1.Tutorial on Acid/Base, Redox, Back Titration Calculation and IA Titration experiments.Prepared by Lawrence Kok http://lawrencekok.blogspot.com

2. Titration for IA (DCP) assessment Acid Base Titration Standardization ExptStandardization NaOH with primary std KHP Click here or here for expt`Titration bet HCI with std NaOH Click here for expt 4.2aStandardization HCI with primary std Na2CO3 Click here for expt 4.2Titration bet NaOH with std HCI Click here for expt 4.2aAcid/Base Titration Expt Determining acetylsalicylic acid in aspirin with std NaOHDetermining ethanoic acid in vinegar with std NaOHClick here or here for expt` Click here for more exptClick here for expt 4.3Determining water crystallization in hydrated Na2CO3 with std HCI Click here for expt 4.4 3. Titration for IA (DCP) assessment Acid Base Titration Standardization ExptStandardization NaOH with primary std KHPAcid/Base Titration ExptStandardization HCI with primary std Na2CO3Click here or here for expt`Determining acetylsalicylic acid in aspirin with std NaOH Click here or here for expt` Click here for more exptClick here for expt 4.2Titration bet HCI with std NaOHClick here for expt 4.3Determining water crystallization in hydrated Na2CO3 with std HCITitration bet NaOH with std HCIClick here for expt 4.2aDetermining ethanoic acid in vinegar with std NaOHClick here for expt 4.4Click here for expt 4.2aRedox TitrationStandardization ExptStandardization KMnO4 with std ammonium iron(II) sulphate Click here for expt 4.5Standardization KI/I2 with std sodium thiosulphate Click here for expt 4.7Hypochlorite (OCI-) in bleach with iodine/thiosulphate Click here for expt 4.8Standardization KI/I2 with std KIO3 Click here for expt 4.7 Click here for more exptIron (II) determination with std KMnO4 Click here for expt 4.6Redox Titration ExptIodine/thiosulphate (iodometric titration)Copper(II) determination in brass with iodine/thiosulphate Click here or here for expt` Click here for more exptVit C determination with iodine/thiosulphate Click here or here for expt Click here more detail expt 4. TitrationRedox TitrationAcid Base TitrationComplexometric titrationNeutralizationCondition for Acid/Alkali Titration -One reactant must be standard (known conc) or capable being standardised - Equivalent point equal amt neutralize each other - End point measurable/detectable by colour change (indicator), pH change /conductivity 5. TitrationRedox TitrationAcid Base TitrationComplexometric titrationNeutralizationCondition for Acid/Alkali Titration -One reactant must be standard (known conc) or capable being standardised - Equivalent point equal amt neutralize each other - End point measurable/detectable by colour change (indicator), pH change /conductivityAccurate known concPrimary standard acids - Potassium hydrogen phthalate 20.4 g in 1LBurette20.4 g KHPVolumetric flaskStandard 0.1M KHPAcid/Base used as primary standard -Stable/solid -Soluble in water -Does not decompose over timeUnknown Conc NaOHAccurate known concPrimary standard bases - Anhydrous sodium carbonate 10.6 gNa2CO310.6g in 1 LUnknown Conc HCIVolumetric flask??Standard 0.1M Na2CO3Burette 6. TitrationRedox TitrationAcid Base TitrationComplexometric titrationNeutralizationCondition for Acid/Alkali Titration -One reactant must be standard (known conc) or capable being standardised - Equivalent point equal amt neutralize each other - End point measurable/detectable by colour change (indicator), pH change /conductivityAccurate known concPrimary standard acids - Potassium hydrogen phthalate 20.4 g in 1LBurette20.4 g KHPVolumetric flaskStandard 0.1M KHPAcid/Base used as primary standard -Stable/solid -Soluble in water -Does not decompose over timeUnknown Conc NaOHAccurate known concPrimary standard bases - Anhydrous sodium carbonate 10.6 gNa2CO310.6g in 1 LUnknown Conc HCIVolumetric flask??Unable to prepare accurate conc of NaOH/HCI due to Hygroscopic nature NaOH absorb water vapour HCI is in vapour state difficult to measure amtStandard 0.1M Na2CO3Burette 7. Standardization of (ACID) with standard (BASE) Data CollectionData Processing Vol Na2CO3Vol Na2CO3 Fin vol = (29.50 0.05) Ini vol = (3.10 0.05)Uncertainty in vol Na2CO3 Add absolute uncertainty for final + initial = (0.05 + 0.05) = 0.10Na2CO3 vol = (26.40 0.10)Average vol Na2CO3 = 26.4 + 26.4 = 26.4cm3 2Average Vol Na2CO3 uncertainty = 26.4 + 26.4 = (26.4 0.10)cm3 2Error Analysis% Uncertainty - burette% Uncertainty - pipetteAbsolute uncertainty vol x 100% Average vol added = 0.10 x 100% 26.4 = 0.38%Na2CO3 0.03Absolute uncertainty vol x 100% Average vol added = 0.03 x 100% 25.00 = 0.12%HCIConc uncertaintyConc HCI = 0.211 0.5% ( % uncertainty)Total % Uncertainty = % uncertainty burette + % uncertainty pipette = 0.38% + 0.12% = 0.5% 0.5 x 0.211 = 0.001 100 (% Absolute uncertainty)Conc HCI=(0.2110.001)M (Absolute