H+] = 4,47 x 10

-10 M

d. Given : pOH = 6,18


-] ... ?

Answer :

pOH = 6,18

pOH = 7 0,82

[OH-] = 7 log 6,61

[OH-] = 6,61 x 10

-7 M

pH = 14 pOH

= 14 6,18

pH = 7,82

pH = 8 0,18

[H+] = 8 log 1,51

[H+] = 1,51 x 10

-8 M

e. Given : pOH = 9,70


-] ... ?

Answer :

pOH = 9,70

pOH = 10 0,30

[OH-] = 10 log 1,995

[OH-] = 1,995 x 10

-10 M

pH = 14 pOH

= 14 9,70

pH = 4,30

pH = 5 0,70

[H+] = 5 log 5,01

[H+] = 5,01 x 10

-5 M

12. Given : Ma = 1 x 10-2

mol/liter

Questions : pH ?

Answer :

[H+] = x . Ma

= 1 . 1 x 10-2

mol/liter

= 1 x 10-2

mol/liter

pH = - log [1 x 10-2

]

= 2

13. Given : M = 0,005M

Questions : pH ?

Answer :

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

+]

= 1 x 5 x 10-3

= - log 5 x 10-3

= 5 x 10-3

= 3 - log 5

= 2,3

14. Given : m NaOH = 6.0 gram

V larutan = 1.00 L

MrNaOH = 40

Question : pOH and pH = .....?

Answer :

[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 of Ba(OH)2 = 0.837 g

Mr Ba(OH)2 = 171

V = 100 mL

Question: pOH and pH of solution = ?

Answer :

16. Given : pH Ca(OH)2 = 11.60

Questions :Molaritas ?

Answer :

pOH = 14 11,60

= 2,4

= 3 11.60

[OH-] =3,98 . 10

-3

[OH-] = M x

3,98 . 10-3

= M . 2

M = 1,99 x 10-3

M

17. Given : pH HCl = 2,50

V = 250 ml = 0,25 L

Questions : grams of HCl?

Answer :

[H+] = 3 0,5

[H+] = 3,16 x 10

-3

[H+] = M .x

3,16 . 10-3

= M . 1

3,16 . 10-3

= M

M = n/V

n = M. V

n = 3,16 . 10-3

. 0,25

n = 0,79 . 10-3

18. Answer :

a. HNO2 H+ + NO2

-

b. H3PO4 H+ + H2PO4

-

c. HAsO42- H+ AsO4

3-

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

19. Answer :

a. HNO2 H+ + NO2

-

b. H3PO4 H+ + H2PO4

-

c. HAsO42- H+ AsO4

3-

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

CH3

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

+ OH-

CH3

b. AsO43-

+ H2O HAsO42-

+ OH-

c. NO2-

+ H2O HNO2 + OH-

CH3

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

CH3

21. Given: a. (CH3)3N

b. AsO43-

c. NO2-

d. (CH3)2N2H2

Asked : Kb expression?

Answer :

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

-

22. C6H5COOH + NaOH C6H5COONa + H2O

C6H5COOH C6H5COO- + H

+

23. C6H5CO2- + H2O C6H5OH + OH

-

Kb =

24. 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. Given : Ka for HF = 6.8x10-4

Questions : What is Kb for F-?

Answer :

HF H+ + F

-

Kw = Ka x Kb

10-14

= 6.8x10-4

x Kb

Kb = 1.47x10-11

26. Kw = Kb x Ka

10-14

= 1,0 x 10 -10

x Ka

Ka =

Ka = 10 -4

28. Given : Kb of CH3NH2 = 4.4 x 10-4

Questions: Ka= ...?

Answer :

29. Given : 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

30. Given : the formula : HIO3 H+ + IO3

-

pKa = 0.77

Questions : a. formula Kb ?

b. Its is conjugate base a stronger or a weaker base than the acetate ion?

Answer :

a. 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 acetate ion.

31. Given : 0.1 M HIO4

[H+] = 3.8 x 10

-2mol L

-1

Question : Ka and pKa.?

Answer :

( 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

32. Given : [HC2H2ClO2] = 0,10 M dan =11 %

Asked : KadanpKa....?

