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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
๐พ๐ = ๐ป3๐
+ [๐๐ปโ]
[๐ป2๐][๐ป2๐]
Oleh karena konsentras sehingga [๐ป2๐]2 harganya juga tetap (konstan), maka
hasil Kc dengan [๐ป2๐]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 :
๐ป2๐ [ ๐ป+] + [๐๐ปโ]
๐พ๐ = ๐ป+ [๐๐ปโ]
[๐ป2๐]
[H+] [OH
-] = Kw
[OH-] = [H
+]
[H+] = ๐พ๐ค
= 2,4 ๐ฅ 10โ14
= 1,54 x 10โ7
pH = - log [H+]
= - log ( 1,54 x 10โ7)
= 7 โ log 1,54
= 7 โ 0,1875
= 6,8124
pOH = - log [OH-]
= - log (1,55 x 107)
= - log 1,55 + ( - log 10-7
)
= 7 โ 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 โ 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 20ยฐC 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
20ยฐC. Calculate also the pD and the pDO.
Answer :
D2O D+
+ OD-
๐พ๐ = ๐ท+ ๐๐ทโ
๐ท2๐
KW = [OD-] [D
+]
[OD-] = [D
+] = ๐พ๐ค
= 8,9 ๐ฅ 10โ16
= 2,98 x 10-8
pD = - log [D+]
= - log (2,98 x 10-8
)
= - log 2,98 + ( - log 10-8
)
= 8 โ log 2,98
= 7,53
pOD = - log [OD-]
= - log (2,98 x 10-8
)
= - log 2,98 + ( - log 10-8
)
= 8 โ 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
-]
10โ14 = [H+] (24 ร 10โ4)
[H+] =
10โ14
24 ร10โ4
[H+] = 4,167 ร 10โ12
b) Kw = [H+] [OH
-]
10โ14 = [H+] (1.4 x 10โ5)
[H+] =
10โ14
1.4 x 10โ5
[H+] = 7,14 ร 10โ10
c) Kw = [H+] [OH
-]
10โ14 = [H+] (5.6 x 10โ9)
[H+] =
10โ14
5.6 x 10โ9
[H+] =0,1785 ร 10โ5
d) Kw = [H+] [OH
-]
10โ14 = [H+] (4.2 x 10โ13)
[H+] =
10โ14
4.2 x 10โ13
[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 โ 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 itโs 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 โ log 1,45
= 5 โ 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 โ 1.62
PH = 12.38
b. POH = 5 โlog 1.4
POH = 4.85
PH = 14- 4.85
= 9.15
c. POH = - log 5.6 x 10-9
POH = 9 โ log 5.6
= 8.25
PH = 14 โ 8.25
= 5.75
d. POH = 13 โ log 4.2
POH = 12.38
PH = 14 โ 12.38
= 1.62
Soal no.6
a. (3.5 x 10-8
)
PH = - log 3.5 x 10-8
PH = 8 โ log 3.5
= 7.45
b. (0.0065)
PH = 6.5 x 10-2
= 2 โ log 6.5
= 1.18
c. 2.5 x 10-13
PH = - log 2.5 x 10 -13
PH = 13 โ log 2.5
PH = 12.6
d. 7.5 x 10 -5
M
PH = - log 7.5 x 10 -5
PH = 5 โ 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 โ 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 โ log 5
= 3 โ 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 = ๐๐
๐๐x
1000
๐
=6,0
40x
1000
๐
= 0,15
NaOH = Na+
+ OH-
OH- = 1 . 0,15
= 0,15
pOH = 1 โ log 1,5
pH = 14 โpOH
= 14 โ ( 1 โ 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 = ๐๐
๐๐ x
1000
๐
= 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 โ pOH
= 14 โ 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 โ 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: ๐) HNO2 aq + H2O l NO2โ
aq + H3O+ aq
๐) H3PO4 aq + H2O l H2PO4โ
aq + H3O+ aq
๐) HAsO4(aq )2โ + H2O l AsO4
3โ aq + H3O+
aq
๐) 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-
๐พ๐ = ๐ป+ [๐๐2
โ]
[๐ป๐๐2]
b. H3PO4 H+ + H2PO4
2-
๐พ๐ = ๐ป+ [๐ป2๐๐4
2โ]
[๐ป3๐๐4]
c. HAsO42-
H+ + AsO4
3+
๐พ๐ = ๐ป+ [๐ด๐ ๐4
3+]
[๐ป๐ด๐ ๐42โ]
d. (CH3)3NH+ H+ + (CH3)3N
๐พ๐ = ๐ป+ [ ๐ถ๐ป3 3๐]
[ ๐ถ๐ป3 3]๐๐ป+
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
+
๐พ๐ = ๐๐ปโ [ ๐ถ๐ป3 3๐๐ป+]
[ ๐ถ๐ป3 3]๐
b. AsO43-
+ H2O OH- + HAsO4
2-
๐พ๐ = ๐๐ปโ [๐ป๐ด๐ ๐4
2โ]
[๐ด๐ ๐43โ]
c. NO2- + H2O OH
- + HNO2
๐พ๐ = ๐๐ปโ [๐ป๐๐2]
[๐๐2โ]
d. (CH3)2N2H2 + H2O OH- + (CH3)3N2H3
+
๐พ๐ = ๐๐ปโ [ ๐ถ๐ป3 3๐2๐ป3
+]
[ ๐ถ๐ป3 3]๐2๐ป2
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 itโs Ka !
