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Tutorial on Acid/Base, Redox, Back Titration. http://lawrencekok.blogspot.com. ADAPTED FROM Lawrence Kok. THANK YOU. Acid/Base Titration Calculation. 1. 25.0 cm 3 of NaOH of unknown conc require 26.5cm 3 of 1.0M sulphuric acid for complete neutralization.

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  1. Tutorial on Acid/Base, Redox, Back Titration http://lawrencekok.blogspot.com ADAPTED FROM Lawrence Kok THANK YOU

  2. Acid/Base Titration Calculation 1 25.0 cm3 of NaOH of unknown conc require 26.5cm3 of 1.0M sulphuric acid for complete neutralization. Find its molarity of NaOH. Find vol of 1.5M aq NH3 required to completely neutralize 30.0cm3 of 0.5M sulphuric acid 2 NH3 / NH4OH M = 1.5M V = ? ml H2SO4 M = 1.00M V = 26.5cm3 H2SO4 M = 0.5M V = 30.0ml NaOH M = ? V = 25.0ml Calculation Calculation 2NaOH + H2SO4→ Na2SO4 + 2H2O M = ? M = 1.00M V = 25.0ml V = 26.5ml 2NH4OH + H2SO4→ (NH4)2SO4 + 2H2O M = 1.5M M = 0.5M V = ? ml V = 30.0ml Mole ratio – 2: 1 Mole ratio – 2: 1

  3. Acid/Base Titration Calculation 1 25.0 cm3 of NaOH of unknown conc require 26.5cm3 of 1.0M sulphuric acid for complete neutralization. Find its molarity of NaOH. Find vol of 1.5M aq NH3 required to completely neutralize 30.0cm3 of 0.5M sulphuric acid 2 NH3 / NH4OH M = 1.5M V = ? ml H2SO4 M = 1.00M V = 26.5cm3 H2SO4 M = 0.5M V = 30.0ml NaOH M = ? V = 25.0ml Calculation Calculation 2NaOH + H2SO4→ Na2SO4 + 2H2O M = ? M = 1.00M V = 25.0ml V = 26.5ml 2NH4OH + H2SO4→ (NH4)2SO4 + 2H2O M = 1.5M M = 0.5M V = ? ml V = 30.0ml Mole ratio – 2: 1 Mole ratio – 2: 1 Using mole ratio Using mole ratio Using formula Using formula • Moles of Acid = MV • = (1.00 x 0.0265) • = 2.65 x 10-2 • Mole ratio (1 : 2) • 1 mole acid neutralize 2 mole base • 2.65 x 10-2 acid neutralize 5.30 x 10-2 base • Moles of Base = M x V • =M x 0.025 • M x 0.025 = 5.30 x 10-2 • M = 2.12M Mb Vb = 2 MaVa 1 M x 25.0 = 2 1.0 x 26.5 1 Mb = 2.12M • Moles of Acid = MV • = (0.5 x 0.030) • = 1.5o x 10-2 • Mole ratio (2 : 1) • 1 mole acid neutralize 2 mole base • 1.50 x 10-2 acid neutralize 3.00 x 10-2 base • Moles of Base = M x V • =1.5 x V • 1.5 x V = 3.00 x 10-2 • Vb = 0.02 dm3 = 20ml M bVb = 2 MaVa 1 1.5 x Vb = 2 0.5x 30.0 1 Vb = 20ml

  4. Acid/Base Titration Calculation Find vol of 2.0M HCI needed to neutralize 2.65g of sodium carbonate (Na2CO3) in 50ml water. 10.0cm3 of 0.200M Na2CO3 needed 25.0cm3 of HCI for neutralization. Find molarity of HCI. 3 4 HCI M = 2.0M V = ? ml Na2CO3 M = 0.200M V = 10.0ml Moles = Mass/M = 2.65 106 = 0.025 mol HCI M = ? V = 25.0ml Na2CO3 2.65g V = 50ml Calculation Calculation Na2CO3 + 2HCI → 2NaCI + CO2 + H2O M = 0.5M M = 2.0M V = 50ml V = ? ml Na2CO3 + 2HCI → 2NaCI + H2O + CO2 M = 0.200M M = ? V = 10.0ml V = 25.0ml Mole ratio – 1: 2 Mole ratio – 1: 2