Answer :

0,11

Ka = (0,11)2 x 0,1

Ka = 1,21 . 10-3

Ka= 3 - log 1,21

pKa = - log Ka

pKa= - log 1,21. 10-3

33. Given : Ethylamine M = 0.1 mol/L

pH = 11.86

Asked : Kb and pKb ?

Answer :

pH = 11.86 so 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. Given : M HONH2 = 0.15 M

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

Asked : Kb ?andpKb ?

Answer : [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. Given : HONH2

M = 0.15 mol/lt

pH = 10.12

Kb = 1.15x10-7

(from exercise 34)

Asked : ?

Answer :

= = 0.000878

36. Given :

Ma = 0.125 mol/liter

Ka = 3,2 x 10-3

Asked : pH?

Answer :

[H+] =

=

=

= 2 x 10-2

pH = - log [2 x 10-2

]

= 2 - log 2

= 2 - 0.301

= 1.699

37. Given : Ma HN3 : 0,15 mol/liter

Ka :1,8 x 10-5

Asked : pH?

Answer : [H+] =

=

=

= 1.64 x 10-3

pH = - log [ H

+]

= -log 1.64 x 10-3

= 3 - log 1.64

38. Given : [H2O2] = 1.0 M

Ka = 1.8 x 10-2

Asked : pH = ....?

Answer : H2O2 O2 + 2H+ + 2e

-

[H+] =

=

= 0.134 M

pH = - log [H+]

= - log 0.134

= 0.87

39. Given : M HC6H5O = 0.050 mol/L

Ka = 1.3 x 10-10

Asked : [H+] = ?

% HC6H5O ioniozed= ?

Answer:

HC6H5O C6H5O + H+

M 0.05 0 0

R 2.55 x 10-6

2.55 x 10-6

2.55 x 10-6

S 0.05 - 2.55 x 10-6

2.55 x 10-6

2.55 x 10-6

40. Given : pKb of Cod = 5,79

M of Cod = 0,020 M

Asked : pH of Cod = ?

Answer :

pKb = - log Kb

Kb = 10 -5,79

= 10 -6

pOH = - log [OH-]

= - log (1,41.10-4

)

= 4 log 1,41

= 4 0,15

= 3,85

pH = pKw pOH

= 14 3.85

= 10,15

41. Given : pKb of ND3= 4,96

Kb = 1,096.10-5

M of ND3 = 0,20 M

T = 25C

Asked : pH of ND3 = ?

Answer :

pOH = - log [OH-]

= - log (1,48.10-3

)

= 3 log 1,48

= 3 0,17

= 2,83

pH = pKw pOH

= 14 2,83

= 11,17

42. Given : pH of CH3COOH = 2,54

Asked : M of CH3COOH = ...?

Answer :

pH = - log [H+]

2,54 = - log [H+]

[H+] = 10

-2,54

[H+] = 2,88 x 10

-3

43. The sodium salt of aspirin is basic, because it from acetylsalyclic acid and sodium hydroxide.

Weak acid with strong base want to produce basic salt or if sodium hydroxide dissolved in

water to produce acetylsalicyclic acid and OH-

NaAcetylsalicyclic + H2O HAcetylsalicyclic + OH-

44. K2C2O4 is base because K2C2O4 a salt formed from a strong base and a weak acid to be base

salt. K2C2O4 can be obtained from the reaction of a strong base and a weak acid that is KOH

H2C2O4.

reaction:

H2C2O4 + 2 KOH K2C2O4 + 2 H2O

45. a. Acidic is (NH4)2SO4 and CsNO3

b. Basic is KF, KCN, KC2H3O2

c. Neutral is NaI and KBr

46. Based onthe Lewis acid-base theory, AlCl3 is acidic compounds because it cannot yet

reached the central atom an octet configuration or can be said to still have vacant orbitals

(empty). Thus, a solution ofAlCl3 can change the color of litmus paper blue to red due to the

acid characteristic.

