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IB Chemistry on Acid Base, Redox, Back Titration calculation and experiments.

IB Chemistry on Acid Base, Redox, Back Titration calculation and experiments.

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- 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 Expt Standardization NaOH with primary std KHP Click here or here for expt` Titration bet HCI with std NaOH Click here for expt 4.2a Standardization HCI with primary std Na2CO3 Click here for expt 4.2 Titration bet NaOH with std HCI Click here for expt 4.2a Acid/Base Titration Expt Determining acetylsalicylic acid in aspirin with std NaOH Determining ethanoic acid in vinegar with std NaOH Click here or here for expt` Click here for more expt Click here for expt 4.3 Determining water crystallization in hydrated Na2CO3 with std HCI Click here for expt 4.4
- 3. Titration for IA (DCP) assessment Acid Base Titration Standardization Expt Standardization NaOH with primary std KHP Acid/Base Titration Expt Standardization HCI with primary std Na2CO3 Click here or here for expt` Determining acetylsalicylic acid in aspirin with std NaOH Click here or here for expt` Click here for more expt Click here for expt 4.2 Titration bet HCI with std NaOH Click here for expt 4.3 Determining water crystallization in hydrated Na2CO3 with std HCI Titration bet NaOH with std HCI Click here for expt 4.2a Determining ethanoic acid in vinegar with std NaOH Click here for expt 4.4 Click here for expt 4.2a Redox Titration Standardization Expt Standardization KMnO4 with std ammonium iron(II) sulphate Click here for expt 4.5 Standardization KI/I2 with std sodium thiosulphate Click here for expt 4.7 Hypochlorite (OCI-) in bleach with iodine/thiosulphate Click here for expt 4.8 Standardization KI/I2 with std KIO3 Click here for expt 4.7 Click here for more expt Iron (II) determination with std KMnO4 Click here for expt 4.6 Redox Titration Expt Iodine/thiosulphate (iodometric titration) Copper(II) determination in brass with iodine/thiosulphate Click here or here for expt` Click here for more expt Vit C determination with iodine/thiosulphate Click here or here for expt Click here more detail expt
- 4. Titration Redox Titration Acid Base Titration Complexometric titration Neutralization Condition 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. Titration Redox Titration Acid Base Titration Complexometric titration Neutralization Condition 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 Accurate known conc Primary standard acids - Potassium hydrogen phthalate 20.4 g in 1L Burette 20.4 g KHP Volumetric flask Standard 0.1M KHP Acid/Base used as primary standard -Stable/solid -Soluble in water -Does not decompose over time Unknown Conc NaOH Accurate known conc Primary standard bases - Anhydrous sodium carbonate 10.6 gNa2CO3 10.6g in 1 L Unknown Conc HCI Volumetric flask ? ? Standard 0.1M Na2CO3 Burette
- 6. Titration Redox Titration Acid Base Titration Complexometric titration Neutralization Condition 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 Accurate known conc Primary standard acids - Potassium hydrogen phthalate 20.4 g in 1L Burette 20.4 g KHP Volumetric flask Standard 0.1M KHP Acid/Base used as primary standard -Stable/solid -Soluble in water -Does not decompose over time Unknown Conc NaOH Accurate known conc Primary standard bases - Anhydrous sodium carbonate 10.6 gNa2CO3 10.6g in 1 L Unknown Conc HCI Volumetric 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 amt Standard 0.1M Na2CO3 Burette