uncertainty)% ErrorLit value - HCI = 0.200M Expt value HCI = 0.211M Difference = 0.011 % Error Difference x 100% Literature value 0.011 x 100% = 5.5%0.200 8. Standardization of (BASE) with standard (ACID) Data CollectionData Processing Vol KHPVol KHP Fin vol = (29.50 0.05) Ini vol = (3.10 0.05)Uncertainty in vol KHP Add absolute uncertainty for final + initial = (0.05 + 0.05) = 0.10KHP vol = (26.40 0.10)Average vol KHP = 26.4 + 26.4 = 26.4cm3 2Average Vol KHP uncertainty = 26.4 + 26.4 = (26.4 0.10)cm3 2Error Analysis% Uncertainty - burette Absolute uncertainty vol x 100% Average vol added = 0.10 x 100% 26.4 = 0.38%% Uncertainty - pipette KHP 0.03Absolute uncertainty vol x 100% Average vol added = 0.03 x 100% 25.00 = 0.12%NaOHConc uncertaintyConc NaOH = 0.106 0.5% ( % uncertainty)Total % Uncertainty = % uncertainty burette + % uncertainty pipette = 0.38% + 0.12% = 0.5% 0.5 x 0.106 = 0.0005 = 0.001 100 (% Abs uncertainty)Conc NaOH=0.106 0.001M (Absolute uncertainty)% ErrorLit value - NaOH = 0.100M Expt value NaOH = 0.106M Difference = 0.006 % Error Difference x 100% Literature value0.006 x 100% = 6%0.100 9. Sample Titration Calculation Standardization of (BASE) with standard (ACID) KHP M = 0.100M V = 26.4 mlStandardization of (ACID) with standard (BASE) Na2CO3 M = 0.100M V = 26.4 ml HCI M=? V = 25.0mlNaOH M=? V = 25.0mlCalculation KHP+M = 0.100M V = 26.40mlNaOHM= ? V = 25.0mlCalculation NaKP+ H2OMole ratio 1: 1Na2CO3 M = 0.100M V = 26.4ml+ 2HCI 2NaCI + H2O + CO2M=? V = 25.0mlMole ratio 1: 2 10. Sample Titration Calculation Standardization of (BASE) with standard (ACID) KHP M = 0.100M V = 26.4 mlStandardization of (ACID) with standard (BASE) Na2CO3 M = 0.100M V = 26.4 ml HCI M=? V = 25.0mlNaOH M=? V = 25.0mlCalculation KHP+M = 0.100M V = 26.40mlNaOHCalculation NaKPM= ? V = 25.0mlUsing mole ratioMoles of Acid = MV = (0.100 x 0.0264) = 2.64 x 10-3 Mole ratio (1 : 1) 1 mole acid neutralize 1 mole base 2.64 x 10-3 acid neutralize 2.64 x 10-3 base Moles of Base = M x V = M x 0.025 M x 0.025 = 2.64 x 10-3 M = 0.106M+ H2OMole ratio 1: 1Using formulaM aVa = 1 Mb Vb 1 0.1 x 26.40 = 1 M x 25.0 1 M b = 0.106MNa2CO3+ 2HCIM = 0.100M V = 26.4ml 2NaCI + H2O + CO2M=? V = 25.0mlUsing mole ratioMoles of Base = MV = (0.100 x 0.0264) = 2.64 x 10-3 Mole ratio (1 : 2) 1 mole base neutralize 2 mole acid 2.64 x 10-3 base neutralize 5.28 x 10-3 acid Moles of Acid = M x V = M x 0.025 M x 0.025 = 5.28 x 10-3 M = 0.211MMole ratio 1: 2Using formulaM bVb = 1 Ma Va 2 0.1 x 26.4 = 1 Ma x 25.0 2 Ma = 0.211M 11. Sample Titration Calculation Standardization of (BASE) with standard (ACID)Standardization of (ACID) with standard (BASE)KHP M = 0.100M V = 26.4 mlNa2CO3 M = 0.100M V = 26.4 ml HCI M=? V = 25.0mlNaOH M=? V = 25.0mlCalculation KHP+M = 0.100M V = 26.40mlNaOHCalculation NaKPM= ? V = 25.0mlUsing mole ratioMoles of Acid = MV = (0.100 x 0.0264) = 2.64 x 10-3 Mole ratio (1 : 1) 1 mole acid neutralize 1 mole base 2.64 x 10-3 acid neutralize 2.64 x 10-3 base Moles of Base = M x V = M x 0.025 M x 0.025 = 2.64 x 10-3 M = 0.106M+ H2ONa2CO3+ 2HCIM = 0.100M V = 26.4mlMole ratio 1: 1 2NaCI + H2O + CO2M=? V = 25.0mlUsing mole ratioUsing formulaM aVa = 1 Mb Vb 1 0.1 x 26.40 = 1 M x 25.0 1 M b = 0.106MMole ratio 1: 2Using formulaMoles of Base = MV = (0.100 x 0.0264) = 2.64 x 10-3 Mole ratio (1 : 2) 1 mole base neutralize 2 mole acid 2.64 x 10-3 base neutralize 5.28 x 10-3 acid Moles of Acid = M x V = M x 0.025 M x 0.025 = 5.28 x 10-3 M = 0.211MM bVb = 1 Ma Va 2 0.1 x 26.4 = 1 Ma x 25.0 2 Ma = 0.211MVideo on TitrationClick here Acid/Base calculationClick here Titration calculationClick here cal Na2CO3/HCIClick here cal NaOH/H2SO4 12. Acid/Base Titration Calculation 125.0 cm3 of NaOH of unknown conc require 26.5cm3 of 1.0M sulphuric acid for complete neutralization. Find its molarity of NaOH.2Find vol of 1.5M aq NH3 required to completely neutralize 30.0cm3 of 0.5M sulphuric acid NH3 / NH4OH M = 1.5M V = ? mlH2SO4 M = 1.00M V = 26.5cm3H2SO4 M = 0.5M V = 30.0mlNaOH M=? V = 25.0mlCalculation 2NaOH M=? V = 25.0ml+ H2SO4 M = 1.00M V = 26.5mlCalculationNa2SO4 + 2H2O Mole ratio 2: 12NH4OH +H2SO4 M = 1.5M V = ? mlM = 0.5M V = 30.0ml(NH4)2SO4 + 2H2O Mole ratio 2: 1 13. Acid/Base Titration Calculation 125.0 cm3 of NaOH of unknown conc require 26.5cm3 of 1.0M sulphuric acid for complete neutralization. Find its molarity of NaOH.2Find vol of 1.5M aq NH3 required to completely neutralize 30.0cm3 of 0.5M sulphuric acid NH3 / NH4OH M = 1.5M V = ? mlH2SO4 M = 1.00M V = 26.5cm3H2SO4 M = 0.5M V = 30.0mlNaOH M=? V = 25.0mlCalculation 2NaOH M=? V = 25.0ml+ H2SO4 M = 1.00M V = 26.5mlCalculationNa2SO4 + 2H2O Mole ratio 2: 1Using mole ratioMoles 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 = 2.12MUsing formulaMb Vb = 2 Ma Va 1 M x 25.0 = 2 1.0 x 26.5 1 Mb = 2.12M2NH4OH +H2SO4 M = 1.5M V = ? mlM = 0.5M V = 30.0ml(NH4)2SO4 + 2H2OUsing mole ratioMoles 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 = 20mlMole ratio 2: 1Using formulaM bVb = 2 Ma Va 1 1.5 x Vb = 2 0.5 x 30.0 1 Vb = 20ml 14. Acid/Base Titration Calculation 3Find vol of 2.0M HCI needed to neutralize 2.65g of sodium carbonate (Na2CO3) in 50ml water. HCI M = 2.0M V = ? ml410.0cm3 of 0.200M Na2CO3 needed 25.0cm3 of HCI for neutralization. Find molarity of HCI. Na2CO3 M = 0.200M V = 10.0mlNa2CO3 2.65g V = 50mlMoles = Mass/M = 2.65 106 = 0.025 molHCI M=? V = 25.0mlCalculation Na2CO3 + M = 0.5M V = 50mlCalculation2HCI 2NaCI + CO2 + H2O M = 2.0M V = ? mlMole ratio 1: 2Na2CO3 M = 0.200M V = 10.0ml+2HCI 2NaCI + H2O + CO2M=? V = 25.0mlMole ratio 1: 2 15. Acid/Base Titration Calculation 3Find vol of 2.0M HCI needed to neutralize 2.65g of sodium carbonate (Na2CO3) in 50ml water. HCI M = 2.0M V = ? ml410.0cm3 of 0.200M Na2CO3 needed 25.0cm3 of HCI for neutralization. Find molarity of HCI. Na2CO3 M = 0.200M V = 10.0mlNa2CO3 2.65g V = 50mlMoles = Mass/M = 2.65 106 = 0.025 molHCI M=? V = 25.0mlCalculation Na2CO3 + M = 0.5M V = 50mlCalculation2HCI 2NaCI + CO2 + H2O M = 2.0M V = ? mlUsing mole ratioMoles 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 = 25cm3Mole ratio 1: 2Using formulaMbVb = 1 Ma Va 2 0.5 x 50.0 = 1 2.0 x V 2 Va = 25cm3Na2CO3+M = 0.200M V = 10.0ml2HCI 2NaCI + H2O + CO2M=? V = 25.0mlUsing mole ratioMoles 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 = 0.160MMole ratio 1: 2Using formulaMbVb = 1 Ma Va 2 0.2 x 10.0 = 1 Ma x 25.0 2 Ma = 0.160M 16. Acid/Base Titration Calculation 5Vol NaOH need to neutralize 1.325 g ethanedioic acid (C2H2O4.2H2O) is 27.52cm3, calculate molarity of NaOH.633.7cm3 of HCI neutralizes 20cm3 of Na2CO3 of 1.37M. Find molarity of acid. HCI M=? V = 33.7mlNaOH M=?M V = 27.52 ml Moles = Mass/M = 1.325 126.08 = 0.0105 molAcid Mass = 1.325gNa2CO3 M = 1.37M V = 20.0mlCalculation C2H2O4+Mass - 1.325g Mole 0.0105Calculation2NaOH Na2C2O4 + 2H2O M= ? V = 27.52mlMole ratio 1: 2Na2CO3 M = 1.37M V = 20.0ml+ 2HCI 2NaCI + H2O + CO2M=? V = 33.7mlMole ratio 1: 2 17. Acid/Base Titration Calculation 5Vol NaOH need to neutralize 1.325 g ethanedioic acid (C2H2O4.2H2O) is 27.52cm3, calculate molarity of NaOH.633.7cm3 of HCI neutralizes 20cm3 of Na2CO3 of 1.37M. Find molarity of acid. HCI M=? V = 33.7mlNaOH M=?M V = 27.52 ml Moles = Mass/M = 1.325 126.08 = 0.0105 molAcid Mass = 1.325gNa2CO3 M = 1.37M V = 20.0mlCalculation C2H2O4+Mass - 1.325g Mole 0.0105Calculation2NaOH Na2C2O4 + 2H2O M= ? V = 27.52mlUsing mole ratioMoles of Acid = Mass/M = (1.325 126.08) = 1.05 x 10-2 Mole ratio (1 : 1) 1 mole acid neutralize 2 mole base 1.05 x 10-2 acid neutralize 2.10 x 10-2 base Mole of Base = M x V = M x 0.02752 M x 0.02752 = 2.10 x 10-2 M = 0.764MMole ratio 1: 2Using formulaM aVa = 1 Mb Vb 2 0.0105 = 1 M x 0.02752 2 Mb = 0.764MNa2CO3+ 2HCIM = 1.37M V = 20.0ml 2NaCI + H2O + CO2M=? V = 33.7mlMole ratio 1: 2Using mole ratioUsing formulaMoles of Base = MV = (1.37 x 0.020) = 2.74 x 10-2 Mole ratio (1 : 2) 1 mole base neutralize 2 mole acid 2.74 x 10-2 base neutralize 5.48 x 10-2 acid Moles of Acid = M x V = M x 0.0337 M x 0.0337 = 5.48 x 10-2 M = 1.63MM bVb = 1 Ma Va 2 1.37 x 20.0 = 1 Ma x 33.7 2 Ma = 1.63M 18. Acid/Base Titration Calculation Ethanoic acid determination in vinegar 725.0ml of conc vinegar (ethanoic acid) was diluted to a total vol of 250 in a volumetric flask. Diluted vinegar was transfer to a burette. 25.0ml, 0.1M NaOH is pipette into a conical flask, with indicator added. End point reached when average 29.2 ml of diluted vinegar added. Find its molarity. Diluted 25xCH3COOH M=? V = 29.2mlNaOH M = 0.1M V = 25.0ml V = 25 ml M=?V = 250ml M = 0.08561NaOH + CH3COOH CH3COONa + H2O M = 0.1M V = 25.0mlM=? V = 29.2mlMole ratio 1: 1 19. Acid/Base Titration Calculation Ethanoic acid determination in vinegar 725.0ml of conc vinegar (ethanoic acid) was diluted to a total vol of 250 in a volumetric flask. Diluted vinegar was transfer to a burette. 25.0ml, 0.1M NaOH is pipette into a conical flask, with indicator added. End point reached when average 29.2 ml of diluted vinegar added. Find its molarity. Diluted 25xCH3COOH M=? V = 29.2mlNaOH M = 0.1M V = 25.0ml V = 25 ml M=?V = 250ml M = 0.08561NaOH + CH3COOH CH3COONa + H2O M = 0.1M V = 25.0mlM=? V = 29.2mlMole ratio 1: 1Using mole ratio 2Mole NaOH = MV = (0.1 x 0.025) = 2.5 x 10-3 Mole ratio (1 : 1) 1 mole NaOH react 1 mole acid 2.5 x 10-3 mole NaOH react 2.5 x 10-3 acid Mole acid = M V = M x 0.0292 M x 0.0292 = 2.5 x 10-3 qcid M = 2.5 x 10-3 0.0292 M = 0.0856MUsing formulaM aVa = 1 Mb Vb 1 0.1 x 0.025 = M x 0.0292 M = 0.0856M1 1 20. Acid/Base Titration Calculation Ethanoic acid determination in vinegar 725.0ml of conc vinegar (ethanoic acid) was diluted to a total vol of 250 in a volumetric flask. Diluted vinegar was transfer to a burette. 