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IB Chemistry on Molarity, Concentration, standard solution serial dilution
 

IB Chemistry on Molarity, Concentration, standard solution serial dilution

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IB Chemistry on Molarity, Concentration, standard solution serial dilution

IB Chemistry on Molarity, Concentration, standard solution serial dilution

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    IB Chemistry on Molarity, Concentration, standard solution serial dilution IB Chemistry on Molarity, Concentration, standard solution serial dilution Presentation Transcript

    • Concentration and Molarity + Solute Solvent Solution Concentration measured measured Grams measured Moles cm3 Conc in mol/dm3 Conc in g/dm3 ↓ ↓ dm3 Moles solute in mol Vol of solution in dm3 Mass solute in grams Vol of solution in dm3 Find conc in g/dm3 and mol/dm3 46 g NaOH in 1.00dm3 1 dm3 Molarity in M 46 g Video on conc/molarity Conc in g/dm3 = mass (g) vol dm3 = 46g 1dm3 = 46g/dm3 = moles (mol) vol dm3 = 1 mol 1dm3 = 1 mol/dm3 =1M Conversion formula Conc in mol/dm3 Moles = mass M = 46g 46g/mol = 1 mol g/dm3 ↔ mol/dm3 g/dm3 ÷ M → mol/dm3 46g/dm3 ÷ 46 = 1 mol/dm3
    • Standard solution Solution of known concentration Preparing standard solution of conc - 1 M NaOH 1 M – 1 mole NaOH in 1 L/dm3 Step 1 Mass of NaOH → 1 mole NaOH x M = 1 x 46g Step 2 Transfer to beaker, add water to dissolve it Step 3 Transfer from beaker to 1L volumetric flask using filter funnel Step 4 Add water until 1L mark with wash bottle Step 5 46 g Mix till it dissolved 1M NaOH – 1 mole of NaOH in total vol of solution (1000ml) 1M NaOH Molarity = 1 mole (1M) 1 L total solution vol (solute + solvent) Molarity = 1 mole (1M) 1 L of vol solvent Video standard solution preparation
    • Diluting a standard solution Adding water 1 dm3 1 dm3 Dilution Solute 5 moles Solvent 1 L/dm3 + Conc Solution 5M or 5 mol/dm3 Diluted solution 2.5M or 2.5 mol/dm3 Dilution 1 NaOH 5 moles NaOH in 1 dm3 5 M or 5mol/dm3 0.5NaOH 1 dm3 5 moles NaOH in 2 dm3 2.5 M or 2.5mol/dm3 1 dm3 Moles bef dilution = 5 mol Vol bef = 1 dm3 Conc = 5M Moles after dilution = 5 mol Vol after = 2 dm3 Conc = 2.5M Moles = Molarity(M) x Vol (dm3) =MxV Before =5x1 = 5 moles Vol Increase ↑, Conc decrease ↓ Number mole NO CHANGE Moles = Molarity(M) x Vol (dm3) =MxV After = 2.5 x 2 = 5 moles Moles before dilution = Moles after dilution M1 V1 = M2V2 M1 = Initial molarity M2 = Final molarity V1 = Initial volume V2 = Final volume M1 V1 = M2V2
    • Diluting a standard solution (1M) → (0.1M) 1 M – 1 mole NaOH in 1 dm3 (stock solution) Standard solution , molarity - 1 M NaOH Step 1 Step 2 Pipette 10cm3 of 1M NaOH with a pipette Step 3 Transfer 10cm3 to a volumetric flask Step 4 Add 90 cm3 water, to a total vol of 100cm3 Step 5 10cm3 1M NaOH in a volumetric flask Mix to dissolve, 0.1M NaOH 10cm3 90cm3 1M Moles before dilution = Moles after dilution M1 V1 = M2V2 M1 = Initial molarity M2 = Final molarity V1 = Initial volume V2 = Final volume 0.1M M1 V1 = M2V2 1M x 10cm3 = 0.1M x 100cm3 (10 + 90) 1000 1000 Video standard solution preparation