47. 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 (NH4NO3)

NH4NO3 NH4+ + NO3

base

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. NaCN Na+ + CN-

Na+ + H2O

CN- + H2O HCN + OH

-

[NaCN] = 0.20 M

Ka = 6.2 x 10-10

[OH-] =

=

=

= 1.796 x10-3

pOH = 3 - Log 1.796

pH = 14 (3 - Log 1.796)

= 11 + Log 1.796

= 11.25

50. KNO2 K+ + NO2

-

K+

+ H2O

NO2- + H2O HNO2 + OH

-

[KNO2] = 0.04 M

Ka = 4.3 x 10-4

[OH-] =

=

=

= 9.64 x10-7

pOH = 7 - Log 9.64

pH = 14 (7 - Log 9.64)

= 7 + Log 9.64

= 7.98

51. CH3NH3Cl CH3NH3+ + Cl

-

Cl-

+ H2O

CH3NH3+ CH3NH2 + H

+

[CH3NH3Cl] = 0.15 M

Kb = 4.4 x 10-4

[H+] =

=

=

= 1.846 x10-6

pH = 6 - Log 1,846

= 5.73

52. Given : 0.15 M BHCl, pH = 4.28

BH+

+ Cl- BHCl

0,15 M 0,15 0,15

Questions :Kb for the base B = ...?

Answer :

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. Given : V H2O = 1.00 L

pH = 5.16

Mr NH4Br = 107

T 25oC

Kb = 1.8 x 10-5

Question : massa of NH4Br . . .?

Answer :

pH =

5,16 = - log

= - log

= - log

=

=

=

=

= 9,218 gram

54. 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. Given : 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

56. Given : Reaction : Qu + H2O H-

+ H3O+

[H- ] = 0.15 M, pKa = 8.52

Questions :pH....?

Answer :pKa = 8.52

pKa = - log Ka

8,52 = - log Ka

Ka = 3,02 x 10-9

[ H+ ] =

=

= 2,13 x 10-5

pH = 5 log 2.13 = 0.33

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

components which react is same.

58. Given : M HF= 0.15 M

Ka HF = 6.5 x 10-4

Asked : and pH ?

Solution : =

=

= 6,58 x 10-2

[H+] =

=

= 9.87 x 10-3

pH = - Log [H+]

= - Log 9.87 x 10-3

= 3 - Log 9.87

= 2.01

59. Given : CH3COOH

M = 0,0010 M

Ka = 1.8 x 10-5

Ask : and pH ?

Solution : =

=

= 1,34 x 10-1

[H+] =

=

= 1,34 x 10-4

pH = - Log [H+]

= - Log 1,34 x 10-4

= 4 - Log 1,34

= 4 0,13

= 3,87

- log 1 = 7

d. The error is produced by incorrectly using the simplifying assumption is 0% because the result

of point B and C is same.

62. pKa = 4,92

Ka = 5-log 1,2 = 1,2 x 10-5

pH = 4 log 6

= 4 0.78

= 3,22

63. pKa = 4.01

Ka = 5 log 9,77

= 9,77 x 10-5

pH = 3 log 2,21

= 3 0,34

= 2,66

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

Ionic equation:

2H+

(aq) + CO32-

(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)

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+

65. Given : a. buffer 0.10 M NH4Cl and 1 M NH3

b. buffer1 M NH4Cl and 0.10 M NH3

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

Answer :

pH solution before added strong base

[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

The pH of a solution after the adding of strong acid

For example the strong acid that 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

the pH of b solution after the adding of strong base

[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

The pH of a solution changes before the adding of strong acid and pH of a solution after the

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

= 1,475

The pH of b solution before the adding of strong acid and pH of b solution after the adding

of strong base = (8,26 - 7,935)

= 0,325

The conclution is buffer solution that the pH is stabile is b solution because the pH changes

only 0,325

66. 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

67. Given : M HC2H3O2 = 0.15 M

M C2H3O2- = 0.25 M

Ka = 1.8 x 10 -5

Question : pH use Kb from acetate ion ?