- 7. Standardization of (ACID) with standard (BASE) Data Collection Data Processing Vol Na2CO3 Vol 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) = ± 0.10 Na2CO3 vol = (26.40 ± 0.10) Average vol Na2CO3 = 26.4 + 26.4 = 26.4cm3 2 Average Vol Na2CO3 ± uncertainty = 26.4 + 26.4 = (26.4± 0.10)cm3 2 Error Analysis % Uncertainty - burette % Uncertainty - pipette Absolute uncertainty vol x 100% Average vol added = 0.10 x 100% 26.4 = 0.38% Na2CO3 ± 0.03 Absolute uncertainty vol x 100% Average vol added = 0.03 x 100% 25.00 = 0.12% HCI Conc ± uncertainty Conc 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=(0.211±0.001)M (Absolute uncertainty) % Error Lit 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 Collection Data Processing Vol KHP Vol 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) = ± 0.10 KHP vol = (26.40 ± 0.10) Average vol KHP = 26.4 + 26.4 = 26.4cm3 2 Average Vol KHP ± uncertainty = 26.4 + 26.4 = (26.4± 0.10)cm3 2 Error Analysis % Uncertainty - burette Absolute uncertainty vol x 100% Average vol added = 0.10 x 100% 26.4 = 0.38% % Uncertainty - pipette KHP ± 0.03 Absolute uncertainty vol x 100% Average vol added = 0.03 x 100% 25.00 = 0.12% NaOH Conc ± uncertainty Conc 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) % Error Lit value - NaOH = 0.100M Expt value– NaOH = 0.106M Difference = 0.006 % Error – Difference x 100% Literature value 0.006 x 100% = 6% 0.100
- 9. Sample Titration Calculation Standardization of (BASE) with standard (ACID) KHP M = 0.100M V = 26.4 ml Standardization of (ACID) with standard (BASE) Na2CO3 M = 0.100M V = 26.4 ml HCI M=? V = 25.0ml NaOH M=? V = 25.0ml Calculation KHP + M = 0.100M V = 26.40ml NaOH → M= ? V = 25.0ml Calculation NaKP + H2O Mole ratio – 1: 1 Na2CO3 M = 0.100M V = 26.4ml + 2HCI → 2NaCI + H2O + CO2 M=? V = 25.0ml Mole ratio – 1: 2
- 10. Sample Titration Calculation Standardization of (BASE) with standard (ACID) KHP M = 0.100M V = 26.4 ml Standardization of (ACID) with standard (BASE) Na2CO3 M = 0.100M V = 26.4 ml HCI M=? V = 25.0ml NaOH M=? V = 25.0ml Calculation KHP + M = 0.100M V = 26.40ml NaOH → Calculation NaKP M= ? V = 25.0ml Using mole ratio Moles 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 + H2O Mole ratio – 1: 1 Using formula M aVa = 1 Mb Vb 1 0.1 x 26.40 = 1 M x 25.0 1 M b = 0.106M Na2CO3 + 2HCI M = 0.100M V = 26.4ml → 2NaCI + H2O + CO2 M=? V = 25.0ml Using mole ratio Moles 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 = 0.211M Mole ratio – 1: 2 Using formula M 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 ml Na2CO3 M = 0.100M V = 26.4 ml HCI M=? V = 25.0ml NaOH M=? V = 25.0ml Calculation KHP + M = 0.100M V = 26.40ml NaOH → Calculation NaKP M= ? V = 25.0ml Using mole ratio Moles 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 + H2O Na2CO3 + 2HCI M = 0.100M V = 26.4ml Mole ratio – 1: 1 → 2NaCI + H2O + CO2 M=? V = 25.0ml Using mole ratio Using formula M aVa = 1 Mb Vb 1 0.1 x 26.40 = 1 M x 25.0 1 M b = 0.106M Mole ratio – 1: 2 Using formula Moles 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 = 0.211M M bVb = 1 Ma Va 2 0.1 x 26.4 = 1 Ma x 25.0 2 Ma = 0.211M Video on Titration Click here Acid/Base calculation Click here Titration calculation Click here cal Na2CO3/HCI Click here cal NaOH/H2SO4
- 12. 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. 2 Find vol of 1.5M aq NH3 required to completely neutralize 30.0cm3 of 0.5M sulphuric acid 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 2NaOH M=? V = 25.0ml + H2SO4 M = 1.00M V = 26.5ml → Calculation Na2SO4 + 2H2O Mole ratio – 2: 1 2NH4OH + H2SO4 → M = 1.5M V = ? ml M = 0.5M V = 30.0ml (NH4)2SO4 + 2H2O Mole ratio – 2: 1
