materi olimpiade, OSN,OSK. semoga bermanfaat

1. KESETIMBANGAN
ASAM & BASA
SMAN PINTAR KUANSING
Oleh : Ahmad mulya

2. Senyawa paling banyak diproduksi
• Sulfuric acid = H2SO4 = Asam sulfat
• Hydrochloric acid = HCl = Asam Klorida
• Nitric acid = HNO3 = Asam Nitrat
• Sodium Hydroxide = NaOH = Basa Natrium
hidroksida
• Calcium hydroxide = Ca(OH)2 = Basa Kalsium
hidroksida
• Ammonia = NH3 = bakal Basa NH4OH

3. Indikator
• Vinegar (cuka) = bersifat asam
• Lemon juice (air jeruk) = bersifat asam
• Air teh = cenderung basa
• Sari jeruk mengubah sifat basa air teh
• Buktinya warna air teh dari coklat menjadi kuning

4. • Senyawa spt dalam teh yang bisa berubah dari
coklat menjadi kuning, ketika terjadi perubahan
ke-asam/basa-an, bisa disebut sebagai indikator
• Indikator menjadi penanda bahwa telah terjadi
perubahan kimia dalam larutan
• Indikator sintetis: phenolphtalein, methylene
blue, bromokresol, dll.

5. Contoh asam yang populer
 Cuka = larutan yang mengandung acetic acid =
CH3COOH = HC2H3O2 = asam cuka

6. Sifat Asam
• Dalam air melepaskan ion H+
• CH3COOH  CH3COO- + H+
• Asam + Logam  Garam + Hidrogen
• CH3COOH + Mg  (CH3COO)2Mg + H2
• Sifat mampu melepaskan Hidrogen ini yang
mendasari sifat asam
• Asam mengubah kertas litmus menjadi merah

7. Citric acid = Asam Sitrat
• Air jeruk = larutan mengandung citric acid =
H3C6H5O7= asam sitrat

8. Sifat Basa
• Dalam air melepaskan ion OH-
• NaOH  Na+ + OH-
• Basa+ Asam  Garam + Air
• NaOH + CH3COOH  CH3COONa + H2O

9. Teori awal Asam Basa
• Tahun 1800-an awal: tiap molekul asam
mengandung minimal satu atom H.
• Th 1887 Svante Arrhenius (Bapak teori ionisasi):
atom H berhubungan dengan sifat keasaman.
Asam = donor H+
• HCl  Cl- + H+
• CH3COOH  CH3COO- + H+ (dalam H2O)
• CH3COOH  CH3COO- + H3O+

10. Listrik
Walau Terlarut
CH3COOH maupun HCl
Pelarut = Benzene = C6H6  Nonelectrolyte

11. Listrik
Terlarut 0,5 M CH3COOH
Asam Lemah
Pelarut = Air Larutan ion dalam air =
electrolyte

12. Listrik
Terlarut 0,5 M HCl
Asam Kuat
Pelarut = Air Larutan ion dalam air =
electrolyte

13. Asam sbg elektrolit
• HCl Cl- + H+
• HCl dalam H2O (air) = strong electrolyte
• CH3COOH CH3COO- + H+
• CH3COOH dalam H2O (air) = weak electrolyte
• HCl dalam C6H6 (benzene) = non electrolyte

14. Basa sebagai elektrolit
• NaOH Na+ + OH-
• NaOH dalam H2O (air) = strong
electrolyte
• NH4OH NH4+ + OH-
• NH4OH dalam H2O (air) = weak electrolyte
• NH4OH dalam C6H6 (benzene) = non
electrolyte

15. Konsep Asam dan Basa
• Svante Arrhenius (1887):
Asam = pemberi H+
Basa = penerima H+
• Gilbert N. Lewis (1916):
Asam  mendapatkan muatan –
Basa  mendapatkan muatan +
• Johannes N. Bronsted & Thomas M. Lowry
(1923):
Asam = pemberi proton
Basa = penerima proton

16.  Arrhenius: HCl asam
 Lewis:  HCl juga asam
 Bronsted & Lowry: HCl = juga asam
 Yang berbeda alasannya
 Saling melengkapi

17. Konstanta Ionisasi :
Asam (Ka) dan Basa (Kb)
 HCl  H+ + Cl-
[H+] [Cl-]
Ka = 107 >102
[HCl]
• Maka HCl digolongkan Strong Acid (asam kuat)