    • Diluting a standard solution Serial Dilution Dilution Serial dilution • Start with Conc solution (stock) • Add water to dilute it down • Difficult to cover a wide range • Time consuming to perform different dilution for different concentration Vs Stock solution 1M NaOH • Easier to make, cover a wide range of conc • Same dilution over again • Using previous dilution in next step 1 in 10 serial dil - 1 part stock – 9 part water - (10x dil), (10 fold), (1 : 10) 1 in 2 serial dil– 1 part stock – 1 part water - (2x dil), (2 fold), (1 : 2) Prepare 0.1M NaOH Prepare a 10 x fold serial dil 1M NaOH Tube 2 1 cm3 1 cm3 Tube 3 1 cm3 Tube 1 1 cm3 Tube 4 1 cm3 9 cm3 9 cm3 Step 1 Pipette 9cm3 water to tube 1 Step 2 Pipette 1 cm3 stock to tube 1 9 cm3 9 cm3 9 cm3 Step 1 Mix well Dilution 1M → 0.1M Step 2 Pipette 1 cm3 stock to tube 1 + Mix well Step 3 Pipette 1 cm3 from tube 1 to 2 + Mix well Step 4 + Pipette 9cm3 water to tube 1, 2, 3, 4 Pipette 1 cm3 from tube 2 to 3 + Mix well + Mix well Step 5 Dilution factor = 1o parts (9 part water +1 part solute) (10x, 1:10) 1 part (1 part solute) Pipette 1 cm3 from tube 3 to 4 Serial dil 10 fold 1M → 0.1M, 0.01M, 0.001M, 0.0001M
    • Dilution factor = 1o parts (5 part water +5 part solute) (2x, 1:2) 5 parts (5 part solute) Serial Dilution Dilution factor = 1o parts (9 part water +1 part solute) (10x, 1:10) 1 part (1 part solute) Stock solution 1M NaOH Prepare 2x fold serial dil 1M NaOH Tube 4 Tube 3 Tube 2 Prepare 10x fold serial dil 1M NaOH Tube 1 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 9 cm3 Pipette 5cm3 water to tube 1, 2, 3, 4 Tube 3 Tube 4 1 cm3 1 cm3 5 cm3 5 cm3 Tube 2 1 cm3 Tube 1 Step 1 9 cm3 1 cm3 9 cm3 9 cm3 Pipette 9cm3 water to tube 1, 2, 3, 4 Pipette 5cm3 stock to tube 1 + Mix well Step 2 Pipette 1cm3 stock to tube 1 + Mix well Pipette 5cm3 from tube 1 to 2 + Mix well Step 3 Pipette 1cm3 from tube 1 to 2 + Mix well Pipette 5cm3 from tube 2 to 3 + Mix well Step 4 Pipette 1cm3 from tube 2 to 3 + Mix well Pipette 5cm3 from tube 3 to 4 + Mix well Step 5 Pipette 1cm3 from tube 3 to 4 + Mix well Serial dil 2 fold 1M → 0.5M, 0.25M, 0.125M, 0.0625M Animation serial dilution Serial dil 10 fold 1M → 0.1M, 0.01M, 0.001M, 0.0001M Video serial dilution
    • Serial Dilution Stock solution 1M NaOH Prepare 2x fold serial dil 1M NaOH Tube 4 Tube 3 Tube 2 Prepare 10x fold serial dil 1M NaOH Tube 1 Tube 1 5 cm3 5 cm3 X 1 16 5 cm3 X 1 8 5 cm3 5 cm3 X 1 4 Tube 3 1 cm3 5 cm3 X Tube 4 1 cm3 5 cm3 5 cm3 Tube 2 9 cm3 1 2 X 1 10 1 cm3 9 cm3 X 1 100 Dilution factor 1 cm3 9 cm3 X 9 cm3 1 1000 X 1 10000 Dilution factor Dilution factor = 1o parts (9 part water +1 part solute) (10x, 1:10) 1 part (1 part solute) Dilution factor = 1o parts (5 part water +5 part solute) (2x, 1:2) 5 part (5 part solute) Pipette 5cm3 water to tube 1, 2, 3, 4 Step 1 Pipette 9cm3 water to tube 1, 2, 3, 4 Pipette 5cm3 stock to tube 1 + Mix well Step 2 Pipette 1cm3 stock to tube 1 + Mix well Pipette 5cm3 from tube 1 to 2 + Mix well Step 3 Pipette 1cm3 from tube 1 to 2 + Mix well Pipette 5cm3 from tube 2 to 3 + Mix well Step 4 Pipette 1cm3 from tube 2 to 3 + Mix well Pipette 5cm3 from tube 3 to 4 + Mix well Step 5 Pipette 1cm3 from tube 3 to 4 + Mix well Serial dil 2 fold 1M → 0.5M, 0.25M, 0.125M, 0.0625M Serial dil 10 fold 1M → 0.1M, 0.01M, 0.001M, 0.0001M