- 13. 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. 2 Find vol of 1.5M aq NH3 required to completely neutralize 30.0cm3 of 0.5M sulphuric acid 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 2NaOH M=? V = 25.0ml + H2SO4 M = 1.00M V = 26.5ml → Calculation Na2SO4 + 2H2O Mole ratio – 2: 1 Using mole ratio 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 Using formula Mb Vb = 2 Ma Va 1 M x 25.0 = 2 1.0 x 26.5 1 Mb = 2.12M 2NH4OH + H2SO4 → M = 1.5M V = ? ml M = 0.5M V = 30.0ml (NH4)2SO4 + 2H2O Using mole ratio 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 Mole ratio – 2: 1 Using formula M bVb = 2 Ma Va 1 1.5 x Vb = 2 0.5 x 30.0 1 Vb = 20ml
- 14. Acid/Base Titration Calculation 3 Find vol of 2.0M HCI needed to neutralize 2.65g of sodium carbonate (Na2CO3) in 50ml water. HCI M = 2.0M V = ? ml 4 10.0cm3 of 0.200M Na2CO3 needed 25.0cm3 of HCI for neutralization. Find molarity of HCI. Na2CO3 M = 0.200M V = 10.0ml Na2CO3 2.65g V = 50ml Moles = Mass/M = 2.65 106 = 0.025 mol HCI M=? V = 25.0ml Calculation Na2CO3 + M = 0.5M V = 50ml Calculation 2HCI → 2NaCI + CO2 + H2O M = 2.0M V = ? ml Mole ratio – 1: 2 Na2CO3 M = 0.200M V = 10.0ml + 2HCI → 2NaCI + H2O + CO2 M=? V = 25.0ml Mole ratio – 1: 2
- 15. Acid/Base Titration Calculation 3 Find vol of 2.0M HCI needed to neutralize 2.65g of sodium carbonate (Na2CO3) in 50ml water. HCI M = 2.0M V = ? ml 4 10.0cm3 of 0.200M Na2CO3 needed 25.0cm3 of HCI for neutralization. Find molarity of HCI. Na2CO3 M = 0.200M V = 10.0ml Na2CO3 2.65g V = 50ml Moles = Mass/M = 2.65 106 = 0.025 mol HCI M=? V = 25.0ml Calculation Na2CO3 + M = 0.5M V = 50ml Calculation 2HCI → 2NaCI + CO2 + H2O M = 2.0M V = ? ml Using mole ratio 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 Mole ratio – 1: 2 Using formula MbVb = 1 Ma Va 2 0.5 x 50.0 = 1 2.0 x V 2 Va = 25cm3 Na2CO3 + M = 0.200M V = 10.0ml 2HCI → 2NaCI + H2O + CO2 M=? V = 25.0ml Using mole ratio 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 Mole ratio – 1: 2 Using formula MbVb = 1 Ma Va 2 0.2 x 10.0 = 1 Ma x 25.0 2 Ma = 0.160M
- 16. Acid/Base Titration Calculation 5 Vol NaOH need to neutralize 1.325 g ethanedioic acid (C2H2O4.2H2O) is 27.52cm3, calculate molarity of NaOH. 6 33.7cm3 of HCI neutralizes 20cm3 of Na2CO3 of 1.37M. Find molarity of acid. HCI M=? V = 33.7ml NaOH M=?M V = 27.52 ml Moles = Mass/M = 1.325 126.08 = 0.0105 mol Acid Mass = 1.325g Na2CO3 M = 1.37M V = 20.0ml Calculation C2H2O4 + Mass - 1.325g Mole – 0.0105 Calculation 2NaOH → Na2C2O4 + 2H2O M= ? V = 27.52ml Mole ratio – 1: 2 Na2CO3 M = 1.37M V = 20.0ml + 2HCI → 2NaCI + H2O + CO2 M=? V = 33.7ml Mole ratio – 1: 2
- 17. Acid/Base Titration Calculation 5 Vol NaOH need to neutralize 1.325 g ethanedioic acid (C2H2O4.2H2O) is 27.52cm3, calculate molarity of NaOH. 6 33.7cm3 of HCI neutralizes 20cm3 of Na2CO3 of 1.37M. Find molarity of acid. HCI M=? V = 33.7ml NaOH M=?M V = 27.52 ml Moles = Mass/M = 1.325 126.08 = 0.0105 mol Acid Mass = 1.325g Na2CO3 M = 1.37M V = 20.0ml Calculation C2H2O4 + Mass - 1.325g Mole – 0.0105 Calculation 2NaOH → Na2C2O4 + 2H2O M= ? V = 27.52ml Using mole ratio Moles 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 = 0.764M Mole ratio – 1: 2 Using formula M aVa = 1 Mb Vb 2 0.0105 = 1 M x 0.02752 2 Mb = 0.764M Na2CO3 + 2HCI M = 1.37M V = 20.0ml → 2NaCI + H2O + CO2 M=? V = 33.7ml Mole ratio – 1: 2 Using mole ratio Using formula Moles 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 = 1.63M M 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 7 25.0ml 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 25x CH3COOH M=? V = 29.2ml NaOH M = 0.1M V = 25.0ml V = 25 ml M=? V = 250ml M = 0.0856 1 NaOH + CH3COOH → CH3COONa + H2O M = 0.1M V = 25.0ml M=? V = 29.2ml Mole ratio – 1: 1