25.0ml, 0.1M NaOH is pipette into a conical flask, with indicator added. End point reached when average 29.2 ml of diluted vinegar added. Find its molarity. Diluted 25xCH3COOH M=? V = 29.2mlNaOH M = 0.1M V = 25.0ml V = 25 ml M=?V = 250ml M = 0.08561NaOH + CH3COOH CH3COONa + H2O M = 0.1M V = 25.0mlM=? V = 29.2mlMole ratio 1: 1Using mole ratioMoles bef dilution = Moles aft dilution M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol3M1 x V1 = M2 x V2 M1 x 25 = 0.0856 x 250 M1 = 0.0856 x 250 25 M1 = 0.856M2Using formulaMole NaOH = MV = (0.1 x 0.025) = 2.5 x 10-3 Mole ratio (1 : 1) 1 mole NaOH react 1 mole acid 2.5 x 10-3 mole NaOH react 2.5 x 10-3 acid Mole acid = M V = M x 0.0292 M x 0.0292 = 2.5 x 10-3 qcid M = 2.5 x 10-3 0.0292 M = 0.0856MM aVa = 1 Mb Vb 1 0.1 x 0.025 = M x 0.0292 M = 0.0856M1 1Video on determination of ethanoic acid in vinegar 21. Acid/Base Titration Calculation - Empirical formula determination for Na2CO3. x H2O 827.82g of hydrated (Na2CO3 . x H2O) dissolved in water, making up to 1L. 25.0ml of solution was neutralized by 48.8ml of 0.100M HCI. Cal molarity and mass of Na2CO3 present in 1L of solution. Find x . HCI M = 0.100 M V = 48.8mlDiuted to 1LNa2CO3 M=?M V = 25.0ml25 ml transfer27.82g Na2CO3. xH2OV = 1L M=?12HCI + Na2CO3 2NaCI +CO2 + H2O M = 0.1M V = 48.8mlM=? V = 25.0mlMole ratio 2: 1 22. Acid/Base Titration Calculation - Empirical formula determination for Na2CO3. x H2O 827.82g of hydrated (Na2CO3 . x H2O) dissolved in water, making up to 1L. 25.0ml of solution was neutralized by 48.8ml of 0.100M HCI. Cal molarity and mass of Na2CO3 present in 1L of solution. Find x . HCI M = 0.100 M V = 48.8mlDiuted to 1LNa2CO3 M=?M V = 25.0ml25 ml transfer27.82g Na2CO3. xH2O1V = 1L M=?2HCI + Na2CO3 2NaCI +CO2 + H2O M = 0.1M V = 48.8mlM=? V = 25.0mlMole ratio 2: 1Using mole ratio 2Mole HCI = MV = (0.1 x 0.0488) = 4.88 x 10-3 Mole ratio (2 : 1) 2 mole HCI react 1 mole base 4.88 x 10-3 mole HCI react 2.44 x 10-3 base Mole base = M V = M x 0.0250 M x 0.0250 = 2.44 x 10-3 base M = 2.44 x 10-3 0.0250 M = 0.0976M3Convert moles to mass in 1Lmol/dm3 g/dm3 mol/dm3 x M g/dm3 0.0976 x 106 = 10.36g/dm3Using formulaM aVa = 2 Mb Vb 1 0.1 x 0.0488 = 2 M x 0.0250 1 M = 0.0976M 23. Acid/Base Titration Calculation - Empirical formula determination for Na2CO3. x H2O 827.82g of hydrated (Na2CO3 . x H2O) dissolved in water, making up to 1L. 25.0ml of solution was neutralized by 48.8ml of 0.100M HCI. Cal molarity and mass of Na2CO3 present in 1L of solution. Find x . HCI M = 0.100 M V = 48.8mlDiuted to 1LNa2CO3 M=?M V = 25.0ml25 ml transfer27.82g Na2CO3. xH2O1V = 1L M=?2HCI + Na2CO3 2NaCI +CO2 + H2O M = 0.1M V = 48.8mlM=? V = 25.0mlMole ratio 2: 1Using mole ratio 24Mass Na2CO3 . x H2O = 27.82 g Mass Na2CO3 = 10.36 g Mass of water = (27.82 10.36) g = 17.46g Empirical formula5Mole HCI = MV = (0.1 x 0.0488) = 4.88 x 10-3 Mole ratio (2 : 1) 2 mole HCI react 1 mole base 4.88 x 10-3 mole HCI react 2.44 x 10-3 base Mole base = M V = M x 0.0250 M x 0.0250 = 2.44 x 10-3 base M = 2.44 x 10-3 0.0250 M = 0.0976MNa2CO317.463RMM10618.02Mole10.34/106 = 0.0977317.46/18.02 = 0.9689mol/dm3 g/dm3 mol/dm3 x M g/dm3 0.0976 x 106 = 10.36g/dm30.09773/0.09733 10.9689/0.09733 10M aVa = 2 Mb Vb 1 0.1 x 0.0488 = 2 M x 0.0250 1 M = 0.0976MH2O10.36Using formulaMass/gLowest ratioEmpirical formula Na2CO3 . 10 H2OConvert moles to mass in 1LVideo on Na2CO3 calculation 24. Redox Titration Calculation- % Iron in iron tablet 9Iron 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. KMnO4 M = 0.002M V = 24.5 ml1.863 g 250mlMnO4M = 0.002M V = 24.5mlFe2+ M=? V = 30ml+ 5Fe2+ + 8H+ Mn2+ + 5Fe2+ + 4H2O M= ?Mole ratio 1: 5 25. Redox Titration Calculation- % Iron in iron tablet 9Iron 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. KMnO4 M = 0.002M V = 24.5 ml1.863 g 250mlMnO4M = 0.002M V = 24.5mlFe2+ M=? V = 30ml+ 5Fe2+ + 8H+ Mn2+ + 5Fe2+ + 4H2O M= ?Mole ratio 1: 5Using mole ratio 1Mole 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 10-5 KMO4-react 2.45 x 10-4 Fe2+Using formulaM aVa = 1 Mb Vb 5 0.002 x 0.0245 = 1 Moles Fe2+ 5 Moles = 2.45 x 10-4 Fe2+ 26. Redox Titration Calculation- % Iron in iron tablet 9Iron 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.Video on % Iron in iron tabletKMnO4 M = 0.002M V = 24.5 ml1.863 g 250mlMnO4M = 0.002M V = 24.5mlFe2+ M=? V = 30ml+ 5Fe2+ + 8H+ Mn2+ + 5Fe2+ + 4H2O M= ?Mole ratio 1: 5Video on Fe2+/KMnO4 titration calculation Using mole ratioUsing formulaMole 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 10-5 KMO4-react 2.45 x 10-4 Fe2+12M aVa = 1 Mb Vb 5 0.002 x 0.0245 = 1 Moles Fe2+ 5 Moles = 2.45 x 10-4 Fe2+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+ + SO4 21 mol 1mol + 1mol 6.125 x 10-3 mol 6.125 x 10-3 mole Fe2+4 3Mole Mass Mole x RMM = Mass FeSO4 6.125 x 10-3 x 278.05 = 1.703g FeSO4Mass of (expt yield) = 1.703g Mass of (Actual tablet) = 1.863g % Fe