  5. Acid/Base Titration Calculation Find vol of 2.0M HCI needed to neutralize 2.65g of sodium carbonate (Na2CO3) in 50ml water. 10.0cm3 of 0.200M Na2CO3 needed 25.0cm3 of HCI for neutralization. Find molarity of HCI. 3 4 HCI M = 2.0M V = ? ml Na2CO3 M = 0.200M V = 10.0ml Moles = Mass/M = 2.65 106 = 0.025 mol HCI M = ? V = 25.0ml Na2CO3 2.65g V = 50ml Calculation Calculation Na2CO3 + 2HCI → 2NaCI + CO2 + H2O M = 0.5M M = 2.0M V = 50ml V = ? ml Na2CO3 + 2HCI → 2NaCI + H2O + CO2 M = 0.200M M = ? V = 10.0ml V = 25.0ml Mole ratio – 1: 2 Mole ratio – 1: 2 Using mole ratio Using mole ratio Using formula Using formula • Moles of Base = Mass/M • = (2.65 ÷ 106) • = 2.5 x 10-2 • Mole ratio (1 : 2) • 1 mole base neutralize 2 mole acid • 2.5 x 10-2 base neutralize 5.0 x 10-2 acid • Moles of Acid = M x V • =2.0 x V • 2.0 x V = 5 x 10-2 • V = 0.25 dm3 = 25cm3 MbVb = 1 MaVa 2 0.5 x 50.0 = 1 2.0 x V 2 Va = 25cm3 • Moles of Base = MV • = (0.200 x 0.010) • = 2.00 x 10-3 • Mole ratio (1 : 2) • 1 mole base neutralize 2 mole acid • 2.00 x 10-3 base neutralize 4.00 x 10-3 acid • Moles of Acid = M x V • =M x 0.025 • M x 0.025 = 4.00 x 10-3 • M = 0.160M MbVb = 1 MaVa 2 0.2 x 10.0 = 1 Max 25.0 2 Ma = 0.160M

  6. Redox Titration Calculation- % Iron in iron tablet Iron tablet contain hydrated iron (II) sulphate (FeSO4.7H2O). One tablet weighing 1.863g crushed, dissolved in water and solution made up to total vol of 250ml. 10ml of this solution added to 20ml of H2SO4 and titrated with 0.002M KMnO4. Average 24.5ml need to reach end point. Cal % iron(II) sulphate in iron tablet. 5 KMnO4 M = 0.002M V = 24.5 ml 1.863 g 250ml Fe2+ mole= ? V = 10ml 10ml transfer MnO4- + 5Fe2+ + 8H+→ Mn2+ + 5Fe+3 + 4H2O M = 0.002M M = ? V = 24.5ml Mole ratio – 1: 5

  7. Redox Titration Calculation- % Iron in iron tablet Iron tablet contain hydrated iron (II) sulphate (FeSO4.7H2O). One tablet weighing 1.863g crushed, dissolved in water and solution made up to total vol of 250ml. 10ml of this solution added to 20ml of H2SO4 and titrated with 0.002M KMnO4. Average 24.5ml need to reach end point. Cal % iron(II) sulphate in iron tablet. 6 KMnO4 M = 0.002M V = 24.5 ml 1.863 g 250ml Fe2+ mole= ? V = 10ml 10ml transfer MnO4- + 5Fe2+ + 8H+→ Mn2+ + 5Fe2+ + 4H2O M = 0.002M M = ? V = 24.5ml Mole ratio – 1: 5 Using mole ratio 1 • Mole KMO4- = MV • = (0.002 x 0.0245) • = 4.90 x 10-5 • Mole ratio (1 : 5) • 1 mole KMO4- react 5 mole Fe2+ • 4.90 x 10-5KMO4-react 2.45 x 10-4Fe2+

