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

4. NaOH M = ? V = 25.0ml H 2 SO 4 M = 1.00M V = 26.5cm 3 2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O M = ? M = 1.00M V = 25.0ml V = 26.5ml Mole ratio – 2: 1 H 2 SO 4 M = 0.5M V = 30.0ml 2NH 4 OH + H 2 SO 4 → (NH 4 ) 2 SO 4 + 2H 2 O M = 1.5M M = 0.5M V = ? ml V = 30.0ml Mole ratio – 2: 1 Acid/Base Titration Calculation Calculation NH 3 / NH 4 OH M = 1.5M V = ? ml 25.0 cm 3 of NaOH of unknown conc require 26.5cm 3 of 1.0M sulphuric acid for complete neutralization. Find its molarity of NaOH. Find vol of 1.5M aq NH 3 required to completely neutralize 30.0cm 3 of 0.5M sulphuric acid 2 2 1 1

5. NaOH M = ? V = 25.0ml H 2 SO 4 M = 1.00M V = 26.5cm 3 2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O M = ? M = 1.00M V = 25.0ml V = 26.5ml Mole ratio – 2: 1 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 M b V b = 2 M a V a 1 M x 25.0 = 2 1.0 x 26.5 1 M b = 2.12M H 2 SO 4 M = 0.5M V = 30.0ml 2NH 4 OH + H 2 SO 4 → (NH 4 ) 2 SO 4 + 2H 2 O M = 1.5M M = 0.5M V = ? ml V = 30.0ml Mole ratio – 2: 1 Using mole ratioUsing formula 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 V b = 0.02 dm 3 = 20ml M b V b = 2 M a V a 1 1.5 x V b = 2 0.5 x 30.0 1 V b = 20ml Using formula Using mole ratio Acid/Base Titration Calculation Calculation NH 3 / NH 4 OH M = 1.5M V = ? ml 25.0 cm 3 of NaOH of unknown conc require 26.5cm 3 of 1.0M sulphuric acid for complete neutralization. Find its molarity of NaOH. Find vol of 1.5M aq NH 3 required to completely neutralize 30.0cm 3 of 0.5M sulphuric acid 2 2 1 1

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

7. Na 2 CO 3 2.65g HCI M = 2.0M V = ? ml Na 2 CO 3 + 2HCI → 2NaCI + CO 2 + H 2 O M = 0.5M M = 2.0M V = 50ml V = ? ml Mole ratio – 1: 2 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 dm 3 = 25cm 3 M b V b = 1 M a V a 2 0.5 x 50.0 = 1 2.0 x V 2 V a = 25cm 3 HCI M = ? V = 25.0ml Na 2 CO 3 + 2HCI → 2NaCI + H 2 O + CO 2 M = 0.200M M = ? V = 10.0ml V = 25.0ml Mole ratio – 1: 2 Using mole ratioUsing formula 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 M b V b = 1 M a V a 2 0.2 x 10.0 = 1 M a x 25.0 2 M a = 0.160M Using formula Using mole ratio Acid/Base Titration Calculation Calculation Na 2 CO 3 M = 0.200M V = 10.0ml Find vol of 2.0M HCI needed to neutralize 2.65g of sodium carbonate (Na 2 CO 3 ) in 50ml water. 3 3 10.0cm 3 of 0.200M Na 2 CO 3 needed 25.0cm 3 of HCI for neutralization. Find molarity of HCI. 4 4 V = 50ml Moles = Mass/M = 2.65 106 = 0.025 mol

9. Titration Example: Acid Base Titration Condition Direct Titration Both titrant and analyte soluble Reaction is fast H 2 SO 4 HCI HNO 3 NaOH NH 4 OH KOH Ba(OH) 2 LiOH Titrant - soluble Analyte - soluble Soluble acid Soluble base (Alkali)

10. And what if the base is insoluble in water ? Ex Ca(OH) 2 (s)

11. Titration Example: Acid Base Titration Condition Direct TitrationBack Titration Both titrant and analyte soluble Reaction is fast H 2 SO 4 HCI HNO 3 NaOH NH 4 OH KOH Ba(OH) 2 LiOH Known conc /vol of acid used Titrant - soluble Analyte - soluble Soluble acid Soluble base (Alkali) Condition CaCO 3 in egg shell Impure limestone added Sparingly soluble acid/base. Reaction is SLOW Ex: Calcium carbonate (egg shell), calcium carbonate in antiacid ( limestone) or calcium hydroxide from antacid table. Impure antacid

12. Titration example : Acid Base Titration Condition Direct TitrationBack Titration Both titrant and analyte soluble Reaction is FAST H 2 SO 4 HCI HNO 3 NaOH NH 4 OH KOH Ba(OH) 2 LiOH Known conc /vol of acid used Amt of excess acid left Titrated with known conc/vol alkali Excess acid left Transfer to flask Titrant - soluble Analyte - soluble Soluble acid Soluble base (Alkali) Condition Left overnight in acid CaCO 3 in egg shell Impure limestone unknown Amt of acid that reacts with insoluble base Amt acid that has reacted = Amt Known – Amt excess acid added acid left added Sparingly soluble acid/base. Reaction is SLOW Ex: Calcium carbonate (egg shell), calcium carbonate in antiacid ( limestone) or calcium hydroxide from antacid table. known Impure antacid Ca(OH) 2 + 2 HCl  CaCl 2 + 2H 2 O

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

14. % Calcium hydroxide in an impure antacid tablet - Back Titration Calculation Back Titration 50.0ml, 0.250M HCI Amt of HCI added Amt of base (solid) Amt HCI left Titrated with known conc/vol NaOH M = 0.1108 V = 33.64ml Amt HCI react = Amt HCI – Amt HCI with Ca(OH )2 add left HCI left Transfer to flask Left overnight in acid 0.5214g impure Ca(OH) 2 added 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. NaOH + HCI → NaCI + H 2 O M = 0.1108M moles = ? V = 33.64ml 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 Mole ratio – 1: 1 Using mole ratio Amt HCI add = M V = 0.250 x 0.050 = 0.01250 mol Amt HCI react = Amt HCI add – Amt HCI left with Ca(OH)2 = 0.01250 – 3.727 x 10 -3 = 0.008773 mol 2HCI + Ca(OH) 2 → CaCI 2 + 2H 2 O Mole Mole 0.008773 ? Mole ratio – 2: 1 Mole ratio (2 : 1) 2 mol HCI react 1 mol Ca(OH) 2 0.008773 mol HCI react o.oo4386 mol Ca(OH) 2 Mass = Mole Ca(OH) 2 x gfm Ca(OH) 2 = 0.004386 x 74.1 = 0.3250g % by mass = mass Ca(OH) 2 x 100% Ca(OH) 2 mass impure = (0.3250/0.5214) x 100% = 62.3% 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 Amt HCI Add Amt HCI Left Amt HCI react with base