Answer :
C6H5CO2H(aq ) + H2O(l) โ C6H5CO2(aq )โ + H3O(aq )
+
๐พ๐ = [ C6H5CO2
โ ][ 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
-
๐พ๐ = ๐๐ปโ [๐ถ6๐ป5๐ถ๐2๐ป ๐ถ6๐ป5๐ถ๐2
โ]
[๐ถ6๐ป5๐ถ๐2โ]
25. The pKa of HCN is 9.21 and that of HF is 3.17. Which is the strong Bronsted base CNโ
or Fโ?
Answer :
pKa HCN = 9,21
-log Ka = 9,21
= 10 โ 0,79
= 10 โ log 6,16
Ka = 6,16 x 10-10
pKa HF = 3,17
-log Ka = 3,17
= 4 โ 0,83
= 4 โ 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 = 10โ14
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 = 10โ14
1,0 ๐ฅ 10โ10
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 = 10โ14
1,8x10โ12
= 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 = 10โ14
4,4 x10โ4
= 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 = 10โ14
1,4 ๐ฅ 10โ4
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
โ]+ [H3O+]
[HIO 3]
IO3- + H2O HIO3+ OH
-
Kb = [HIO 3]+ [OH โ]
[IO 3โ]
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+] = ๐พ๐ ๐ฅ ๐
3,8 x 10-2
= ๐พ๐ ๐ฅ 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 โ log 14,44
pKa = 3 โ 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 :
โ = ๐พ๐
๐
(0, 11)2 =
๐พ๐
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 :
๐๐๐ป = 14 โ 11,86
= 2,14
โ๐๐๐ ๐๐ปโ = 3 โ 0,86
= 3 โ log 7,24
๐๐ปโ = 7,24 . 10โ3
๐๐ปโ = ๐พ๐ . ๐
๐๐ปโ 2 = ๐พ๐ . ๐
7,24 . 10โ3 2 = ๐พ๐ . 0.1
5,24. 10โ4 = ๐พ๐
๐๐พ๐ = โ๐๐๐๐พ๐
= 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 :
๐๐๐ป = 14 โ 10,12
= 3,88
โ๐๐๐ ๐๐ปโ = 4 โ 0,12
= 4 โ log 1,138
๐๐ปโ = 1,138 . 10โ4
๐๐ปโ = ๐พ๐ . ๐
๐๐ปโ 2 = ๐พ๐ . ๐
1,138 . 10โ4 2 = ๐พ๐ . 0,15
1,158 . 10โ7 = ๐พ๐
๐๐พ๐ = โ log ๐พ๐
= 6,936
36. Refer to data in the preceding question to calculate the percentage ionization of the base
in 0.15 M HONH2.