- 19. Acid/Base Titration Calculation – Ethanoic acid determination in vinegar 7 25.0ml 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 25x CH3COOH M=? V = 29.2ml NaOH M = 0.1M V = 25.0ml V = 25 ml M=? V = 250ml M = 0.0856 1 NaOH + CH3COOH → CH3COONa + H2O M = 0.1M V = 25.0ml M=? V = 29.2ml Mole ratio – 1: 1 Using mole ratio 2 Mole 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 = 0.0856M Using formula M aVa = 1 Mb Vb 1 0.1 x 0.025 = M x 0.0292 M = 0.0856M 1 1
- 20. Acid/Base Titration Calculation – Ethanoic acid determination in vinegar 7 25.0ml 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 25x CH3COOH M=? V = 29.2ml NaOH M = 0.1M V = 25.0ml V = 25 ml M=? V = 250ml M = 0.0856 1 NaOH + CH3COOH → CH3COONa + H2O M = 0.1M V = 25.0ml M=? V = 29.2ml Mole ratio – 1: 1 Using mole ratio Moles bef dilution = Moles aft dilution M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol 3 M1 x V1 = M2 x V2 M1 x 25 = 0.0856 x 250 M1 = 0.0856 x 250 25 M1 = 0.856M 2 Using formula Mole 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 = 0.0856M M aVa = 1 Mb Vb 1 0.1 x 0.025 = M x 0.0292 M = 0.0856M 1 1 Video on determination of ethanoic acid in vinegar
- 21. Acid/Base Titration Calculation - Empirical formula determination for Na2CO3. x H2O 8 27.82g 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 = 48.8ml Diuted to 1L Na2CO3 M=?M V = 25.0ml 25 ml transfer 27.82g Na2CO3. xH2O V = 1L M=? 1 2HCI + Na2CO3 → 2NaCI +CO2 + H2O M = 0.1M V = 48.8ml M=? V = 25.0ml Mole ratio – 2: 1
- 22. Acid/Base Titration Calculation - Empirical formula determination for Na2CO3. x H2O 8 27.82g 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 = 48.8ml Diuted to 1L Na2CO3 M=?M V = 25.0ml 25 ml transfer 27.82g Na2CO3. xH2O 1 V = 1L M=? 2HCI + Na2CO3 → 2NaCI +CO2 + H2O M = 0.1M V = 48.8ml M=? V = 25.0ml Mole ratio – 2: 1 Using mole ratio 2 Mole 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 = 0.0976M 3 Convert moles to mass in 1L mol/dm3 ↔ g/dm3 mol/dm3 x M → g/dm3 0.0976 x 106 = 10.36g/dm3 Using formula M 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 8 27.82g 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 = 48.8ml Diuted to 1L Na2CO3 M=?M V = 25.0ml 25 ml transfer 27.82g Na2CO3. xH2O 1 V = 1L M=? 2HCI + Na2CO3 → 2NaCI +CO2 + H2O M = 0.1M V = 48.8ml M=? V = 25.0ml Mole ratio – 2: 1 Using mole ratio 2 4 Mass Na2CO3 . x H2O = 27.82 g Mass Na2CO3 = 10.36 g Mass of water = (27.82 – 10.36) g = 17.46g Empirical formula 5 Mole 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 = 0.0976M Na2CO3 17.46 3 RMM 106 18.02 Mole 10.34/106 = 0.09773 17.46/18.02 = 0.9689 mol/dm3 ↔ g/dm3 mol/dm3 x M → g/dm3 0.0976 x 106 = 10.36g/dm3 0.09773/0.09733 1 0.9689/0.09733 10 M aVa = 2 Mb Vb 1 0.1 x 0.0488 = 2 M x 0.0250 1 M = 0.0976M H2O 10.36 Using formula Mass/g Lowest ratio Empirical formula Na2CO3 . 10 H2O Convert moles to mass in 1L Video on Na2CO3 calculation
- 24. Redox Titration Calculation- % Iron in iron tablet 9 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. KMnO4 M = 0.002M V = 24.5 ml 1.863 g 250ml MnO4M = 0.002M V = 24.5ml Fe2+ M=? V = 30ml + 5Fe2+ + 8H+ → Mn2+ + 5Fe2+ + 4H2O M= ? Mole ratio – 1: 5
- 25. Redox Titration Calculation- % Iron in iron tablet 9 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. KMnO4 M = 0.002M V = 24.5 ml 1.863 g 250ml MnO4M = 0.002M V = 24.5ml Fe2+ M=? V = 30ml + 5Fe2+ + 8H+ → Mn2+ + 5Fe2+ + 4H2O M= ? 