  8. Redox Titration Calculation- % Iron in iron tablet Iron tablet contain hydrated iron (II) sulphate (FeSO4.7H2O). One tablet weighing 1.863g crushed, dissolved in water and solution made up to total vol of 250ml. 10ml of this solution added to 20ml of H2SO4 and titrated with 0.002M KMnO4. Average 24.5ml need to reach end point. Cal % iron(II) sulphate in iron tablet. 6 Video on % Iron in iron tablet KMnO4 M = 0.002M V = 24.5 ml 1.863 g 250ml Fe2+ Mole ? V = 10ml 10ml transfer MnO4- + 5Fe2+ + 8H+→ Mn2+ + 5Fe2+ + 4H2O M = 0.002M M = ? V = 24.5ml Mole ratio – 1: 5 Video on Fe2+/KMnO4 titration calculation Using mole ratio 1 • Mole KMO4- = MV • = (0.002 x 0.0245) • = 4.90 x 10-5 • Mole ratio (1 : 5) • 1 mole KMO4- react 5 mole Fe2+ • 4.90 x 10-5KMO4-react 2.45 x 10-4Fe2+ 10ml sol contain - 2.45 x 10-4 Fe2+ 250ml sol contain - 250 x 2.45 x 10-4 Fe2+ 10 = 6.125 x 10-3 mole Fe2+ FeSO4.7H2O  FeSO4 + 7H2O 1 mol 1 mol + 7 mol FeSO4 Fe2+ + SO42- 1 mol 1mol + 1mol 6.125 x 10-3 mol6.125 x 10-3 mole Fe2+ 2 4 Mass of (expt yield) = 1.703g Mass of (Actual tablet) = 1.863g % Fe in iron tablet = 1.703 x 100% 1.863 = 91.4% 3 Mole  Mass Mole x RMM = Mass FeSO4 6.125 x 10-3 x 278.05 = 1.703g FeSO4

  9. Titration Direct Titration Condition • Both titrant and analyte soluble • Reaction is fast Titrant - soluble Analyte - soluble Acid Base Titration Redox Titration H2SO4 HCI HNO3 Soluble acid NaOH NH4OH KOH Ba(OH)2 LiOH Soluble base (Alkali)

  10. Titration Direct Titration Back Titration Condition Condition • Both titrant and analyte soluble • Reaction is fast Titrant - soluble • Sparingly soluble acid/base. • Reaction is SLOW • Ex: Calcium carbonate (egg shell) , calcium carbonate in antiacid ( limestone) or calcium hydroxide from antacid table. Analyte - soluble added Impure limestone CaCO3 in egg shell Redox Titration Acid Base Titration Impure antacid Known conc /vol of acid used H2SO4 HCI HNO3 Soluble acid NaOH NH4OH KOH Ba(OH)2 LiOH Soluble base (Alkali)

  11. Titration Direct Titration Back Titration Condition Condition • Both titrant and analytesoluble • Reaction is FAST Titrant - soluble • Sparingly soluble acid/base. • Reaction is SLOW • Ex: Calcium carbonate (egg shell) , calcium carbonate in antiacid ( limestone) or calcium hydroxide from antacid table. Analyte - soluble added Impure limestone CaCO3 in egg shell Redox Titration Acid Base Titration Impure antacid Known conc /vol of acid used Left overnight in acid H2SO4 HCI HNO3 Soluble acid Amt of known acid added Titrated with known conc/vol alkali NaOH NH4OH KOH Ba(OH)2 LiOH Soluble base (Alkali) known unknown Amt of excess acid left Excess acid left Transfer to flask Amt of base (solid) react with acid Amt base(solid) = Amt Known – Amt excess acid added acid left

  12. % Calcium hydroxide in an impure antacid tablet - Back Titration Calculation Back Titration 0.5214g of impure Ca(OH)2 from antacid was dissolved in 50.0ml, 0.250M HCI. 33.64ml, 0.1108M NaOH needed to neutralize excess acid. Cal percentage sample containing Ca(OH)2 in a tablet. added 0.5214g impure Ca(OH)2 50.0ml, 0.250M HCI Titrated with known conc/vol NaOH M = 0.1108 V = 33.64ml Left overnight in acid Amt of HCI added Transfer to flask Amt HCI left HCI left Amt of base (solid) Amt HCI react = Amt HCI – Amt HCI with NaOH add left