in iron tablet = 1.703 x 100% 1.863 = 91.4% 27. Redox Titration Calculation CIO- in Bleach 1010.0ml of bleach (CIO-) diluted to a total vol of 250ml. 20.0ml is added to 1g of KI (excess) and iodine produced is titrated with 0.0206M Na2S2O3.Using starch indicator, end point was 17.3ml. Cal molarity of CIO in bleach. Diuted 25x 10.0ml CIOtransferWater added till 250ml20ml transferNa2S2O3 M = 0.0206M V = 17.3ml1g KI excess addedI2 M=?titrated V = 10 M=?V = 250ml M = 1.78 x 10-2 M 12S2O32M = 0.0206 V = 17.3ml+I2 Mole = ? V = 0.02S4O62- + 2IMole ratio 2: 1 28. Redox Titration Calculation CIO- in Bleach 1010.0ml of bleach (CIO-) diluted to a total vol of 250ml. 20.0ml is added to 1g of KI (excess) and iodine produced is titrated with 0.0206M Na2S2O3.Using starch indicator, end point was 17.3ml. Cal molarity of CIO in bleach. Diuted 25x 10.0ml CIOtransferWater added till 250mlNa2S2O3 M = 0.0206M V = 17.3ml1g KI excess added20ml transferI2 M=?titrated V = 10 M=?V = 250ml M = 1.78 x 10-2 M 2S2O32-1M = 0.0206 V = 17.3ml+I2 Mole = ? V = 0.02S4O62- + 2IMole ratio 2: 1Using direct formula 2Mole S2O32- = MV = (0.0206 x 0.0173) = 3.56 x 10-4 Mole ratio (2 : 1) 2 mole S2O32- react 1 mole I2 3.56 x 10-4 S2O32--react 1.78 x 10-4 I22 mol 2CIO- +2I- +I2 + 1 mol2S2O322 mol1 mol 2H+ I2 + 2CI- + H2O S4O62- + 2I-Mole ratio ( 1 : 1) 2 mole CIO- : 1 mole I2 : 2 mole S2O322 mole CIO2 mole S2O32- 29. Redox Titration Calculation CIO- in Bleach 1010.0ml of bleach (CIO-) diluted to a total vol of 250ml. 20.0ml is added to 1g of KI (excess) and iodine produced is titrated with 0.0206M Na2S2O3.Using starch indicator, end point was 17.3ml. Cal molarity of CIO in bleach. Diuted 25x 10.0ml CIOtransferVideo on CIO- in bleachWater added till 250mlNa2S2O3 M = 0.0206M V = 17.3ml1g KI excess added20ml transferI2 M=?titrated V = 10 M=?V = 250ml M = 1.78 x 10-2 M 2S2O32-1 6M = 0.0206 V = 17.3mlMole bef dilution = Mole aft dilution M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol M1 V1 = M2 V2 M1 x 10 = 1.78 x 10-2 x 250 M1 = 1.78 x 10-2 x 250 10 M1 = 0.445M+I2 Mole = ? V = 0.022Mole ratio 2: 1Mole S2O32- = MV = (0.0206 x 0.0173) = 3.56 x 10-4 Mole ratio (2 : 1) 2 mole S2O32- react 1 mole I2 3.56 x 10-4 S2O32--react 1.78 x 10-4 I22CIO- + 2I- + 2H+ I2 + 2CI- + H2O 2CIOI2 Mole ratio 2: 1Mole = ? Moles of CIO- = M x V MxV = 3.56 x 10-4 M x 0.02 = 3.56 x 10-4 M = 3.56 x 10-4 002 M = 1.78 x 10-2 M diluted 25xS4O62- + 2IUsing direct formula354Mole ratio (2 : 1) 2 mole CIO 3.56 x 10-4 CIO-2 mol 2CIO- +2I- +I2 + 1 mol2S2O322 mol1 mol 2H+ I2 + 2CI- + H2O S4O62- + 2I-Mole ratio ( 1 : 1) 2 mole CIO- : 1 mole I2 : 2 mole S2O322 mole CIO2 mole S2O32-Mole = 1.78 x 10-41 mole I2 1.78 x 10-4 I2M V (CIO+) = 2 = 1 M V (S2032-) 2 1 Moles of CIO+ = 1 0.0206 x 0.0173 1 Moles of CIO- = 3.56 x 10-4 30. Redox Titration Calculation Vitamin C quantification 11Iodometric titration was performed on Vit C, (C6H8O6). 25ml Vit C is titrated with 0.002M KIO3 from burette, using excess KI and starch. Average vol KIO3 is 25.5ml. Cal molarity of Vit C. KIO3 M = 0.002M V = 25.5ml1g KI excess + starch added25ml transferVitamin CVit C M=? V = 25mltitrated V = 25ml M= ? 1KIO3 + 5KI + 6H+ M = 0.002M V = 25.5ml3I2 +Mole = ?3H2O + 6K= Mole ratio 1: 3 31. Redox Titration Calculation Vitamin C quantification 11Iodometric titration was performed on Vit C, (C6H8O6). 25ml Vit C is titrated with 0.002M KIO3 from burette, using excess KI and starch. Average vol KIO3 is 25.5ml. Cal molarity of Vit C. KIO3 M = 0.002M V = 25.5ml1g KI excess + starch added25ml transferVitamin CVit C M=? V = 25mltitrated V = 25ml M= ? 1KIO3 + 5KI + 6H+ M = 0.002M V = 25.5ml3I2 +Mole = ?3H2O + 6K= Mole ratio 1: 3Using mole ratio 2Mole KIO3 = MV = (0.002 x 0.0255) = 5.10 x 10-5 Mole ratio (1 : 3) 1 mole KIO3 produce 3 mole I2 5.10 x 10-5 KIO3 produce 1.53 x 10-4 I2Using formula1 mol 3 mol KIO3 + 5KI + 6H+ 3I2 + 6K+ + 3H2O 3C6H8O6 + 3I2 3C6H6O6 + 6I- + 6H+ 3 mol 3 molMole ratio (1 :3) 1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6 1 mol KIO3 3 mol C6H8O6 32. Redox Titration Calculation Vitamin C quantification 11Iodometric titration was performed on Vit C, (C6H8O6). 25ml Vit C is titrated with 0.002M KIO3 from burette, using excess KI and starch. Average vol KIO3 is 25.5ml. Cal molarity of Vit C. KIO3 M = 0.002M V = 25.5ml1g KI excess + starch added25ml transferVitamin CVit C M=? V = 25mltitrated V = 25ml M= ? 1KIO3 + 5KI + 6H+ M = 0.002M V = 25.5ml3I2 +Mole = ?3H2O + 6K= Mole ratio 1: 3Using mole ratio 23Mole KIO3 = MV = (0.002 x 0.0255) = 5.10 x 10-5 Mole ratio (1 : 3) 1 mole KIO3 produce 3 mole I2 5.10 x 10-5 KIO3 produce 1.53 x 10-4 I23C6H8O6 Mole = ?