Answer :

68. Given : pH buffer = 4.98

mol of HCl = 0.05 mol

volume = 1.00 L

mol of HC2H3O2 = 0.15 mol

mol of C2H3O2-

= 0.25 mol

Ka HC2H3O2 = 1.8 x 10-5

Asked : the pH change after 0.050 mol of HCl is added to the buffer

Answer :

HCl(aq) + C2H3O2-

(aq) HC2H3O2 (aq) + Cl- (aq)

m 0.05 mol 0.25 mol 0.15 mol 0

r -0.05 mol -0.05 mol +0.05 mol +0.05 mol

s 0 0.20 mol 0.20 mol 0.05 mol

[H+] = Ka . pH = log [H+]

= 1.8 x 10-5

x = log 1.8 x 10-5

= 1.8 x 10-5

= 4.5

Change of pH = 4.98 4.5 = 0.48

69. Given : pH buffer = 4.98

mol of NaOH = 0.005 mol

volume = 0.50 L

Ka HC2H3O2 = 1.8 x 10-5

Asked : the pH change after 0.005 mol of NaOH is added to the buffer

Answer :

mol of HC2H3O2 = 0.15 M x 0.5 L = 0.075 mol

mol of C2H3O2-

= 0.25 M x 0.5 L = 0.125 mol

NaOH(aq) + HC2H3O2(aq) C2H3O2- (aq) + Cl

- (aq)

m 0.005 mol 0.075 mol 0.125 mol 0

r -0.005 mol -0.005 mol +0.005 mol +0.005 mol

s 0 0.070 mol 0.130 mol 0.005 mol

[H+] = Ka . pH = log [H+]

= 1.8 x 10-5

x = log 9.69 x 10-6

= 9.69 x 10-6

= 5

Change of pH = 5 4.98 = 0.02

70. Given : Kb NH3 = 1.8 x 10-5

We assumed the volume of solution : 1L

M NH3 : 0.25 M

MNH4+

: 0.14 M

Question : pH ?

Answer :

mol of NH3 : 0.25 M x 1 = 0.25 mol

mol of NH4+

: 0.14 M x 1 = 0.14 mol

a. NH3(aq) + H2O(l) NH4+

(aq) + OH-

[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 = 9.5

b. Ka = = 5.56 x 10 -10

[H+] = Ka .

= 5.56 x 10 -10

x

= 5.56 x 10 -10

x 0.56

= 3.11 x 10-10

pH = 10 - log 3.11

= 9.5

71. Given : Buffer = 0.25 M NH3 and 0.14 M NH4+ 1.00 L

pH Buffer : 9.5

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

Answered : the mol of NH3= 0.25 M x 1.00 L = 0.25 mol

The mol of NH4+

= 0.14 M x 1.00 L = 0.14 mol

*If using Kb of NH3

If added 0.020 mol HCl

NH3 (aq) + H+

(aq) NH4+

B 0.25mol 0.020mol 0.14 mol

C -0.02mol -0.02mol +0.02mol

A 0.23mol - 0.16mol

[OH-] = Kb .

= 1.8 x 10-5

x

= 2.6 x 10-5

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+

If added 0.020mol HCl

[H+] = Ka .

= 5.56 x 10 -10

x

= 5.56 x 10 -10

x 0.7

= 3.892 x 10 -10

pH = 10 - log 3.892

pH = 9.41

Change of pH = 9.5 - 9.41 = 0.09

72. Given : 0,02 ml HCl 0,1 M

0,25 M NH3 1,00 L

1,00 L NH4+ 0,14M

Question: the change of pH??

Answer: 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

With NH3

[OH-] = Kb .

= 1,8 x10-5

.

= 3,2 x10-5

pOH = -log 3,2 x 10

-5

= 5 log 3,2

= 5 0,50 = 4,5

pH = 14 - 4,5= 9,5

Mol HCl 0,02 ml x 1,00 M

HCl + NH3 NH4Cl

M 0,020 mmol 0,25 mmol 0,14 mmol

R -0,020 mmol -0,020 mmol +0,020 mmol

S - 0,023 mmol 0,160 mmol

[OH-] = Kb .

= 1,8 x10-5

.

= 2,58 x10-5

pOH = -log 2,58 x 10

-5

= 5 log 2,58

= 5 0,41

= 4,59

pH = 14 - 4,59 = 9,41

the change of pH = 9,5 9,41= 0,09

With NH4+

[H+] = Ka .

= 10-5

.

= 0,56 x10-5

pH = -log 0,56 x 10

-6

= 6 log 5,6

= 6 0,75 = 5,75

[H+] = Ka .