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 10-5 KMO4-react 2.45 x 10-4 Fe2+ Using formula M 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 9 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. Video on % Iron in iron tablet KMnO4 M = 0.002M V = 24.5 ml 1.863 g 250ml MnO4M = 0.002M V = 24.5ml Fe2+ M=? V = 30ml + 5Fe2+ + 8H+ → Mn2+ + 5Fe2+ + 4H2O M= ? Mole ratio – 1: 5 Video on Fe2+/KMnO4 titration calculation Using mole ratio Using formula 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 10-5 KMO4-react 2.45 x 10-4 Fe2+ 1 2 M 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 3 Mole Mass Mole x RMM = Mass FeSO4 6.125 x 10-3 x 278.05 = 1.703g FeSO4 Mass 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 10 10.0ml 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 CIOtransfer Water added till 250ml 20ml transfer Na2S2O3 M = 0.0206M V = 17.3ml 1g KI excess added I2 M=? titrated V = 10 M=? V = 250ml M = 1.78 x 10-2 M 1 2S2O32M = 0.0206 V = 17.3ml + I2 Mole = ? V = 0.02 → S4O62- + 2IMole ratio – 2: 1
- 28. Redox Titration Calculation – CIO- in Bleach 10 10.0ml 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 CIOtransfer Water added till 250ml Na2S2O3 M = 0.0206M V = 17.3ml 1g KI excess added 20ml transfer I2 M=? titrated V = 10 M=? V = 250ml M = 1.78 x 10-2 M 2S2O32- 1 M = 0.0206 V = 17.3ml + I2 Mole = ? V = 0.02 → S4O62- + 2IMole ratio – 2: 1 Using direct formula 2 Mole 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 I2 2 mol 2CIO- + 2I- + I2 + 1 mol 2S2O322 mol 1 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 10 10.0ml 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 CIOtransfer Video on CIO- in bleach Water added till 250ml Na2S2O3 M = 0.0206M V = 17.3ml 1g KI excess added 20ml transfer I2 M=? titrated V = 10 M=? V = 250ml M = 1.78 x 10-2 M 2S2O32- 1 6 M = 0.0206 V = 17.3ml Mole 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 = 0.02 2 Mole ratio – 2: 1 Mole 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 I2 2CIO- + 2I- + 2H+ → I2 + 2CI- + H2O 2CIOI2 Mole ratio – 2: 1 Mole = ? 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 25x S4O62- + 2IUsing direct formula 3 5 → 4 Mole ratio (2 : 1) • 2 mole CIO• 3.56 x 10-4 CIO- 2 mol 2CIO- + 2I- + I2 + 1 mol 2S2O322 mol 1 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 10-4 1 mole I2 1.78 x 10-4 I2 M 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 11 Iodometric 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 = 25.5ml 1g KI excess + starch added 25ml transfer Vitamin C Vit C M=? V = 25ml titrated V = 25ml M= ? 1 KIO3 + 5KI + 6H+ → M = 0.002M V = 25.5ml 3I2 + Mole = ? 3H2O + 6K= Mole ratio – 1: 3
- 31. Redox Titration Calculation – Vitamin C quantification 11 Iodometric 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 = 25.5ml 1g KI excess + starch added 25ml transfer Vitamin C Vit C M=? V = 25ml titrated V = 25ml M= ? 1 KIO3 + 5KI + 6H+ → M = 0.002M V = 25.5ml 3I2 + Mole = ? 3H2O + 6K= Mole ratio – 1: 3 Using mole ratio 2 Mole 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 I2 Using formula 1 mol 3 mol KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O 3C6H8O6 + 3I2 → 3C6H6O6 + 6I- + 6H+ 3 mol 3 mol Mole 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 11 Iodometric 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 = 25.5ml 1g KI excess + starch added 25ml transfer Vitamin C Vit C M=? V = 25ml titrated V = 25ml M= ? 1 KIO3 + 5KI + 6H+ → M = 0.002M V = 25.5ml 3I2 + Mole = ? 3H2O + 6K= Mole ratio – 1: 3 Using mole ratio 2 3 Mole 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 I2 3C6H8O6 Mole = ? + 3I2 → 3C6H6O6 + 6I- + 6H+ 1.53 x 10-4 Mole ratio – 3: 3 Using formula 1 mol 3 mol KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O 3C6H8O6 + 3I2 → 3C6H6O6 + 6I- + 6H+ 3 mol 3 mol Mole ratio (1 :3) 1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6 1 mol KIO3 3 mol C6H8O6 Using formula 4 Mole 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 M M 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 12 2.5g 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 added Pour into Volumetric flask 25ml transfer 2.5g brass 10 ml HNO3 Na2S2O3 M = 0.1M V = 28.2ml V = 250ml M=? 1 I2 M=? titrated 2S2O32M = 0.1M V = 28.2ml + I2 Mole = ? → S4O62- + 2IMole ratio – 2: 1