18. Ka Asam Lemah
 CH3COOH  CH3COO- + H+
[CH3COO- ] [H+ ]
Ka = = 1,8 x 10-5
[CH3COOH ]
10-8 < Ka CH3COOH < 10-3
• Maka CH3COOH digolongkan Weak Acid (asam lemah)

19. Strong Acid (Ka > 102)
• Perchloric acid = HClO4
• Sulfuric acid = H2SO4
• Iodide acid = Hydrogen Iodide = HI
• Bromide acid = Hydrogen Bromide = HBr
• Chloride acid = Hydrogen Chloride = HCl
• Nitric acid = HNO3

20. Weak Acid (10-8 < Ka < 10-3)
 Acetic acid = CH3C00H
 Carbonic acid = H2CO3
 Hydrogen Sulfide = H2S
 Nitrous acid = HNO2
 H3PO4 , H2SO3

21. Ka Meningkat seiring Bil. Oksidasi
Acid Formula Oxidatio Ka
n
Number
Hypochlorous HClO 1+ 5 x 10-8
Chlorous HClO2 3+ 1 x 10-2
Chloric HClO3 5+ 1 x 1010
Perchloric HClO4 7+ 1 x 1011
Catatan: Ka HCl 107

22. Ka Air
 H2O  H+ + OH-
[H+] [OH- ]
Ka = = 1 x 10-14
[H2O]
Pada [H2O]=1 mole/L,
Didapatkan [H+] [OH- ]= 1 x 10-14 mole/L
Karena [H+] = [OH-] maka [H+] [OH-] = [H+] [H+] = [H+]2
Jadi [H+]2 = 1 x 10-14
dan [H+] = (1 x 10-14) = 1 x 10-7
dipermudah penulisannya: - log 1 x 10-7 = - log 10-7 = -(-
7) = 7
Maka pH air murni = 7

23. Selanjutnya Ka (a=acid) Air menjadi Kw (w=water)
Kw = [H+] [OH- ] = 10-14
Kw = [H+] [H+ ] = 10-14
[H+ ]2 = 10-14
[H+ ] = 10-7
pH = -Log[H+ ] = -Log(10-7)
pH air = 7
Kesetimbangan ion-ion dalam larutan didasarkan pada Kw air ini

24. Ka Air  Kw
 Air murni terurai sebagian  H+ maupun OH-
• Dalam suatu larutan ------dengan pelarut air,
• ada juga ion H+ maupun OH-
• Jika [H+] > [OH-], sifat asam,  pH < 7
• Jika [H+] = [OH-], sifat netral, pH = 7
• Jika [H+] < [OH-], sifat basa,  pH > 7
• Range pH
• pH=1....................pH=7....................pH=14
• Asam...................Netral...................Basa

25. pH (Potential of Hydrogen)
 pH = - log [H+]
 Problem:
 Suatu larutan, volume 200 mL, diukur dg alat pH-
meter ternyata pH-nya = 5
 Jika larutan diencerkan 10 x menjadi 2000 mL,
berapa pH-nya sekarang?

26. • pH = - log [H+]
• 5 = - log [H+]
• 5 = - (-5)
• 5 = - (log 10-5)
• Jadi [H+] = 10-5 mole/Liter

27. • Diencerkan 10 x, 
• [H+] menjadi 10-5 x 10-1 mole/Liter 
• = 10-6 mole/Liter
• pH sekarang = - (log 10-6) = -(-6) = 6
• Pengenceran membuat pH mendekati netral

28. pH Asam Kuat
• Berapa pH dari larutan 0,01 M HCl?
• HCl  H+ + Cl-
• Dalam air, Asam Kuat diasumsikan terdisosiasi
sepenuhnya menjadi ion-ion, maka:
• [H+] = [HCl] = 0,01 mole/L = 10-2 mole/L
• pH = -Log(10-2) = -(-2) = 2

29. pH Basa Kuat
 Berapa pH dari larutan 0,01 M NaOH?
 NaOH  Na+ + OH-
 Dalam air, Basa Kuat diasumsikan terdisosiasi
sepenuhnya menjadi ion-ion, maka:
 [OH-] = [NaH] = 0,01 mole/L = 10-2 mole/L

30.  Rumus kesetimbangan ion dalam air:
 Kw = [H+] [OH-] = 10-14
 [H+] (10-2) = 10-14
 [H+] = 10-14/(10-2) = 10-12
 pH = -Log(10-12) = -(-12) = 12