= 10-5

.

= 0,7 x10-5

pOH = -log 0,7 x 10

-5

= 5 log 0,7

= 5 0,15 = 4,84

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

Question : The Change of pH??

Answer :

Mol of KOH 75 mlx 0,1 M = 7,5 mmol

Mol of NH3 200 ml x 0,25 M = 50 mmol

Mol of NH4+

200 ml x 0,14 M = 28 mmol

[OH-] = Kb .

= 1,8 x10-5

.

= 1,78 x10-5

pOH = -log 1,78 x 10

-5

= 5 log 1,78

= 5 0,25 = 4,75

pH = 14 - 4,5 = 9,25

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 .

= 1,8 x10-5

.

= 5,05 x10-5

pOH = -log 5,05 x 10

-5

= 5 log 5,05

= 5 0,70 = 4,30

pH = 14 - 4,30= 9,70

the change of pH is 9,70-9,25 = 0,45

74. Given : 1.0 L of 0.15 M acetic acid (pKa 4.74 = 1,8 x 10-5

)

A buffer pH is 5,00 = 10-5

Question : how many grams od sodium asetate??

Answer :

Mol of asetic acid 1.0 L x 0.15 M= 0,15 mol

[H+] = Ka .

10-5

= 1,8 x10-5

.

10-5

x = 0,27 x 10-5

x = 0,27 mol

Mass of sodium asetate = mol x mr of sodium asetate

massa = 0,27mol x 82

massa = 22,14 gram

75. Given : 1.0 L of 0.12 M formic acid (Pka 3.74 = 1,8 x 10-4

)

a buffer for pH 3.80 = 0,2 x 10-4

Question : how many grm of sodium formate??

Answer :

Mol of formic acid = 1.0 L x 0.12 M = 0,12 mol

[H+] = Ka .

0,2 x 10-5

= 1,8 x10-4

.

10-5

x = 1,08 x 10-4

x = 0,108 mol

Mass of sodium asetate = mol x mr of sodium formate

massa = 0,108mol x 68

massa = 7.344 gram

75. Given : buffer a solution have pH 9.25

Question : what the ratio of NH4Cl to NH3??

Answer:

[H+] = Kb.

5,6 x 10-10

= 1,8 x10-5

.

=

=

76. Known : 500 ml of NH3 0,20 M

Mole of NH3 = V x M

= 500 ml x 0,20

= 100 mmole

pH buffer = 10,00

Kb NH3 = 1,8 x 10-5

Mr NH4Cl = 53,5

Question : mass of NH4Cl = ...?

Answer :

pH = 10 pOH = 4 [OH-] = 10

-4

[OH-] =

10-4

= 10-5

x

Mole of NH4Cl =

Mole of NH4Cl =

Mole of NH4Cl = 18 mmol

Mass of NH4Cl = mole x Mr NH4Cl

= 18 x 53,5

= 963 mgram

= 0,963 gram

77. Known : 125 ml of NH3 0,10 M

Mole of NH3 = V x M

= 125 x 0,1

= 12,5mmol

Kb NH3 = 1,8 x 10-5

pH = 9,15

Question : mass of NH4Cl = ...?

Answer :

pH = 9,15 pOH = 4,85 [OH-] = 1,4 x 10

-5

[OH-] =

1,4 x 10-5

= 10-5

x

Mole of NH4Cl =

Mole of NH4Cl =

Mole of NH4Cl = 16,07mmol

Mass of NH4Cl = mole x Mr NH4Cl

= 16,07 x 53,5

= 859,75mgram

= 0,859 gram

78. Known : 25 ml of HCl 0,1 mol

Mole of HCl = V x M

= 25 x 0,1

= 2,5mmol

= 0,0025mol

Mole of CH3COOH = 0,1mol

Mole of CH3COONa = 0,11mol

Ka CH3COOH = 1,82 x 10-5

Question :

a. The initial of pH =...?

The final of pH =...?

b. the pH if the same amount of HCl solution were added to 125 ml of pure water =...?