- 34. Redox Titration Calculation - % Cu in Brass 12 2.5g 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 added Pour into Volumetric flask 25ml transfer 2.5g brass 10 ml HNO3 Na2S2O3 M = 0.1M V = 28.2ml V = 250ml M=? I2 M=? titrated 2S2O32- 1 M = 0.1M V = 28.2ml + I2 Mole = ? → S4O62- + 2IMole ratio – 2: 1 Using mole ratio 2 Mole 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 I2 Using formula 2 mol 2Cu2+ + I2 + 1 mol 4I- → 2S2O32- → 2 mol 1 mol I2 + 2CuI S4O62- + 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 12 2.5g 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 added Pour into Volumetric flask 25ml transfer 2.5g brass 10 ml HNO3 Na2S2O3 M = 0.1M V = 28.2ml V = 250ml M=? I2 M=? titrated 2S2O32- 1 M = 0.1M V = 28.2ml + I2 Mole = ? → S4O62- + 2IMole ratio – 2: 1 Using mole ratio 2 Using formula Mole 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 I2 2Cu2+ + 4I- → I2 + 2CuI Mole = ? 1.41 x 10-3 I2 Mole ratio – 2: 1 3 2 mol 2Cu2+ + I2 + 1 mol 4I- → 2S2O32- → 2 mol 1 mol I2 + 2CuI S4O62- + 2I- Mole ratio ( 1 : 1) 2 mole Cu2+ : 1 mole I2 : 2 mole S2O322 mole Cu2+ 2 mole S2O32Using formula 5 Mole 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 250ml 4 6 Mole 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. Titration Direct Titration Condition • Both titrant and analyte soluble • Reaction is fast Titrant - soluble Analyte - soluble Redox Titration Acid Base Titration Complexometric Titration H2SO4 HCI HNO3 Soluble acid NaOH NH4OH Soluble base KOH (Alkali) Ba(OH)2 LiOH
- 37. Titration Direct Titration Back Titration Condition Condition • 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 - soluble Analyte - soluble added Redox Titration Acid Base Titration Complexometric Titration Impure antacid Known conc /vol of acid used H2SO4 HCI HNO3 Soluble acid NaOH NH4OH Soluble base KOH (Alkali) Ba(OH)2 LiOH Impure limestone CaCO3 in egg shell
- 38. Titration Direct Titration Back Titration Condition Condition • 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 - soluble Analyte - soluble added Redox Titration Acid Base Titration Complexometric Titration Impure limestone CaCO3 in egg shell Impure antacid Known conc /vol of acid used Left overnight in acid H2SO4 HCI HNO3 Soluble acid NaOH NH4OH Soluble base KOH (Alkali) Ba(OH)2 LiOH Amt of known acid added Titrated with known conc/vol alkali known unknown Amt of excess acid left Transfer to flask Amt of base (solid) react with acid Video on back titration calculation Amt base(solid) = Amt Known – Amt excess acid added acid left Excess acid left
- 39. Back Titration % Calcium carbonate in egg shell - Back Titration Calculation 25.0g 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. added 25.0g impure CaCO3 in egg shell 250.0ml, 2.0M HNO3 Left overnight in acid Amt of HNO3 added Titrated with known conc/vol NaOH M = 1.0 V = 17.0ml Transfer to flask Amt of HNO3 left Amt of base (egg) Amt HNO3 react = Amt HNO3 – Amt HNO3 with base add left HNO3 left
- 40. Back Titration % Calcium carbonate in egg shell - Back Titration Calculation 25.0g 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. 1 added 25.0g impure CaCO3 in egg shell 2 250.0ml, 2.0M HNO3 Left overnight in acid Amt of HNO3 added Titrated with known conc/vol NaOH M = 1.0 V = 17.0ml Transfer to flask Amt of HNO3 left Amt of base (egg) Amt HNO3 react = Amt HNO3 – Amt HNO3 with base add left HNO3 left Amt HNO3 add = M V = 2.0 x 0.250 = 0.50 mol NaOH + HNO3 M = 1.00M V = 17.0ml moles = ? → NaNO3 + H2O Mole ratio – 1: 1 Amt HNO3 add