+ 3I2 3C6H6O6 + 6I- + 6H+ 1.53 x 10-4 Mole ratio 3: 3Using formula1 mol 3 mol KIO3 + 5KI + 6H+ 3I2 + 6K+ + 3H2O 3C6H8O6 + 3I2 3C6H6O6 + 6I- + 6H+ 3 mol 3 molMole ratio (1 :3) 1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6 1 mol KIO3 3 mol C6H8O6 Using formula4Mole ratio (1 : 3) 1 mol KIO3 react 3 mol C6H8O6 5.10 x 10-5 KIO3 react 1.53 x 10-4 C6H8O6 Mole C6H8O6 = M x V MxV = 1.53 x 10-4 M x 0.025 = 1.53 x 10-4 M = 1.53 x 10-4 0025 M = 6.12 x 10-3 MM aVa (KIO3) = 1 Mb Vb (C6H8O6) 3 0.002 x 0.0255 = 1 Mole C6H8O6 3 Mole C6H8O6 = 1.53 x 10-4 Mole C6H8O6 = M x V M x V = 1.53 x 10-4 M x 0.025 = 1.53 x 10-4 M = 1.53 x 10-4 0025 M = 6.12 x 10-3 M 33. Redox Titration Calculation - % Cu in Brass 122.5g brass react with 10ml conc HNO3 producing Cu2+ ions. Solution made up to 250ml using water in a volumetric flask. Pipette 25ml of solution into conical flask. Na2CO3 added to neutralize excess acid. 1g KI (excess) and few drops of starch added to flask. Titrate with 0.1M S 2O32- and end point, reached when 28.2ml added. Find molarity copper ions and % copper found in brass. Water added till 250ml 1g KI excess + starch addedPour into Volumetric flask25ml transfer 2.5g brass10 ml HNO3Na2S2O3 M = 0.1M V = 28.2mlV = 250ml M=?1I2 M=?titrated2S2O32M = 0.1M V = 28.2ml+I2Mole = ?S4O62- + 2IMole ratio 2: 1 34. Redox Titration Calculation - % Cu in Brass 122.5g brass react with 10ml conc HNO3 producing Cu2+ ions. Solution made up to 250ml using water in a volumetric flask. Pipette 25ml of solution into conical flask. Na2CO3 added to neutralize excess acid. 1g KI (excess) and few drops of starch added to flask. Titrate with 0.1M S 2O32- and end point, reached when 28.2ml added. Find molarity copper ions and % copper found in brass. Water added till 250ml 1g KI excess + starch addedPour into Volumetric flask25ml transfer 2.5g brass10 ml HNO3Na2S2O3 M = 0.1M V = 28.2mlV = 250ml M=?I2 M=?titrated2S2O32-1M = 0.1M V = 28.2ml+I2Mole = ?S4O62- + 2IMole ratio 2: 1Using mole ratio 2Mole S2O32- = MV = (0.1 x 0.0282) = 2.82 x 10-3 Mole ratio (2 : 1) 2 mole S2O32- react 1 mole I2 2.82 x 10-3 S2O32-- react 1.41 x 10-3 I2Using formula2 mol 2Cu2+ +I2 + 1 mol4I-2S2O32- 2 mol1 mol I2 + 2CuIS4O62- + 2I-Mole ratio ( 1 : 1) 2 mole Cu2+ : 1 mole I2 : 2 mole S2O322 mole Cu2+ 2 mole S2O32- 35. Redox Titration Calculation - % Cu in Brass 122.5g brass react with 10ml conc HNO3 producing Cu2+ ions. Solution made up to 250ml using water in a volumetric flask. Pipette 25ml of solution into conical flask. Na2CO3 added to neutralize excess acid. 1g KI (excess) and few drops of starch added to flask. Titrate with 0.1M S 2O32- and end point, reached when 28.2ml added. Find molarity copper ions and % copper found in brass. Water added till 250ml 1g KI excess + starch addedPour into Volumetric flask25ml transfer 2.5g brass10 ml HNO3Na2S2O3 M = 0.1M V = 28.2mlV = 250ml M=?I2 M=?titrated2S2O32-1M = 0.1M V = 28.2ml+I2Mole = ?S4O62- + 2IMole ratio 2: 1Using mole ratio 2Using formulaMole S2O32- = MV = (0.1 x 0.0282) = 2.82 x 10-3 Mole ratio (2 : 1) 2 mole S2O32- react 1 mole I2 2.82 x 10-3 S2O32-- react 1.41 x 10-3 I22Cu2+ + 4I- I2 + 2CuI Mole = ? 1.41 x 10-3 I2 Mole ratio 2: 132 mol 2Cu2+ +I2 + 1 mol4I-2S2O32- 2 mol1 mol I2 + 2CuIS4O62- + 2I-Mole ratio ( 1 : 1) 2 mole Cu2+ : 1 mole I2 : 2 mole S2O322 mole Cu2+ 2 mole S2O32Using formula5Mole of Cu2+ = M x V MxV = 2.82 x 10-3 M x 0.025 = 2.82 x 10-3 M = 2.82 x 10-3 0.025 M = 1.13 x 10-1 M Mass Cu = Molarity Cu x M Mass Cu = (1.13 x 63.5)g Cu in 1000ml = 7.18g Cu in 1000ml = 1.79g Cu in 250ml46Mole ratio (2 : 1) 2 mole Cu2+ 2.82 x 10-3 Cu2+1 mole I2 1.41 x 10-3 I2% Cu in brass = mass Cu x 100% mass brass = 1.79 x 100% 2.5 = 71.8%M V (Cu2+) = 2 = 1 M V (S2032-) 2 1 Moles of Cu2+ = 1 0.1 x 0.0282 1 Moles of Cu2+ = 2.82 x 10-3 36. TitrationDirect Titration Condition Both titrant and analyte soluble Reaction is fastTitrant - solubleAnalyte - solubleRedox TitrationAcid Base TitrationComplexometric TitrationH2SO4 HCI HNO3Soluble acidNaOH NH4OH Soluble base KOH (Alkali) Ba(OH)2 LiOH 37. TitrationDirect TitrationBack Titration ConditionCondition Both titrant and analyte soluble Reaction is fast Sparingly soluble acid/base. Rxn too slow, unable to dissolve/react Calcium carbonate (egg shell) and impurities (base) from antacid tablet/impure limestone. Salicyclic acid in aspirin tablet.Titrant - solubleAnalyte - solubleaddedRedox TitrationAcid Base TitrationComplexometric TitrationImpure antacid Known conc /vol of acid usedH2SO4 HCI HNO3Soluble acidNaOH NH4OH Soluble base KOH (Alkali) Ba(OH)2 LiOHImpure limestoneCaCO3 in egg shell 38. TitrationDirect TitrationBack Titration ConditionCondition Both titrant and analyte soluble Reaction is fast Sparingly soluble acid/base. Rxn too slow, unable to dissolve/react Calcium carbonate (egg shell) and impurities (base) from antacid tablet/impure limestone. Salicyclic acid in aspirin tablet.Titrant - solubleAnalyte - solubleaddedRedox TitrationAcid Base TitrationComplexometric TitrationImpure limestoneCaCO3 in egg shell Impure antacid