Answer :

a. [H+] =

= 10-5

x

= 16,54 x 10-6

= 1,65 x 10-5

pH = 5 log 1,65

= 5 0,21

= 4,79

So, the initial of pH = 4,79

HCl + CH3COONa CH3COOH + NaCl

M 0,0025 0,11 0,1 0

R 0,0025 0,0025 0,0025 0,0025

S 0 0,1075 0,1025 0,0025

[H+] =

= 10-5

x

= 10-5

x

= 1,735 x 10-5

pH = 5 log 1,735

= 5 0,239

= 4,761

So, the final pH = 4,761

b. The pH if the same amount of HCl solution were added to 125 ml of pure water is

same 4,761. Addition of water in the buffer solution, the pH will be constant bacause

the determinant of pH is the number of mol not the concentration of buffer.

79. Known : HCl = 0,15 M

Volume of buffer = 100 ml

pH buffer from no.78 = 4,79

Question :

a. Volume of HCl to make the pH decrease by 0,05 pH unit = ...?

b. volume of the same HCl solution would, if added to 100 ml of pure water, make the

pH decrease by 0,05 pH unit = ...?

Answer :

HCl + CH3COONa CH3COOH + NaCl

M a 0,11 0,1 0

R a a a a

S 0 0,11 - a 0,1 + a a

a. pH buffer from no.78 = 4,79

pH after = 4,79 - 0,05

= 4,74

[H+] = 1,82 x 10

-5

[H+] =

1,82 x 10-5

= 10-5

x

1,82 x 10-5

x (0,11 a) = 10-5 x (0,1 + a)

(2 x 10-6

) - (1,82 x 10-5

a) = (1,82 x 10-6

) + (1,82 x 10-5

a)

(2 x 10-6

) - (1,82 x 10-6

) = (1,82 x 10-5

a) + (1,82 x 10-5

a)

1,8 x 10-7

= 3,64 x 10-5

a

a =

a = 5x 10-3

= 0,005 mol

Mole of HCl = 0,005 mol

Volume of HCl =

=

= 0,033 litre

= 33 ml

b. Amount of HCl which added was same is 33 ml. The buffer solution is added with

water the pH is constant not change.

80. In acid base titration, equivalence point can reach at pH is not 7, its mean that one of

solution is same or its weak base. The indicator choosing is very important to effort the

titration process because to know the end point of titration with the change color of the

solution, with the Ph range that estimated until the indicator choosing is not false.

81. 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. For titrating potassium hydroxide with hydrobromic acid we can use few indicators. The

indicators are metil red indicator, bromtimol blue indicator, and phenolphthalein indicator.

Titrating potassium hydroxide with hydrobromic acid is example of titration strong base and

strong acid. The equivalen point is occur in value pH 7 (neutral). The color of indicators can

be change around the equivalen point. Therefore the change of color in the phenolphthalein

indicator is sharper (more noticeable), so the phenolphthalein indicator is more frequently

used.

83. The quantities of reactans ought to be true at the equivalence point when the equivalence

mole of acid is as same as the equivalence mole of base in the titration.

84. Given : M formic acid : 0.10 M

V formic acid : 50 ml

M NaOH : 0.10 M

Asked : what is the pH at the equivalence point?

Answer :

Mol HCOOH = MolNaOH

M1 . V1= M2 . V2

0.10 . 50 = 0.10 V2

V2 = 50 ml

MolNaOH = M. V

= 0.10 . 50

= 5 mmol

Mol HCOOH = M. V

= 0.10 . 50

= 5 mmol

HCOOH + NaOH NaCOOH + H2O

m 5 5 0 0

r -5 -5 +5 +5

s 0 0 5 5

Ka = 1.8 x 10-5

85. Given: V NH3 = 25 mL

M NH3 = 0.10 mol/L

Kb NH3= 1.8 x 10-5

M HBr = 0.10 mol/L

Asked : pH at equivalence point and 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

86. Given : [HCl] = 0.1000 M

[NaOH] = 0.1000 M

V HCl = 25 mL

MolHCl = 25 ml x 0.1000 M = 2.5 mol

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

Solution :

a. V NaOH = 0 mL

MolNaOH = 0 x 0.1000M = 0

HCl + NaOH NaCl + H2O

M 2.5 mol 0 0 0

R 0 0 0 0

S 2.5 mol 0 0 0

[HCl] =

= 0.1 M

[H+] = valence x [HCl]