- 41. Back Titration % Calcium carbonate in egg shell - Back Titration Calculation 25.0g 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. 1 added 25.0g impure CaCO3 in egg shell 2 250.0ml, 2.0M HNO3 Left overnight in acid Amt of HNO3 added Titrated with known conc/vol NaOH M = 1.0 V = 17.0ml Transfer to flask Amt of HNO3 left Amt of base (egg) Amt HNO3 react = Amt HNO3 – Amt HNO3 with base add left HNO3 left Amt HNO3 add = M V = 2.0 x 0.250 = 0.50 mol NaOH + HNO3 M = 1.00M V = 17.0ml moles = ? Amt HNO3 add → NaNO3 + H2O Mole ratio – 1: 1 Using mole ratio 3 Mole 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 mol Using formula 3 MbVb = 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 Calculation 25.0g 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. 1 added 25.0g impure CaCO3 in egg shell 2 250.0ml, 2.0M HNO3 Left overnight in acid Amt of HNO3 added Titrated with known conc/vol NaOH M = 1.0 V = 17.0ml HNO3 left 3 4 6 → NaNO3 + H2O Mole ratio – 1: 1 Using formula Mole 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 mol 3 Mole ? Amt HNO3 react with base 7 Mass = Mole CaCO3 x RMM CaCO3 = 0.242 x 100 = 24.2g 8 % by mass = mass CaCO3 x 100% CaCO3 mass impure = (24.2/25.0) x 100% = 96.8% Mole ratio – 2: 1 Mole ratio (2 : 1) • 2 mol HNO3 react 1 mol CaCO3 • 0.483 mol HNO3 react o.242 mol CaCO3 MbVb = 1 Ma Va 1 1.00 x 0.017 = 1 moles 1 mole HNO3 = 1.7 x 10-2 Amt HNO3 left Amt HNO3 react = Amt HNO3 add – Amt HNO3 left with NaOH = 0.50 – 1.7 x 10-2 = 0.483 mol 2HNO3 + CaCO3 → (CaNO3)2 + H2O + 2CO2 Mole 0.483 Amt HNO3 react = Amt HNO3 – Amt HNO3 with base add left moles = ? Amt HNO3 add Using mole ratio Amt of base (egg) 5 NaOH + HNO3 M = 1.00M V = 17.0ml Transfer to flask Amt of HNO3 left Amt HNO3 add = M V = 2.0 x 0.250 = 0.50 mol
- 43. Back Titration % Calcium hydroxide in an impure antacid tablet - Back Titration Calculation 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 Left overnight in acid Titrated with known conc/vol NaOH M = 0.1108 V = 33.64ml Amt of HCI added Transfer to flask Amt HCI left Amt of base (solid) Amt HCI react = Amt HCI – Amt HCI with NaOH add left HCI left
- 44. Back Titration % Calcium hydroxide in an impure antacid tablet - Back Titration Calculation 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. 1 added 0.5214g impure Ca(OH)2 2 50.0ml, 0.250M HCI Left overnight in acid Titrated with known conc/vol NaOH M = 0.1108 V = 33.64ml Amt of HCI added Transfer to flask Amt HCI left Amt of base (solid) Amt HCI react = Amt HCI – Amt HCI with NaOH add left HCI left Amt HCI add = M V = 0.250 x 0.050 = 0.01250 mol NaOH + M = 0.1108M V = 33.64ml HCI → moles = ? NaCI + H2O Mole ratio – 1: 1 Amt HCI Add
- 45. Back Titration % Calcium hydroxide in an impure antacid tablet - Back Titration Calculation 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. 1 added 0.5214g impure Ca(OH)2 2 50.0ml, 0.250M HCI Left overnight in acid Titrated with known conc/vol NaOH M = 0.1108 V = 33.64ml Amt of HCI added Transfer to flask Amt HCI left Amt of base (solid) Amt HCI react = Amt HCI – Amt HCI with NaOH add left HCI left Amt HCI add = M V = 0.250 x 0.050 = 0.01250 mol NaOH + M = 0.1108M V = 33.64ml HCI → moles = ? Amt HCI Add NaCI + H2O Mole ratio – 1: 1 Using mole ratio 3 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 Using formula 3 MbVb = 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 Calculation 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. 