    • Concept Map Moles = Mass Molar mass 1 mole gas Mole = Vol gas (stp) 22.4 Moles in solution = M x V M = Molarity -M or mol/dm3 V = Vol in dm3 mole changes Moles bef dilution = Moles aft dilution M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol Moles in solution = M x V 1000 Moles = M(Molarity) x V(Vol) M = Molarity- M or mol/dm3 V = Vol in cm3 mole NO change concentration change
    • Moles vs Concentration/Molarity 1 dm3 1 dm3 1 dm 3 1 dm3 Molarity = 5M 1 dm3 5 moles Molarity = 5M Moles bef dilution = Moles aft dilution M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol Moles = M x V = 5M x 1 dm3 = 5 mol 1 dm3 1 dm3 Mole = M x V = 5M x 1dm3 = 5 mol 2 dm3 2 dm3 Mole = M x V = 5M x 2dm3 = 10 mol mole change Mole = M x V Conc = Mole V = 10mol 2dm3 Conc = 5M Concentration NO change Moles bef dilution = Moles aft dilution Conc = 5M M1 V1 = M2V2 5 x 1 = M2 x 2 M2 = 5/2 M2 = 2.5M Moles = M x V = 5M x 1 dm3 = 5 mol concentration Conc = 2.5M Mole = M x V = 2.5M x 2dm3 = 5 mol change mole NO change Mole = M x V Conc = Mole V = 5mol 1dm3 Conc = 5M
    • Moles vs Concentration/Molarity 1 dm3 1 dm3 1 dm 1 dm3 3 1 dm3 5 moles Molarity = 10M Molarity = 5M Total = 5 + 10 = 15 mol moles Moles bef dilution = Moles aft dilution M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol Moles = M x V = 5M x 1 dm3 = 5 mol 1 dm3 1 dm3 Mole = M x V = 5M x 1 dm3 = 5 mol 2 dm3 2 dm3 Mole = M x V = 7.5M x 2dm3 = 15 mol mole change Mole = M x V Conc = Mole V = 15mol 2dm3 = 7.5M Concentration change Moles bef dilution = Moles aft dilution Conc = 5M M1 V1 = M2V2 5 x 1 = M2 x 2 M2 = 5/2 M2 = 2.5M Moles = M x V = 5M x 1 dm3 = 5 mol concentration Conc = 2.5M Mole = M x V = 2.5M x 2dm3 = 5 mol change mole NO change Mole = M x V Conc = Mole V = 5mol 1dm3 = 5M
    • IB Questions on Conc and Molarity 1 5.00g of copper(II) sulphate dissolve in water form 500cm3 solution. Cal conc in g/dm3 and mol/dm3 Cal mass of Na2CO3 require to dissolve in water to prepare 200cm3 solution, containing 50 g/dm3 2 Answer: Vol = 200cm3 → 0.2dm3 Answer: Mass copper(II) sulphate = 5.00g Vol solution = 500cm3 → 500/1000 → 0.5dm3 Conc = Mass Vol 5.00g → 0.5dm3 = 5.00g 0.5dm3 = 10.0g/dm3 Alternative 0.5dm3 → 5.00g 5.00 g 1 dm3 → 5.00 x 1 0.5 = 10g/dm3 0.5 dm3 Conc (g/dm3) = mass(g) vol(dm3) Mass = conc(g/dm3) x vol(dm3) = 50 x 0.2 = 10g Alternative 1 dm3 → 50g 0.2dm3 → 50 x 0.2 = 10g 10g 0.2 dm3 Convert g/dm3 to mol/dm3 RMM copper(II) sulphate = 160 mol/dm3 → g/dm3 RMM = 10 160 = 0.0625M 3 Cal vol in dm3 of 0.8M H2SO4 which contain 0.2 mol. Answer: Moles = M x V V = mol M V = 0.2 mol 0.8mol/dm3 V = 0.25dm3 0.2mol 0.8M 4 Cal moles of NH3 in 150cm3 of 2M aqueous NH3 solution. Answer: Moles = M x V 1000 = 2 X 150 1000 = 0.3 mol 2M 150cm3
