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IB Chemistry Serial Dilution, Molarity and Concentration
The document provides information on concentration, molarity, standard solutions, and serial dilution preparation and calculations. It defines key terms like solute, solvent, and solution. It explains how to calculate concentration in g/dm3 and mol/dm3. It also discusses dilution techniques like serial dilution, where subsequent dilutions use the previous dilution. Serial dilution allows preparing a wide range of concentrations more easily compared to separate dilutions. The document includes examples of calculations for molarity, moles, mass, and volume in dilution problems.
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IB Chemistry Serial Dilution, Molarity and Concentration
- 1. http://lawrencekok.blogspot.com Prepared by Lawrence Kok IBChemistry Molarity, Concentration, Standard Solution and Serial Dilution Preparation .
- 2. Concentration and Molarity SoluteSolvent Solution + Solute/Solvent/Solution measured Find conc in g/dm3 and mol/dm3 46 g NaOH in 1 dm3 46 g 1 dm3 Conversion formula ).( 3 dmVol Mole Conc ).( ).( 3 dmVol gMass Conc ÷ RMM x RMM 3 3 1 1 1 ).( moldmConc dmVol Mole Conc 3 3 46 1 46 ).( ).( gdmConc dmVol gMass Conc 5 moles 1 L/dm3 Conc Sol 5M or 5 mol/dm3 Dilution 1 M NaOH 1 dm3 5 mole in 1 dm3 5 M or 5 mol/dm3 •••• • ••••• 1 dm3 0.5 M NaOH Mole bef dil = 5 mol Vol bef = 1 dm3 Conc = 5 M Mole aft dil = 5 mol Vol aft = 2 dm3 Conc = 2.5 M water 2 dm3 Vol increase ↑ Conc decrease ↓ ••• • • VMMole dmVolMMolarityMole )()( 3 molMole VMMole 515 molMole VMMole 525.2 Moles bef dil = Moles aft dil M1 V1 = M2V2 M1 = Ini conc M2 = Final conc V1 = Ini vol V2 = Final vol Animation/video serial dilution
- 3. 1M NaOH –1 mole of NaOH in total vol of solution (1L) Solution Preparation Mass of NaOH → 1 mole NaOH x M = 1 x 46 gStep 1 Step 2 Pour to 1L volumetric flaskStep 3 Add water until 1L mark Transfer to beaker, add water to dissolve Step 4 Step 5 46 g Molarity = 1 mole (1M) 1 L total vol (solute + solvent) Conc NaOH Preparing sol – 1 M NaOH – 1 mole NaOH in 1 L Diluting a std sol (1M) → (0.1M) Moles before dilution = Moles after dilution M1 V1 = M2V2 M1 = Initial molarity M2 = Final molarity V1 = Initial volume V2 = Final volume M1 V1 = M2V2 1M x 10 cm3 = 0.1M x 100 cm3 (10 + 90) 1000 1000 10 cm3 90 cm3 water 1M 0.1M 1M, 10 ml 0.1M, 100 ml Diluting a std sol
- 4. Stock solution 1MNaOH vs Prepare 0.1M NaOH Diluting a std sol Serial Dilution Prepare 10 x fold serial dil 9 cm3 9 cm3 9 cm39 cm39 cm3 Pipette 9 cm3 water to tube 1, 2, 3, 4 Pipette 1 cm3 stock to tube 1 Pipette 1 cm3 from tube 1 to 2 Pipette 1 cm3 from tube 2 to 3 Pipette 1 cm3 from tube 3 to 4 Tube 1 Tube 2 Tube 3 Tube 4 1 cm3 + mix well Serial dil 10 fold 1M → 0.1M, 0.01M, 0.001M, 0.0001M Step 1 Pipette 9 cm3 water to tube 1 1 cm3 Pipette 1 cm3 stock to tube 1 Dilution 1M → 0.1M Dilution factor = 1o part = (9 part water +1 part solute) (10x, 1:10) 1 part (1 part solute) 1 cm3 1 cm3 1 cm3 Dilution • Start with Conc sol (stock) • Add water to dilute it down • Diff to cover a wide range • Time consuming to perform diff dilution for diff conc Serial dilution • 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) Step 2 Step 1 Step 2 Step 3 Step 4 Step 5 + mix well + mix well + mix well + mix well Tube 1 + mix well Dilution factor = 1o part = (5 part water +5 part solute) (2x, 1:2) 5 part (5 part solute)