Answer : ๐ผ = ๐พ๐
๐ . 100%
= 1,158 . 10โ7
0,15. 100%
= 0,0878%
37. What is the pH of 0.125 M pyruvic acid ? Itโs Ka is 3.2 x 10-3
Answer :
๐ป+ = ๐พ๐ . ๐
= 3,2 . 10โ3 . 0,125
= 0,02
๐๐ป = โ log 2 . 10โ2
= 1,69
38. What is pH of 0.15 M HN3 ? for HN3, Ka = 1.8 x 10-5
Answer :
๐ป+ = ๐พ๐ . ๐
= 1,8 . 10โ5 . 0,15
= 1,643 . 10โ3
๐๐ป = โ log 1,643 . 10โ3
= 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 :
๐ป+ = ๐พ๐ . ๐
= 1,8. 10โ2 . 1
= 1,3416 . 10โ1
๐๐ป = โ log 1,3416 . 10โ1
= 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 :
๐ผ = ๐พ๐
๐. 100%
= 1,3. 10โ10
0,050. 100%
= 5, 099 . 10โ3%
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 :
๐๐พ๐ = 5,79
= 6 โ 0,21
๐พ๐ = 6 โ log 1,62
๐พ๐ = 1,62 . 10โ6
๐๐ปโ = ๐พ๐ . ๐
๐๐ปโ = 1,8 . 10โ4
๐๐๐ป = 4 โ ๐๐๐1,8
= 3,745
๐๐ป = 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 โ 0,04
-log Kb = 5 โ log 1,09
Kb = 1,09 x 10-5
[OH-] = ๐พ๐ ๐ฅ ๐๐
[OH-] = 1,09 ๐ฅ 10โ5๐ฅ 0,2
[OH-] = 2,18 ๐ฅ 10โ6
[OH-] = 1,47 x 10
-3
pOH = - log [OH-]
pOH = - log (1,47 x 10-3
)
pOH = 3 โ 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 โ log 2,88
[H+] = 2,88 x 10
-3
[H+] = ๐พ๐ ร ๐๐
[H+]
2 = Ka x Ma
8,29 x 10-6
= 1,8 x 10-5
x Ma
Ma = ๐,๐๐ ๐ฑ ๐๐โ๐
๐,๐ ๐ฑ ๐๐โ๐
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 = ๐ฒ๐
๐ฒ๐ Ka > Kb , maka larutan ini bersifat ASAM
= ๐๐โ๐๐
๐,๐๐ ๐ ๐๐โ๐
= 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
10โ14 = 1,9 . 10โ10 . Ka
Ka = 5,26 . 10โ5
Ka > Kb
5,26 . 10โ5 > 1,9 . 10โ10 (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-]
= ๐๐ค
๐๐ . ๐
= 10โ14
1,8 ๐ฅ 10โ5 . 2.10โ1
= 2. 10โ10
[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-]
= ๐๐ค
๐๐ . ๐
= 10โ14
1,8 ๐ฅ 10โ5 . 4 ๐ฅ 10โ2
= 2,22. 10โ11
[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+]
=
๐๐ค
๐๐ . ๐
= 10โ14
4,4.10โ4 . 1,5.10โ1
= 3,4 10 โ10
= 1,8 x 10-5
pH = - log [H+]
pH = - log ( 1.8 x 10-5
)
pH = 5 โ 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 โ 0,72
-log [H+] = 5 โ log 5,25
[H+]
= 5,25 x 10-5
[H+]
= ๐๐ค
๐๐ . ๐บ
5,25 x 10-5
= 10โ14
๐๐ ๐ฅ 1,5 ๐ฅ 10โ1
(5,25 x 10-5
)2 =
10โ14
๐พ๐ x 1,5 x 10
-1
2,75 x 10-9
x Kb = 1,5.10โ15
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+]
= ๐๐ค
๐๐ . [๐บ]
0,84 . 10-6
= 10โ14
1,8 .10 โ5 ๐ฅ [๐บ]
Masing-masing ruas dikuadratkan
0,7056 . 10-12
= ๐๐โ๐๐
๐,๐ .๐๐โ๐ x [G]
[G] = 1,27008 .10-3
n = M.V
= 1,27008 .10-3
mol/L . 1 L
= 1,27008 .10-3
mol
n = ๐
๐ด๐ด
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
-] = 10โ14
10โ5 [Y-] = 10โ14
[Y-] = 10โ9
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 โ log 7.41
Kb = 7.41 x 10-7
Kw = Ka x Kb
10-14
= Ka x 7.41 x 10-7
Ka = ๐๐โ๐๐
๐.๐๐ ๐ ๐๐โ๐
Ka = 0.135 x 10-7
[H+] = ๐พ๐ ๐ฅ ๐
[H+] = 0.135 x 10โ7 ๐ฅ 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-๐๐ข+. What is the calculate pHof a 0,15 M
solution of H-๐๐ข+ ? Its pKa is 8,52 at 25 0C.