Known conc /vol of acid used Left overnight in acidH2SO4 HCI HNO3Soluble acidNaOH NH4OH Soluble base KOH (Alkali) Ba(OH)2 LiOHAmt of known acid added Titrated with known conc/vol alkali known unknownAmt of excess acid left Transfer to flaskAmt of base (solid) react with acid Video on back titration calculationAmt base(solid) = Amt Known Amt excess acid added acid leftExcess acid left 39. Back Titration% Calcium carbonate in egg shell - Back Titration Calculation25.0g of egg shell (CaCO3) dissolved in 250.0ml, 2.0M HNO3. Solution was titrated with NaOH. 17.0ml, 1.00M NaOH needed to neutralize excess acid. Cal percentage CaCO3 by mass in egg shell.added25.0g impure CaCO3 in egg shell 250.0ml, 2.0M HNO3 Left overnight in acidAmt of HNO3 addedTitrated with known conc/vol NaOH M = 1.0 V = 17.0mlTransfer to flaskAmt of HNO3 left Amt of base (egg)Amt HNO3 react = Amt HNO3 Amt HNO3 with base add leftHNO3 left 40. Back Titration% Calcium carbonate in egg shell - Back Titration Calculation25.0g of egg shell (CaCO3) dissolved in 250.0ml, 2.0M HNO3. Solution was titrated with NaOH. 17.0ml, 1.00M NaOH needed to neutralize excess acid. Cal percentage CaCO3 by mass in egg shell.1added25.0g impure CaCO3 in egg shell 2250.0ml, 2.0M HNO3 Left overnight in acidAmt of HNO3 addedTitrated with known conc/vol NaOH M = 1.0 V = 17.0mlTransfer to flaskAmt of HNO3 left Amt of base (egg)Amt HNO3 react = Amt HNO3 Amt HNO3 with base add leftHNO3 leftAmt HNO3 add = M V = 2.0 x 0.250 = 0.50 molNaOH + HNO3 M = 1.00M V = 17.0mlmoles = ? NaNO3 + H2O Mole ratio 1: 1Amt HNO3 add 41. Back Titration% Calcium carbonate in egg shell - Back Titration Calculation25.0g of egg shell (CaCO3) dissolved in 250.0ml, 2.0M HNO3. Solution was titrated with NaOH. 17.0ml, 1.00M NaOH needed to neutralize excess acid. Cal percentage CaCO3 by mass in egg shell.1added25.0g impure CaCO3 in egg shell 2250.0ml, 2.0M HNO3 Left overnight in acidAmt of HNO3 addedTitrated with known conc/vol NaOH M = 1.0 V = 17.0mlTransfer to flaskAmt of HNO3 left Amt of base (egg)Amt HNO3 react = Amt HNO3 Amt HNO3 with base add leftHNO3 leftAmt HNO3 add = M V = 2.0 x 0.250 = 0.50 molNaOH + HNO3 M = 1.00M V = 17.0mlmoles = ?Amt HNO3 add NaNO3 + H2O Mole ratio 1: 1Using mole ratio 3Mole NaOH = MV = (1.00 x 0.017) = 1.7 x 10-2 Mole ratio (1 : 1) 1 mole NaOH react 1 mole HNO3 1.7 x 10-2 mole NaOH react 1.7 x 10-2 HNO3 HNO3 left = 1.7 x 10-2 molUsing formula 3MbVb = 1 Ma Va 1 1.00 x 0.017 = 1 moles 1 mole HNO3 = 1.7 x 10-2 Amt HNO3 left 42. Back Titration% Calcium carbonate in egg shell - Back Titration Calculation25.0g of egg shell (CaCO3) dissolved in 250.0ml, 2.0M HNO3. Solution was titrated with NaOH. 17.0ml, 1.00M NaOH needed to neutralize excess acid. Cal percentage CaCO3 by mass in egg shell.1added25.0g impure CaCO3 in egg shell 2250.0ml, 2.0M HNO3 Left overnight in acidAmt of HNO3 addedTitrated with known conc/vol NaOH M = 1.0 V = 17.0mlHNO3 left346 NaNO3 + H2O Mole ratio 1: 1Using formulaMole NaOH = MV = (1.00 x 0.017) = 1.7 x 10-2 Mole ratio (1 : 1) 1 mole NaOH react 1 mole HNO3 1.7 x 10-2 mole NaOH react 1.7 x 10-2 HNO3 HNO3 left = 1.7 x 10-2 mol3Mole ?Amt HNO3 react with base7Mass = Mole CaCO3 x RMM CaCO3 = 0.242 x 100 = 24.2g8% by mass = mass CaCO3 x 100% CaCO3 mass impure = (24.2/25.0) x 100% = 96.8%Mole ratio 2: 1Mole ratio (2 : 1) 2 mol HNO3 react 1 mol CaCO3 0.483 mol HNO3 react o.242 mol CaCO3MbVb = 1 Ma Va 1 1.00 x 0.017 = 1 moles 1 mole HNO3 = 1.7 x 10-2 Amt HNO3 leftAmt HNO3 react = Amt HNO3 add Amt HNO3 left with NaOH = 0.50 1.7 x 10-2 = 0.483 mol2HNO3 + CaCO3 (CaNO3)2 + H2O + 2CO2 Mole 0.483Amt HNO3 react = Amt HNO3 Amt HNO3 with base add leftmoles = ?Amt HNO3 addUsing mole ratioAmt of base (egg)5NaOH + HNO3 M = 1.00M V = 17.0mlTransfer to flaskAmt of HNO3 leftAmt HNO3 add = M V = 2.0 x 0.250 = 0.50 mol 43. Back Titration% Calcium hydroxide in an impure antacid tablet - Back Titration Calculation0.5214g 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.added0.5214g impure Ca(OH)2 50.0ml, 0.250M HCI Left overnight in acidTitrated with known conc/vol NaOH M = 0.1108 V = 33.64mlAmt of HCI addedTransfer to flaskAmt HCI left Amt of base (solid)Amt HCI react = Amt HCI Amt HCI with NaOH add leftHCI left 44. Back Titration% Calcium hydroxide in an impure antacid tablet - Back Titration Calculation0.5214g 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. 1added0.5214g impure Ca(OH)2 250.0ml, 0.250M HCI Left overnight in acidTitrated with known conc/vol NaOH M = 0.1108 V = 33.64mlAmt of HCI addedTransfer to flaskAmt HCI left Amt of base (solid)Amt HCI react = Amt HCI Amt HCI with NaOH add leftHCI leftAmt HCI add = M V = 0.250 x 0.050 = 0.01250 molNaOH + M = 0.1108M V = 33.64mlHCImoles = ?NaCI + H2O Mole ratio 1: 1Amt HCI Add 45. Back Titration% Calcium hydroxide in an impure antacid tablet - Back Titration Calculation0.5214g 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. 