  13. % Calcium hydroxide in an impure antacid tablet - Back Titration Calculation Back Titration 0.5214g of impure Ca(OH)2 from antacid was dissolved in 50.0ml, 0.250M HCI. 33.64ml, 0.1108M NaOH needed to neutralize excess acid. Cal percentage sample containing Ca(OH)2 in a tablet. Amt HCI add = M V = 0.250 x 0.050 = 0.01250 mol 1 added 0.5214g impure Ca(OH)2 Amt HCI Add 50.0ml, 0.250M HCI 2 NaOH + HCI → NaCI + H2O M = 0.1108M moles = ? V = 33.64ml Titrated with known conc/vol NaOH M = 0.1108 V = 33.64ml Mole ratio – 1: 1 Left overnight in acid Amt of HCI added Transfer to flask Amt HCI left HCI left Amt of base (solid) Amt HCI react = Amt HCI – Amt HCI with NaOH add left

  14. % Calcium hydroxide in an impure antacid tablet - Back Titration Calculation Back Titration 0.5214g of impure Ca(OH)2 from antacid was dissolved in 50.0ml, 0.250M HCI. 33.64ml, 0.1108M NaOH needed to neutralize excess acid. Cal percentage sample containing Ca(OH)2 in a tablet. Amt HCI add = M V = 0.250 x 0.050 = 0.01250 mol 1 added 0.5214g impure Ca(OH)2 Amt HCI Add 50.0ml, 0.250M HCI 2 NaOH + HCI → NaCI + H2O M = 0.1108M moles = ? V = 33.64ml Titrated with known conc/vol NaOH M = 0.1108 V = 33.64ml Mole ratio – 1: 1 Left overnight in acid Using mole ratio • Mole NaOH = MV • = (0.1108 x 0.03364) • = 3.727 x 10-3 • Mole ratio (1 : 1) • 1 mole NaOH react 1 mole HCI • 3.727 x 10-3 mole NaOH react 3.727 x 10-3 HCI • HCI left = 3.727 x 10-3 mol 3 Amt of HCI added Transfer to flask Amt HCI Left Amt HCI left HCI left Amt of base (solid) Amt HCI react = Amt HCI – Amt HCI with NaOH add left

  15. % Calcium hydroxide in an impure antacid tablet - Back Titration Calculation Back Titration 0.5214g of impure Ca(OH)2 from antacid was dissolved in 50.0ml, 0.250M HCI. 33.64ml, 0.1108M NaOH needed to neutralize excess acid. Cal percentage sample containing Ca(OH)2 in a tablet. Amt HCI add = M V = 0.250 x 0.050 = 0.01250 mol 1 added 0.5214g impure Ca(OH)2 Amt HCI Add 50.0ml, 0.250M HCI 2 NaOH + HCI → NaCI + H2O M = 0.1108M moles = ? V = 33.64ml Titrated with known conc/vol NaOH M = 0.1108 V = 33.64ml Mole ratio – 1: 1 Left overnight in acid Using mole ratio • Mole NaOH = MV • = (0.1108 x 0.03364) • = 3.727 x 10-3 • Mole ratio (1 : 1) • 1 mole NaOH react 1 mole HCI • 3.727 x 10-3 mole NaOH react 3.727 x 10-3 HCI • HCI left = 3.727 x 10-3 mol 3 Amt of HCI added Transfer to flask Amt HCI Left Amt HCI left HCI left 4 Amt HCI react = Amt HCI add – Amt HCI left with NaOH = 0.01250 – 3.727 x 10-3 = 0.008773 mol Amt HCI react with base Amt of base (solid) 2HCI + Ca(OH)2→ CaCI2 + 2H2O Mole Mole 0.008773 ? 5 7 Mass = Mole Ca(OH)2 x RMM Ca(OH)2= 0.004386 x 74.1 = 0.3250g Mole ratio – 2: 1 Amt HCI react = Amt HCI – Amt HCI with NaOH add left 6 • Mole ratio (2 : 1) • 2 mol HCI react 1 molCa(OH)2 • 0.008773 mol HCI react o.oo4386 molCa(OH)2 8 % by mass = mass Ca(OH)2 x 100% Ca(OH)2 mass impure = (0.3250/0.5214) x 100% = 62.3%

  16. Acknowledgements Thanks to source of pictures and video used in this presentation Thanks to Creative Commons for excellent contribution on licenses http://creativecommons.org/licenses/ Prepared by Lawrence Kok Check out more video tutorials from my site and hope you enjoy this tutorial http://lawrencekok.blogspot.com

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