Answer :
pKa = 8.52
Ka = 9 โ log 3.01
Ka = 3.01 x 10-9
[H+] = ๐พ๐ ๐ฅ ๐
[H+] = 3.01 x 10โ9 ๐ฅ 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)
ฮฑ = ๐พ๐
๐
ฮฑ = 7.2 ๐ฅ 10โ4
0.15
ฮฑ = 0.069
ฮฑ = 6.9 %
Ka = 7.2 x 10-4
[H+] = ๐พ๐ ๐ฅ ๐
[H+] = 7.2 x 10โ4 ๐ฅ 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 :
ฮฑ = ๐พ๐
๐
ฮฑ = 1.8 ๐ฅ 10โ5
0.001
ฮฑ = 0.018
ฮฑ = 1.8 %
Ka = 1.8 ๐ฅ 10โ5
[H+] = ๐พ๐ ๐ฅ ๐
[H+] = 1.8 ๐ฅ 10โ5 ๐ฅ 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 โ (HCl) (H
+) โ Kw = 0
(H+)2 โ 10
-7 (H
+) โ 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
๐พ๐ = ๐ป+ (๐๐4
โ)
(๐ป๐๐4โ)
b. b. Dengan menggunakan persamaan kuadrat, dapat memecahkan persamaan
berikut:
x2+10
-2x-10
-4 = 0
Dengan menggunakan rumus kuadrat
x = โ๐ ยฑ ๐2โ4๐๐
2๐
= โ10โ2 ยฑ 10โ22โ4.10โ4
2
=โ10โ2ยฑ10โ2
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 โx +x +x
S (0,001-x) x x
Ka = H+ [SO 4โ]
[HSO 4โ] = 1.0x10
-2
๐ฅ2
0,001โx= 1.0x10
-2
Dengan menggunakan pendekatan 0,001-x = 0,001
๐ฅ2
0,001โxโ
๐ฅ2
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 10โ3
1,0 ๐ฅ 10โ2 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 โ 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 โ 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 โcarbonateโ buffer in blood)
b. NaH2PO4 and Na2HPO4 (the โphosphateโ 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
๐๐ปโ = ๐พ๐ ๐๐๐ ๐
๐๐๐๐๐
= 1,8 ร 10โ5 1๐
0,1๐
= 1,8 ร 10โ4
๐๐๐ป = 4 โ ๐๐๐1,8
= 4 โ 0,2553
= 3,7447
๐๐ป = 14 โ 3,7447
= 10,2553
๐๐ข๐๐๐๐ 2
Basa : NH3 0,1 M
Asam konjugasi : NH4Cl 1 M
๐๐ปโ = ๐พ๐ ๐๐ ๐๐
๐๐๐๐๐
= 1,8 ร 10โ50,1 ๐
1 ๐
= 1,8 ร 10โ6
๐๐๐ป = 6 โ ๐๐๐1,8
= 5,7447
๐๐ป = 14 โ 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 โ H+
+ C2H3O2โ
(buffer asam)
[H+] = Ka ร
[CH 3COOH ]
[CH 3COO โ]
= 1,8 ร 10โ5
ร 0,15 ๐
0,25 ๐
= 1,8 ร 10โ5
ร 0,6
= 1,08 ร 10โ5
pH = โlog [H+]
= โlog 1,08 ร 10โ5
= 5 โ log 1,08
= 5 โ 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
๐.๐๐
๐.๐๐ 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 = โlog Kb x ๐๐๐๐
๐๐๐๐๐
= - log 1.8 x 10-5
x ๐ ๐๐๐๐๐
๐.๐๐ ๐๐๐๐๐
= 3.7
pH = 14 โ pOH
= 14 โ 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 = โlog Kb x ๐๐๐ ๐
๐๐๐๐๐
= - log 1.8 x 10-5
x 0.005 ๐๐๐
0.045 ๐๐๐
= 5.69
pH = 14 โ pOH
= 14 โ 5.69
= 8.31
Perubahan pH = 10.25 โ 8.31
= 1.94
Buffer 2
pOH = โlog Kb x ๐๐๐ ๐
๐๐๐๐๐
= - log 1.8 x 10-5
x 0.1 ๐๐๐
1 ๐๐๐
= 5.74
pH = 14 โ pOH
= 14 โ 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
pOH = โlog Kb x ๐๐๐ ๐
๐๐๐๐๐
= - log 1.8 x 10-5
x 0.005 ๐๐๐
0.495 ๐๐๐
= 6.74
pH = 14 โ pOH
= 14 โ 8.74
= 7.26
Perubahan pH = 8.25 โ 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 :