1added0.5214g impure Ca(OH)2 250.0ml, 0.250M HCI Left overnight in acidTitrated with known conc/vol NaOH M = 0.1108 V = 33.64mlAmt of HCI addedTransfer to flaskAmt HCI left Amt of base (solid)Amt HCI react = Amt HCI Amt HCI with NaOH add leftHCI leftAmt HCI add = M V = 0.250 x 0.050 = 0.01250 molNaOH + M = 0.1108M V = 33.64mlHCImoles = ?Amt HCI AddNaCI + H2O Mole ratio 1: 1Using mole ratio 3Mole 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 molUsing formula 3MbVb = 1 Ma Va 1 0.1108 x 0.03364 = 1 moles 1 mole HCI = 3.727 x 10-3 Amt HCI Left 46. Back Titration% Calcium hydroxide in an impure antacid tablet - Back Titration Calculation0.5214g 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. 1added0.5214g impure Ca(OH)2 250.0ml, 0.250M HCI Left overnight in acidTitrated with known conc/vol NaOH M = 0.1108 V = 33.64mlAmt of HCI addedHCImoles = ?Amt HCI AddNaCI + H2O Mole ratio 1: 1Using formulaUsing mole ratio 3Mole 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 mol3MbVb = 1 Ma Va 1 0.1108 x 0.03364 = 1 moles 1 mole HCI = 3.727 x 10-3 Amt HCI LeftHCI left 4Amt of base (solid)56Amt HCI react = Amt HCI add Amt HCI left with NaOH = 0.01250 3.727 x 10-3 = 0.008773 mol2HCI + Ca(OH)2 Mole 0.008773Amt HCI react = Amt HCI Amt HCI with NaOH add leftNaOH + M = 0.1108M V = 33.64mlTransfer to flaskAmt HCI leftAmt HCI add = M V = 0.250 x 0.050 = 0.01250 molMole ? CaCI2 + 2H2O7Mass = Mole Ca(OH)2 x RMM Ca(OH)2= 0.004386 x 74.1 = 0.3250g8% by mass = mass Ca(OH)2 x 100% Ca(OH)2 mass impure = (0.3250/0.5214) x 100% = 62.3%Mole ratio 2: 1Mole ratio (2 : 1) 2 mol HCI react 1 mol Ca(OH)2 0.008773 mol HCI react o.oo4386 mol Ca(OH)2Amt HCI react with base 47. Molar mass of insoluble acid in a tablet -Back Titration CalculationBack Titration2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acidadded2.04g impure dibasic acid H2A 20.0ml, 2.00M NaOH Left overnightTitrated with known conc/vol HCI M = 0.50 V = 17.6mlAmt NaOH addedTransfer to flaskAmt of NaOH leftNaOH leftAmt of acid (solid)Amt NaOH react = Amt NaOH Amt NaOH with acid add left 48. Molar mass of insoluble acid in a tablet -Back Titration CalculationBack Titration2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acid1added2.04g impure dibasic acid H2A 220.0ml, 2.00M NaOH Left overnightTitrated with known conc/vol HCI M = 0.50 V = 17.6mlAmt NaOH addedTransfer to flaskAmt of NaOH leftNaOH leftAmt of acid (solid)Amt NaOH react = Amt NaOH Amt NaOH with acid add leftAmt NaOH add = M V = 2.00 x 0.02 = 0.040 molHCI+M = 0.50M V = 17.6mlNaOH moles = ?NaCI + H2O Mole ratio 1: 1Amt NaOH add 49. Molar mass of insoluble acid in a tablet -Back Titration CalculationBack Titration2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acid1added2.04g impure dibasic acid H2A 220.0ml, 2.00M NaOH Left overnightTitrated with known conc/vol HCI M = 0.50 V = 17.6mlAmt NaOH addedTransfer to flaskAmt of NaOH leftNaOH leftAmt of acid (solid)Amt NaOH react = Amt NaOH Amt NaOH with acid add leftAmt NaOH add = M V = 2.00 x 0.02 = 0.040 molHCI+M = 0.50M V = 17.6mlNaOH moles = ?NaCI + H2O Mole ratio 1: 1Using mole ratio 3Amt NaOH addMole HCI = MV = (0.50 x 0.0176) = 8.8 x 10-3 Mole ratio (1 : 1) 1 mol HCI react 1 mol NaOH 8.8 x 10-3 mol HCI react 8.8 x 10-3 NaOH NaOH left = 8.8 x 10-3 molUsing formula 3MaVa = 1 Mb Vb 1 0.50 x 0.0176 = 1 moles 1 mole HCI = 8.8 x 10-3 Amt NaOH Left 50. Molar mass of insoluble acid in a tablet -Back Titration CalculationBack Titration2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acid1added2.04g impure dibasic acid H2A 220.0ml, 2.00M NaOH Left overnightTitrated with known conc/vol HCI M = 0.50 V = 17.6mlAmt NaOH addedHCI+M = 0.50M V = 17.6mlNaOH moles = ?Amt NaOH addNaCI + H2O Mole ratio 1: 1Using formulaUsing mole ratio 3Transfer to flaskAmt of NaOH leftAmt NaOH add = M V = 2.00 x 0.02 = 0.040 molMole HCI = MV = (0.50 x 0.0176) = 8.8 x 10-3 Mole ratio (1 : 1) 1 mol HCI react 1 mol NaOH 8.8 x 10-3 mol HCI react 8.8 x 10-3 NaOH NaOH left = 8.8 x 10-3 mol3MaVa = 1 Mb Vb 1 0.50 x 0.0176 = 1 moles 1 mole HCI = 8.8 x 10-3 Amt NaOH LeftNaOH left 4Amt of acid (solid)5Amt NaOH = Amt NaOH Amt NaOH react with acid add left = 0.040 8.8 x 10-3 = 0.0312 mol 2NaOH + H2A Na2 A + 2H2O Mole 0.00312Amt NaOH react = Amt NaOH Amt NaOH with acid add left 6Amt NaOH react with acidMole ?Mole ratio 2: 1Mole ratio (2 : 1) 2 mol NaOH react 1 mol H2 A 0.0312 mol NaOH react o.o156 mol H2A7Molar Mass 0.0156 mol H2A - 2.04g 1 mol H2A - 2.04 0.0156 = 131 51. Video on TitrationClick here titration NaOH /HCIClick here titration videoSimulation on TitrationClick here titration simulationClick here titration simulationClick here on titration simulationClick here titration simulationClick here on titration simulation 52. 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