1 added 0.5214g impure Ca(OH)2 2 50.0ml, 0.250M HCI Left overnight in acid Titrated with known conc/vol NaOH M = 0.1108 V = 33.64ml Amt of HCI added HCI → moles = ? Amt HCI Add NaCI + H2O Mole ratio – 1: 1 Using formula Using mole ratio 3 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 MbVb = 1 Ma Va 1 0.1108 x 0.03364 = 1 moles 1 mole HCI = 3.727 x 10-3 Amt HCI Left HCI left 4 Amt of base (solid) 5 6 Amt HCI react = Amt HCI add – Amt HCI left with NaOH = 0.01250 – 3.727 x 10-3 = 0.008773 mol 2HCI + Ca(OH)2 Mole 0.008773 Amt HCI react = Amt HCI – Amt HCI with NaOH add left NaOH + M = 0.1108M V = 33.64ml Transfer to flask Amt HCI left Amt HCI add = M V = 0.250 x 0.050 = 0.01250 mol Mole ? → CaCI2 + 2H2O 7 Mass = Mole Ca(OH)2 x RMM Ca(OH)2= 0.004386 x 74.1 = 0.3250g 8 % by mass = mass Ca(OH)2 x 100% Ca(OH)2 mass impure = (0.3250/0.5214) x 100% = 62.3% 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 Amt HCI react with base
- 47. Molar mass of insoluble acid in a tablet -Back Titration Calculation Back Titration 2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acid added 2.04g impure dibasic acid H2A 20.0ml, 2.00M NaOH Left overnight Titrated with known conc/vol HCI M = 0.50 V = 17.6ml Amt NaOH added Transfer to flask Amt of NaOH left NaOH left Amt 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 Calculation Back Titration 2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acid 1 added 2.04g impure dibasic acid H2A 2 20.0ml, 2.00M NaOH Left overnight Titrated with known conc/vol HCI M = 0.50 V = 17.6ml Amt NaOH added Transfer to flask Amt of NaOH left NaOH left Amt of acid (solid) Amt NaOH react = Amt NaOH – Amt NaOH with acid add left Amt NaOH add = M V = 2.00 x 0.02 = 0.040 mol HCI + M = 0.50M V = 17.6ml NaOH moles = ? → NaCI + H2O Mole ratio – 1: 1 Amt NaOH add
- 49. Molar mass of insoluble acid in a tablet -Back Titration Calculation Back Titration 2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acid 1 added 2.04g impure dibasic acid H2A 2 20.0ml, 2.00M NaOH Left overnight Titrated with known conc/vol HCI M = 0.50 V = 17.6ml Amt NaOH added Transfer to flask Amt of NaOH left NaOH left Amt of acid (solid) Amt NaOH react = Amt NaOH – Amt NaOH with acid add left Amt NaOH add = M V = 2.00 x 0.02 = 0.040 mol HCI + M = 0.50M V = 17.6ml NaOH moles = ? → NaCI + H2O Mole ratio – 1: 1 Using mole ratio 3 Amt NaOH add Mole 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 mol Using formula 3 MaVa = 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 Calculation Back Titration 2.04g insoluble dibasic acid dissolve in 20.0ml, 2.00M NaOH. Excess NaOH require 17.6ml, 0.5M HCI to neutralize it. Find molar mass acid 1 added 2.04g impure dibasic acid H2A 2 20.0ml, 2.00M NaOH Left overnight Titrated with known conc/vol HCI M = 0.50 V = 17.6ml Amt NaOH added HCI + M = 0.50M V = 17.6ml NaOH moles = ? → Amt NaOH add NaCI + H2O Mole ratio – 1: 1 Using formula Using mole ratio 3 Transfer to flask Amt of NaOH left Amt NaOH add = M V = 2.00 x 0.02 = 0.040 mol Mole 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 mol 3 MaVa = 1 Mb Vb 1 0.50 x 0.0176 = 1 moles 1 mole HCI = 8.8 x 10-3 Amt NaOH Left NaOH left 4 Amt of acid (solid) 5 Amt 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 0.00312 Amt NaOH react = Amt NaOH – Amt NaOH with acid add left 6 Amt NaOH react with acid Mole ? Mole ratio – 2: 1 Mole ratio (2 : 1) • 2 mol NaOH react 1 mol H2 A • 0.0312 mol NaOH react o.o156 mol H2A 7 Molar Mass 0.0156 mol H2A - 2.04g 1 mol H2A - 2.04 0.0156 = 131
- 51. Video on Titration Click here titration NaOH /HCI Click here titration video Simulation on Titration Click here titration simulation Click here titration simulation Click here on titration simulation Click here titration simulation Click 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

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