๐. ๐๐ป3 + ๐ป2๐ โ ๐๐ป4+ + ๐๐ปโ
๐พ๐ = ๐๐ป4
+ ๐๐ปโ
๐๐ป3
1.8๐ฅ10โ5 =0.14 ๐๐ปโ
0.25
๐๐ปโ = 3.214๐ฅ10โ5
๐๐๐ป = โ log ๐๐ปโ
= โ๐๐๐3.214 โ 10โ5
= 5 โ log 3.214
๐๐ป = 9 + log 3.214
= 9.507
b. ๐พ๐ =๐พ๐ค
๐พ๐
=1๐ฅ10โ14
1.8๐ฅ10โ5
= 5.5๐ฅ10โ10
๐ป+ = ๐พ๐๐ฅ๐
๐
= 5.5๐ฅ10โ10๐ฅ0.14
0.25
= 3.08๐ฅ10โ10
๐๐ป = โ log ๐ป+
= โ log 3.08 โ 10โ10
= 10 โ 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 :
๐๐ป3 + ๐ป+ โ ๐๐ป4+
๐ โถ 0.25 0.02 0.14
๐ โถ 0.02 0.02 0.02
๐ โถ 0.23 โ 0.16
๐๐ปโ = ๐พ๐
๐
๐
= 1.8๐ฅ10โ50.23
0.16
= 2.5875๐ฅ10โ5
๐๐๐ป = 5 โ log 2.5875
= 4.58
๐๐ป = 14 โ 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 :
๐๐ป4+ + ๐๐ปโ โ ๐๐ป3 + ๐ป2๐
๐ โถ 0.028 0.0075 0.05 โ
๐ โถ 0.0075 0.0075 0.0075 โ
๐ โถ 0.0205 โ 0.0575 โ
๐๐ปโ = ๐พ๐
๐
๐
= 1.8๐ฅ10โ50.0575
0.0205
= 2.804๐ฅ10โ5
๐๐๐ป = 5 โ log 2.804
= 4.55
๐๐ป = 14 โ 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 โ 0,26 = - log Ka
Ka = 5 โ log 1,82
Ka = 1,82 x 10-5
Larutan Buffer memiliki pH = 5, maka [H+] = 10
-5
[H+] = Ka x
๐ .๐๐ ๐๐
๐ .๐๐๐ ๐ ๐๐๐๐๐ข๐๐๐ ๐๐๐ฆ๐
10-5
= 1,82 x 10-5
x 0,15
๐ .๐๐๐ ๐ ๐๐๐๐๐ข๐๐๐ ๐
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 โ 0,26 = - log Ka
Ka = 4 โ 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
๐ .๐๐ ๐๐
๐ .๐๐๐ ๐ ๐๐๐๐๐ข๐๐๐ ๐๐๐ฆ๐
1,6 x 10-4
= 1,82 x 10-4
x 0,12
๐ .๐๐๐ ๐ ๐๐๐๐๐ข๐๐๐ ๐
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
๐ .๐๐๐ ๐
๐ .๐๐ ๐๐ ๐๐๐๐๐ข๐๐๐ ๐๐๐ฆ๐
1,78 x 10-5
= 10-5
x ๐ .๐๐ป3
๐ .๐๐ป4+
10โ5
1,78 ๐ฅ 10 โ5 = ๐ . ๐๐ป4+
๐ .๐๐ป3
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
๐ .๐๐๐ ๐
๐ .๐๐ ๐๐ ๐๐๐๐๐ข๐๐๐ ๐๐๐ฆ๐
10-4
= 1,8 x 10-5
x 0,1
๐ .๐๐ ๐๐ ๐๐๐๐๐ข๐๐๐ ๐
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
๐ .๐๐๐ ๐
๐ .๐๐ ๐๐ ๐๐๐๐๐ข๐๐๐ ๐๐๐ฆ๐
1,4 x 10-5
= 1,8 x 10-5
x 0,0125
๐ .๐๐ ๐๐ ๐๐๐๐๐ข๐๐๐ ๐
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+] = 1,8x10โ5 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+] = 1,8x10โ5 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+] = 1,8x10โ5 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+] = 1,8X10โ5 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 โ 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 10โ5 = 1,8X10โ5 x 0,02 + 0,15X mol
0,022 โ 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 โ log 1,836
[H+] = 1,836 x 10
-5
[H+] = Ka .a
g
[H+] = 1,8x10โ5 x 0,035 + 0,15X mol
0,007 + 0,15X mol
1,836 x 10-5
= 1,8x10โ5 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 โ 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 โ 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 โ 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.