- 5. Stock sol 1M NaOH Stocksolution 1M NaOH vs Diluting a std sol Serial Dilution Prepare 10 x fold serial dil 9 cm3 9 cm3 9 cm3 9 cm3 Pipette 9 cm3 water to tube 1, 2, 3, 4 Pipette 1 cm3 stock to tube 1 Pipette 1 cm3 from tube 1 to 2 Pipette 1 cm3 from tube 2 to 3 Pipette 1 cm3 from tube 3 to 4 Tube 1 Tube 2 Tube 3 Tube 4 1 cm3 + mix well Serial dil 10x 1M → 0.1M, 0.01M, 0.001M, 0.0001M Step 1 1 cm3 1 cm3 1 cm3 Dilution • Start with Conc solution (stock) • Add water to dilute it down • Diff to cover a wide range • Time consuming to perform diff dilution for diff conc Serial dilution • 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) Step 2 Step 1 Step 2 Step 3 Step 4 Step 5 + mix well + mix well + mix well + mix well Prepare 2 x fold serial dil Pipette 5 cm3 water to tube 1, 2, 3, 4 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 5 cm35 cm3 5 cm3 Tube 1 Tube 2 Tube 3 Tube 4 Serial dil 2x 1M → 0.5M, 0.25M, 0.125M, 0.0625M + mix well Pipette 5 cm3 stock to tube 1 + mix well Pipette 5 cm3 from tube 1 to 2 Pipette 5 cm3 from tube 2 to 3 Pipette 5 cm3 from tube 3 to 4 + mix well + mix well + mix well Step 3 Step 4 Step 5 Stock sol 1M NaOH
- 6. Stock sol 1M NaOH vs Dilutinga std sol Serial Dilution Prepare 10 x fold serial dil 9 cm3 9 cm3 9 cm3 9 cm3 Tube 1 Tube 2 Tube 3 Tube 4 1 cm3 Serial dil 10x 1M → 0.1M, 0.01M, 0.001M, 0.0001M 1 cm3 1 cm3 1 cm3 Dilution • Start with Conc solution (stock) • Add water to dilute it down • Diff to cover a wide range • Time consuming to perform diff dilution for diff conc Serial dilution • 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 2 x fold serial dil 5 cm3 5 cm3 5 cm3 5 cm3 5 cm3 5 cm35 cm3 5 cm3 Tube 1 Tube 2 Tube 3 Tube 4 Serial dil 2x 1M → 0.5M, 0.25M, 0.125M, 0.0625M X 1 2 X 1 4 X 1 8 X 1 16 X 1 10 X 1 100 X 1 1000 X 1 10000 Dilution factor = 1o part = (5 part water +5 part solute) (2x, 1:2) 5 part (5 part solute) Dilution factor = 1o part = (9 part water +1 part solute) (10x, 1:10) 1 part (1 part solute) Stock sol 1M NaOH
- 7. 10 mole in2 dm3 (5 M) 5 mole in 1 dm3 (5 M) 5 mole in 1 dm3 (5 M) Concentration and Molarity Solute Solvent Solution + Solute/Solvent/Solution measured ).( 3 dmVol Mole Conc ).( ).( 3 dmVol gMass Conc ÷ RMM x RMM 5 moles 1 L/dm3 Conc Sol 5M or 5 mol/dm3 Dilution 1 M NaOH 1 dm3 5 mole in 1 dm3 5 M or 5 mol/dm3 •••• • ••••• 1 dm3 0.5 M NaOH Mole bef dil = 5 mol Vol bef = 1 dm3 Conc = 5 M Mole aft dil = 5 mol Vol aft = 2 dm3 Conc = 2.5 M water 2 dm3 Vol increase ↑ Conc decrease ↓ ••• • • molMole VMMole 515 molMole VMMole 525.2 Moles bef dil = Moles aft dil M1 V1 = M2V2 M1 = Ini conc M2 = Final conc V1 = Ini vol V2 = Final vol Amt (mole) - NO CHANGE Conc - Change 1 dm3 2 dm3 1 dm3 ••••• ••• •••• • •• ••••• Vol increase ↑ ↓ Amt (mole) ↑ ↓ Conc remain same molMole VMMole 515 molMole VMMole 1025 Amt (mole) – CHANGE Conc – NO CHANGE