=1 x 0.1 M

= 0.1 M

pH = - log [H+]

= - log 0.1

= 1

b. V NaOH = 10.00 ml

MolNaOH = 10.00 ml x 0.1000 M

= 1 mol


M 2.5 mol 1mol 0 0

R 1mol 1mol 1mol 1mol

S 1.5mol 0 1mol 1mol

[HCl] =

= 0.043 M

[H+] = valence x [HCl

= 1 x 0.043 M

= 0.043 M

pH = - log [H+]

= - log 0.043

= 1.37

c. V NaOH = 24.90 mL

MolNaOH = 24.90 mL x 0.1000 M

= 2.49 mol


M 2.5mol 2.49mol

R 2.49mol 2.49 mol 2.49mol 2.49mol

S 0.01mol 0 2.49mol 2.49mol

[HCl] =

= 2.0 x 10-4

M

[H+] = valence x [HCl]

= 1 x (2.0 x 10-4

)

= 2.0 x 10-4

M

pH = - log [H+]

= - log 2.0 x 10-4

= 3.698

d. V NaOH = 24.99 mL

MolNaOH = 24.99 ml x 0.1000 M

= 2.499 mol


M 2.5mol 2.499mol 0 0

R 2.499mol 2.499mol 2.499mol 2.499mol

S 1x10-3

mol 0 2.499mol 2.499mol

[HCl] =

= 2.0 x 10-5

M

[H+] = valence x [HC]

=1 x (2.0x10-5

)

= 2.0 x 10-5

M

pH = - log [H+]

= - log 2.0 x 10-5

= 4.7

e. V NaOH = 25.00 mL

MolNaOH = 25 ml x 0.1000 M

= 2.5 mol


M 2.5mol 2.5mol 0 0

R 2.5mol 2.5mol 2.5mol 2.5mol

S 0 0 2.5mol 2.5mol

Titration in equivalent point

pH = 7 (neutral)

f. V NaOH = 25.01 mL

MolNaOH = 25.01 ml x 0.1000M

= 2.501 mol


M 2.5mol 2.501mol 0 0


S 0 1x10-3

mol 2.5mol 2.5mol

[NaOH] =

= 1.9996 x 10-5

M

[OH-] = valence x [NaOH]

= 1 x (1.9996 x 10-5

)

= 1.9996 x 10-5

M

pOH = - log [OH-]

= - log 1.9996 x 10-5

= 4.699

pH = 14 - 4.699

= 9.3

g. V NaOH = 25.10 mL

MolNaOH = 25.10 ml x 0.1000 M

= 25.1 mol


M 2.5mol 2.51mol 0 0

R 2.5mol 2.5 mol 2.5mol 2.5 mol

S 0 0.01mol 2.5mol 2.5 mol

[NaOH] =

= 1.996 x 10-4

M


= 1 x (1.996 x 10-4

)

= 1.996 x 10-4

M

pOH = - log [OH-]

= - log 1.996 x 10-4

= 3.7

pH = 14 - 3.7

= 10.3

h. V NaOH = 26.00 mL

MolNaOH= 26 ml x 0.1000 M

= 2.6 mol


M 2.5mol 2.6mol 0 0

R 2.5mol 2.5 mol 2.5mol 2.5mol

S - 0.1 mol 2.5mol 2.5mol

[NaOH] =

= 1.96 x 10-3

M


= 1 x (1.96 x 10-3

)

= 1.96 x 10-3

M

pOH = - log [OH-]

= - log 1.96 x 10-3

= 2.7

pH = 14 - 2.7

= 11.3

i. V NaOH = 50.00 mL

MolNaOH = 50 ml x 0.1000 M

= 5 mol


M 2.5mol 5mol 0 0


S 0 2.5mol 2.5mol 2.5mol

[NaOH] =

= 0.033 M


= 1 x (0.033)

= 0.033 M

pOH = - log [OH-]

= - log 0.033

= 1.477

pH = 14 - 1.477

= 12.523

87. pKa = 1. 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

88. 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 are11

a. 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

b. 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

c. 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

004 2024 Nurullah Aulia Sugiarti Rombel 01 Tugas Ke 01

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