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Kesetimbangan ion dalam Asam dan Basa

Kesetimbangan ion dalam Asam dan Basa. Senyawa paling banyak diproduksi. Sulfuric acid = H 2 SO 4 = Asam sulfat Hydrochloric acid = HCl = Asam Klorida Nitric acid = HNO 3 = Asam Nitrat Sodium Hydroxide = NaOH = Basa Natrium hidroksida

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Kesetimbangan ion dalam Asam dan Basa

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  1. Kesetimbangan iondalamAsam dan Basa

  2. Senyawa paling banyakdiproduksi • 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. Akan ada soal test • Saya titipkan file • http://bambangtriatma.wikispaces.com

  4. 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

  5. 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.

  6. Contoh asam yang populer • Cuka = larutan yang mengandung acetic acid = CH3COOH = HC2H3O2 = asam cuka

  7. 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

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

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

  10. 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+

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

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

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

  14. 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

  15. 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

  16. 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

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

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

  19. 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)

  20. 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

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

  22. Ka Meningkat seiring Bilangan Oksidasi Ka= Konstanta ionisasi asam. Catatan: Ka HCl  107

  23. Ka Air H2O  H+ + OH- [H+] [OH-] Ka = = 1 x 10-14 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 [H2O]

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

  25. 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

  26. pH (Power 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?

  27. pH (Power of Hydrogen) • pH = - log [H+] • 5 = - log [H+] • 5 = - (-5) • 5 = - (log 10-5) • Jadi [H+] = 10-5 mole/Liter

  28. 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

  29. 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

  30. 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

  31. 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

  32. pH beberapa zat

  33. The importance of pHPentingnya 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)

  34. 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.

  35. [CH3COO-] [H+] Ka = = 1,75 x 10-5 [CH3COOH] 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

  36. (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 (y) (y) Ka = = 1,75 x 10-5 (0,1 – y) y2 = 1,75 x 10-6

  37. 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

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

  39. 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]

  40. = 5,65 x 10-10 (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 (y) (y) Kb = (0,2 – y) y2 = (5,65 x 10-10) 0,2

  41. 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

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

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

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

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

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