Kimia. kesetimbangan asam dan basa

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Kimia. kesetimbangan asam dan basa

  1. 1. KESETIMBANGAN ASAM & BASA SMAN PINTAR KUANSING Oleh : Ahmad mulya
  2. 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. 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. 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. 5. Contoh asam yang populer Cuka = larutan yang mengandung acetic acid = CH3COOH = HC2H3O2 = asam cuka
  6. 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. 7. Citric acid = Asam Sitrat• Air jeruk = larutan mengandung citric acid = H3C6H5O7= asam sitrat
  8. 8. Sifat Basa• Dalam air melepaskan ion OH-• NaOH  Na+ + OH-• Basa+ Asam  Garam + Air• NaOH + CH3COOH  CH3COONa + H2O
  9. 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. 10. Listrik Walau Terlarut CH3COOH maupun HClPelarut = Benzene = C6H6  Nonelectrolyte
  11. 11. Listrik Terlarut 0,5 M CH3COOH Asam Lemah Pelarut = Air Larutan ion dalam air = electrolyte
  12. 12. Listrik Terlarut 0,5 M HCl Asam Kuat Pelarut = Air Larutan ion dalam air = electrolyte
  13. 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. 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. 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. 16.  Arrhenius: HCl asam Lewis:  HCl juga asam Bronsted & Lowry: HCl = juga asam Yang berbeda alasannya Saling melengkapi
  17. 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. 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. 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. 20. Weak Acid (10-8 < Ka < 10-3) Acetic acid = CH3C00H Carbonic acid = H2CO3 Hydrogen Sulfide = H2S Nitrous acid = HNO2 H3PO4 , H2SO3
  21. 21. Ka Meningkat seiring Bil. OksidasiAcid Formula Oxidatio Ka n NumberHypochlorous HClO 1+ 5 x 10-8Chlorous HClO2 3+ 1 x 10-2Chloric HClO3 5+ 1 x 1010Perchloric HClO4 7+ 1 x 1011 Catatan: Ka HCl 107
  22. 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/LKarena [H+] = [OH-] maka [H+] [OH-] = [H+] [H+] = [H+]2Jadi [H+]2 = 1 x 10-14dan [H+] = (1 x 10-14) = 1 x 10-7dipermudah penulisannya: - log 1 x 10-7 = - log 10-7 = -(-7) = 7Maka pH air murni = 7
  23. 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 = 7Kesetimbangan ion-ion dalam larutan didasarkan pada Kw air ini
  24. 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. 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. 26. • pH = - log [H+]• 5 = - log [H+]• 5 = - (-5)• 5 = - (log 10-5)• Jadi [H+] = 10-5 mole/Liter
  27. 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. 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. 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. 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. 31. pH beberapa zatNama Zat pHGastric juice (cairan lambung) 1 – 3 (Sangat asam)Lemons (bangsa Jeruk) 2,2 – 2,4 (Sangat asam)Vinegar (Cuka) 2,4 – 3,4Wine (Anggur) 2,8 – 3,8Apples 2,9 – 3,3Oranges (bangsa Jeruk) 3-4Beer 4-5Urine (air kencing) 5-8Water saturated with CO2 (bangsa minuman Sprite) 6Cow’s milk (susu sapi) 6,3 – 6,6Saliva (Air liur) 6,5 – 7,5 (netral)Blood (darah) 7,3 – 7,5Household ammonia (NH4OH) 12 (Sangat basa)
  32. 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. 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. 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. 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. 36. y = (1,75 x 10-6) 1/2y = (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. 37. pH = - Log1,322876 +3pH = - 0,121519024 + 3pH = - 0,1 + 3pH = 2,9
  38. 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. 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. 40. y2 = (1,13 x 10-10)y = (1,13 x 10-10)1/2y = (1,13 1/2) x (10-5)y = 1.063014581 x 10-5Karena NH4OH =Basa, maka y = [OH-][OH-] = 1.063014581 x 10-5
  41. 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. 42. [H+] = (0,940720869) x 10-9)
  43. 43. pH = - Log(0,940720869) - Log(10-9)pH = - (-0,026539221) +9pH = +0,0 + 9pH = 9,0
  44. 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 8.373. Calculate the pH of a strong base 1.0 x 10-3 M NaOH4. Calculate the pH of a strong base 5.0 x 10-3 M Ba(OH)25. Calculate the pH of a weak acid 2.0 x 10-3 M H2CO3 (Ka = 5.64 x 10-11)
  45. 45. Semoga bermanfaat daaaahhh....ASSALAMU’ALAIKUM WR.WB

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