    • IB Questions on Conc and Molarity 5 4.0g of Na2CO3 dissolved in water, making up 250cm3 Calculate its molarity. Answer: Moles Na2CO3 = Mass M = 16 106 = 0.0377 mol Moles = M x V 0.0377 = M x 0.25 M = 0.0377 0.25 M = 0.15M Alternative Conc in g/dm3 = Mass Vol = 4g 0.25dm3 = 16g/dm3 3 to mol/dm3 Convert g/dm mol/dm3 → g/dm3 RMM = 16 106 = 0.15M 8 6 Answer: 0.4mol Moles = M x V 1000 0.4 = M X 250 1000 M = 0.4 x 1000 250 = 1.6M 0.25 dm3 Alternative 0.25 dm3 → 0.4 mol HNO3 1 dm3 → 0.4 x 1 mol HNO3 0.25 = 1.6M 250cm 3 → 0.25dm3 0.0377mol 0.25 dm3 7 RMM = 106 Cal moles of hydrogen ions H+ in 200cm3 of 0.5M H2SO4 Answer: Moles = M x V 1000 = 0.5 X 200 1000 = 0.1 mol H2SO4 → 2 H+ + SO42H2SO4 diprotic produce 2 mol H+ ions Moles H+ = 2 x 0.1 mol = 0.2 mol Moles? 0.5M 0.2 dm3 250cm3 of HNO3 contain 0.4moles. Cal its molarity. HCI acid has conc of 2.0M. Cal the mass of hydrogen chloride gas in 250cm3 in HCI. Mass ? Answer: Moles = M x V 1000 = 2.0 X 250 1000 2.0M = 0.5 mol 0.25 dm3 RMM HCI= 36.5 Mass of HCI = Moles x RMM = o.5 x 36.5 = 18.25g
    • IB Questions on Conc and Molarity 9 Cal molarity of KOH when 750cm3 water added to 250cm3, 0.8M KOH 10 Cal vol of water added to 60cm3, 2.0M of H2SO4 to produce 0.3M H2SO4 Answer Answer 250cm 1000cm3 3 750cm3 0.8M M1 V1 = M2V2 M1 V1 = M2V2 2.0 x 60 = 0.3 x V2 V2 = 2.0 x 60 0.3 V2 = 400cm3 (final vol) Vol water = 400 – 60 = 340cm3 added 0.8 x 250 = M2 x 1000 M2 = 0.8 x 250 Final vol 1000 750 + 250 = 1000cm3 M2 = 0.2M Cal molarity of NaOH produce when 500cm3 , 2mol NaOH added to 1500cm3, 4mol NaOH 12 Cal molarity of HCI produce when 200cm3 , 2M HCI added to 300cm3, 0.5M of HCI Answer Answer A 2 mol + 6 mol 4 mol 2M 200cm3 500cm 0.3M Moles bef dilution = Moles aft dilution Moles bef dilution = Moles aft dilution 11 ? cm3 60cm3 2M 3 1500cm B + 0.5M 300cm3 C ?M 500cm3 3 Total moles = (2.0 + 4.0) = 6mol Total vol = (500 + 1500) = 2000cm3 → 2dm3 Moles = M x V M = Moles V = 6 mol 2 dm3 = 3.0 mol/dm3 2000cm3 Mole = M x V A 1000 = 2.0 x 200 = 0.4mol 1000 Total moles A + B = 0.4 + 0.15 = 0.55 mol Total vol = (200 + 300) = 500cm3 → 0.5dm3 Moles = M x V M = Moles = 0.55 = 1.1M V 0.5 Mole = M x V B 1000 = 0.5 x 300 = 0.15mol 1000
    • IB Questions on Conc and Molarity 13 Prepare 250cm3, 0.1M HCI using concentrated HCI, 1.63M. What vol of concentrated acid must be diluted. 14 Answer ? cm3 Answer 1.63M 2.00g 0.1M 250cm 250cm3 3 Moles KCI = mass (solid) M = 2.00 = 0.02683 mol 74.55 250 cm3 → 0.250dm3 Moles KCI = M x V (solution) M = moles = 0.02683 V 0.250 M = 0.107 mol/dm3 Moles bef dilution = Moles aft dilution M1 V1 = M2V2 Moles aft dilution = M x V 250 cm3 → 0.250dm3 = 0.1 x 0.25 = 0.025mol Moles bef dilution = 0.025 M x V = 0.025 V = 0.025 = 0.025 = 0.0154 dm3 or 15.4 cm3 M 1.63 15 How to prepare 500cm3 of 0.1M NaCI solution ? Answer Cal conc formed when 2.00g KCI dissolved in 250cm3 of solution 16 How to prepare 1.2dm3 of 0.4M HCI solution starting from 2.0M HCI ? Answer 2..92g NaCI 0.1 M 240 cm3 0.4 M 2M 500cm3 500 cm3 → 0.5dm3 Moles NaCI = M x V = 0.1 x 0.5dm3 = 0.05 mol needed Moles NaCI = Mass RMM Mass = Moles x RMM = 0.05 x 58.5 = 2.92g Weigh 2.92g of NaCI (0.05mol) and make up to 500cm3 solution in a 500cm3 volumetric flask 1.2 dm3 Moles of HCI = M x V = 0.4 x 1.2 = 0.48 mol Moles bef dilution = Moles aft dilution M1 V1 = M2V2 2 x V1 = 0.4 x 1.2 V1 = 0.4 x 1.2 2 V = 0.24 dm3 or 240 cm3 cm3 Measure 240 of 2M HCI and make up to 1.2 dm3 with water using volumetric flask.