[OHโ] = Kw
Ka [M]
[OHโ] = Kw
Ka
mol
V campuran
[OHโ] = 10โ14
1,8 x 10โ5 5 x 10โ3
10โ1
= 2,77 ๐ฅ 10โ11
= 5,26 x 10-6
[OH-] = -log [OH
-]
= -log 5,26 x 10-6
= 5,279
pH = 14 โ 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 โ 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 = ๐
๐ ( pKw - pKa - log G )
= ๐
๐( 14 - 3.745 - log
๐ ๐๐๐๐
๐๐๐ ๐๐ )
= ๐
๐ ( 14 - 3.745 โ (- 1.301 )
=๐
๐ ( 14 โ 3.745 + 1.301 )
= 5.278
Indikator yang paling tepat digunakan adalah indikator Metil merah yang memiliki trayek
pH 4,4 โ 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 = ๐ฆ๐จ๐ฅ
๐ ๐๐๐ฆ๐ฉ๐ฎ๐ซ๐๐ง
= ๐,๐ . ๐๐โ๐
๐,๐ .๐๐โ๐
= 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
HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
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 . 10โ3
3,5 .10โ2
= 4,2857 x 10-2
M
[H+] = M x valensi [H
+]
= 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
HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
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
= 10โ5
4,99 .10โ2
= 2 x 10-4
M
[H+] = M x valensi [H
+]
= 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
HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
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
= 10โ6
4,99 .10โ2
= 2 x 10-5
M
[H+] = M x valensi [H
+]
= 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
HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
mol 2,5 x 10-3
mol - -
Bereaksi : 2,5 x 10-3
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
HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
mol 2,501 x 10-3
mol - -
Bereaksi : 2,5 x 10-3
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
= 10โ6
5,001.10โ2
= 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 โ pOH
= 14 โ 4,701
= 9,299
g. 25,10 mL = 0,0251 L
HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
mol 2,51 x 10-3
mol - -
Bereaksi : 2,5 x 10-3
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
= 10โ5
5,01.10โ2
= 1,99 x 10-4
M
[OH-] = M x valensi [OH
-]
= 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 โ pOH
= 14 โ 3,701
= 10,299
h. 26 mL = 0,026 L
HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
mol 2,6 x 10-3
mol - -
Bereaksi : 2,5 x 10-3
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
= 10โ4
5,1.10โ2
= 1,96 x 10-3
M
[OH-] = M x valensi [OH
-]
= 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 โ pOH
= 14 โ 2,707
= 11,293
i. 50 mL = 0,05 L
HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
mol 5 x 10-3
mol - -
Bereaksi : 2,5 x 10-3
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 . 10โ3
7,5.10โ2
= 3,33 x 10-2
M
[OH-] = M x valensi [OH
-]
= 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 โ pOH
= 14 โ 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 . 10โ5 . 10โ1
= 1,8 . 10โ6
= 1,34 . 10-3
pH = -log [H+]
= -log 1,34 . 10-3
= 2,87
b.
CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
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 . 10โ3
10โ3
= 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 . 10โ3
1,25 . 10โ3
= 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
CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)
Mula-mula : 2,5 x 10-3
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. ๐๐ปโ = ๐พ๐ . ๐
= 1,8 . 10โ5 . 0,1
= 1,3 . 10โ3
๐๐๐ป = โ log 1,3 . 10โ3
= 3 โ log 1,3
= 2,8860
๐๐ป = 14 โ 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 . 10โ5 . 1,5
1
=2,7 . 10-5
๐๐๐ป = โ log 2,7 . 10โ5
= 5 โ log 2,7
= 4,5686
๐๐ป = 14 โ 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 . 10โ5 . 1,25
1,25
=1,8 . 10-5
๐๐๐ป = โ log 1,8 . 10โ5
= 5 โ log 1,8
= 4,74
๐๐ป = 14 โ 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-] =
๐พ๐ค
๐พ๐ . [๐๐๐๐๐]
= 10โ14
10โ5. [
2,5
50]
= 10โ9
1,8. [
1
20]
= 1
36. 10โ9
= 0,0277 . 10 โ9
= 27,7 . 10 โ12
= 5,19 . 10โ6
๐๐๐ป = โ log 5,19. 10โ6
= 6 โ log 5,19
= 5,284
๐๐ป = 14 โ 5,284
= 10,715
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