31. pH beberapa zat
Nama Zat pH
Gastric juice (cairan lambung) 1 – 3 (Sangat asam)
Lemons (bangsa Jeruk) 2,2 – 2,4 (Sangat
asam)
Vinegar (Cuka) 2,4 – 3,4
Wine (Anggur) 2,8 – 3,8
Apples 2,9 – 3,3
Oranges (bangsa Jeruk) 3-4
Beer 4-5
Urine (air kencing) 5-8
Water saturated with CO2 (bangsa minuman Sprite) 6
Cow’s milk (susu sapi) 6,3 – 6,6
Saliva (Air liur) 6,5 – 7,5 (netral)
Blood (darah) 7,3 – 7,5
Household ammonia (NH4OH) 12 (Sangat basa)

32. The importance of pH
 The effectiveness of enzymes depends very much
on pH
 Efektivitas enzim- sangat tergantung pada pH
tertentu
 Plants grow best in soil in the right pH range
(slightly basic or acidic) depending on the plant
 Tanaman tumbuh baik pada kisaran pH tertentu
(bisa basa maupun asam)

33.  The rate of deterioration of metals, stone and
concrete is determined largely by pH of the
water to which they are exposed
 Laju kerusakan logam, batu dan batu beton
sangat ditentukan oleh air yang menerpa mereka
 Rain water has been becoming more acidic
because of increasing pollution of the
atmosphere by SO2, NO2, etc.
 Air hujan menjadi lebih asam karena polusi gas-
gas SO2, NO2, dll.

34. pH Asam Lemah
 Acetic acid CH3COOH memiliki Ka 1,75 x 10-5
 Jika konsentrasinya 0,1 M, berapa pHnya?
 CH3COOH CH3COO- + H+
 (0,1-y) y y
[CH3COO-] [H+ ]
Ka = = 1,75 x 10-5
[CH3COOH]

35. (y) (y)
Ka = = 1,75 x 10-5
(0,1 – y)
(y2)
= 1,75 x 10-5
(0,1 – y)
Karena diasumsikan y sangat kecil, maka (0,1-y)
dianggap = (0,1-0) = 0,1 sehingga
y2 = 1,75 x 10-6

36. y = (1,75 x 10-6) 1/2
y = (1,751/2 x 10-6/2)
-Log (y) = -Log(1,751/2 x 10-6/2)
-Log (y) = -Log1,75) ½ + (-Log10-6/2)
pH = -Log1,75½ +3
pH = -Log1,322875656 +3

37. pH = - Log1,322876 +3
pH = - 0,121519024 + 3
pH = - 0,1 + 3
pH = 2,9

38. pH Basa Lemah
• Ammonia NH4OH memiliki Kb 5,65 x 10-10
• Jika konsentrasinya 0,2 M, berapa pHnya?
• NH4OH NH4+ + OH-
• (0,2-y) y y
[NH4+ ] [OH- ]
Kb = = 5,65 x 10-10
[NH4OH]

39. (y) (y)
Kb = = 5,65 x 10-10
(0,2 – y)
(y2)
= 5,65 x 10-10
(0,2 – y)
Karena diasumsikan y sangat kecil, maka (0,2 - y)
dianggap = (0,2 - 0) = 0,2 sehingga
y2 = (5,65 x 10-10) 0,2

40. y2 = (1,13 x 10-10)
y = (1,13 x 10-10)1/2
y = (1,13 1/2) x (10-5)
y = 1.063014581 x 10-5
Karena NH4OH =Basa, maka y = [OH-]
[OH-] = 1.063014581 x 10-5

41. Kw = [H+] [OH- ] = 10-14
10-14
[H+] =
[OH- ]
10-14
[H+] =
1.063014581 x 10-5
[H+] = (1/1,063014581) x 10-14-(-5)

42. [H+] = (0,940720869) x 10-9)

43. pH = - Log(0,940720869) - Log(10-9)
pH = - (-0,026539221) +9
pH = +0,0 + 9
pH = 9,0

44. Self-Test:
1. Find the pH of solution in which [H+] = 6.38 x
10-6 mol/L.
2. Calculate [H+] for a solution of pH 8.37
3. Calculate the pH of a strong base 1.0 x 10-3 M
NaOH
4. Calculate the pH of a strong base 5.0 x 10-3 M
Ba(OH)2
5. Calculate the pH of a weak acid 2.0 x 10-3 M
H2CO3 (Ka = 5.64 x 10-11)

46. Semoga bermanfaat
daaaahhh....
ASSALAMU’ALAIKUM WR.WB