- 8. 15 mole in2 dm35 mole in 1 dm3 (5 M) 10 mole in 2 dm3 (5 M) 5 mole in 1 dm3 (5 M) 5 mole in 1 dm3 (5 M) Concentration and Molarity Solute Solvent Solution + Solute/Solvent/Solution measured ).( 3 dmVol Mole Conc ).( ).( 3 dmVol gMass Conc ÷ RMM x RMM 5 moles 1 L/dm3 Conc Sol 5M or 5 mol/dm3 1 dm3 •••• • ••••• 1 dm3 water 2 dm3 • •• • • 1 dm3 2 dm3 1 dm3 ••••• ••• •••• • •• ••••• Vol increase ↑ ↓ Amt (mole) ↑ ↓ Conc remain same molMole VMMole 515 molMole VMMole 1025 Amt (mole) – CHANGE Conc – NO CHANGE 10 mole in 1 dm3 (10 M) 1 dm3 ••••• • •• •• ••• •• ••• • • molMole VMMole 515 MConc dmVol Mole Conc 5 1 5 ).( 3 MConc dmVol Mole Conc 5.7 2 15 ).( 3 molMole VMMole 1525.7 Amt (mole) – CHANGE Conc – CHANGE
- 9. Cal mass ofNa2CO3 require to prepare 200 ml sol, containing 50 g/dm3 Mass CuSO4 = 5 g, Vol sol = 500 cm3 → 0.5 dm3 5 g of CuSO4 dissolve in water form 500 ml sol Cal conc in g/dm3 and mol/dm3 Cal moles of NH3 in 150 ml of 2M NH3 sol. Cal vol in dm3 of 0.8M H2SO4 which contain 0.2 mol. 5 g CuSO4 0.2 dm3 10g 0.8 M 0.2mol 150 cm3 2M 3 3 10 5.0 5 ).( ).( gdmConc dmVol gMass Conc MConc dmVol Mole Conc 0625.0 5.0 03125.0 ).( 3 molMole dmVol gMass Mole 03125.0 160 5 ).( ).( 3 gMass VolConcMass dmVol gMass Conc 102.050 ).( ).( 3 Conc = 50 g/dm3 3 25.0 8.0 2.0 dm Conc Mole Vol VolConcMole molMole VolConcMole 3.0150.02
- 10. 4 g ofNa2CO3 dissolved in 250cm3 water. Cal its molarity. 250cm3 of HNO3 contain 0.4 mol. Cal its molarity. 0.25 dm3 4 g Na2CO3 0.4 mol 0.25 dm3 HCI has conc of 2M. Find mass of HCI gas in 250cm3 in HCI. 2.0M 0.25 dm3 Mass ? Cal moles of H+ ion in 200 cm3 of 0.5M H2SO4 H2SO4 → 2 H+ + SO4 2- 0.5 M, 0.2 dm3 0.1 mol 2 mol H+ Moles? molMole M gMass Mole r 0377.0 106 4 ).( MConc dmVol Mole Conc 15.0 25.0 0377.0 ).( 3 M Vol Mole Conc VolConcMole 6.1 250.0 4.0 molMole VolConcMole 1.02.05.0 H2SO4 diprotic produce 2 mol H+ molMole VolConcMole 5.0250.02 gMass RMMMoleMass 25.185.365.0 RMM HCI = 36.5
- 11. Cal molarity ofKOH when 750 cm3 water added to 250cm3 , 0.8M KOH 750cm3 water 250cm3 Cal vol water added to 60 cm3, 2M of H2SO4 to produce 0.3M H2SO4 ? cm3 60cm3 2M Cal molarity of NaOH when 500cm3 , 2 mol NaOH added to 1500 cm3 , 4 mol NaOH 2 mol 2000 cm3 500cm3 1500cm3 4 mol+ 6 mol Cal molarity of HCI produce when 200cm3 , 2M HCI added to 300 cm3 , 0.5M of HCI A B C +2 M 0.5M ? M Mole B = M x V 1000 = 0.5 x 300 1000 = 0.15 mol 200cm3 300cm3 500cm3 0.8M Mol bef dilution = Mol aft dilution M1 V1 = M2V2 0.8 x 250 = M2 x 1000 M2 = 0.8 x 250 1000 M2 = 0.2M Mol bef dilution = Mol aft dilution M1 V1 = M2V2 2 x 60 = 0.3 x V2 V2 = 2.0 x 60 0.3 V2 = 400 cm3 (final vol) Vol water = 400 – 60 = 340cm3 added Total mol = (2 + 4) = 6mol Total vol = (500 + 1500) = 2000 cm3 Moles = M x V M = Moles V = 6 mol 2 dm3 = 3.0 mol/dm3 Mole A = M x V 1000 = 2 x 200 1000 = 0.4 mol Total moles A + B = 0.4 + 0.15 = 0.55 mol Total vol = (200 + 300) = 500 cm3 Moles = M x V Conc = Moles = 0.55 = 1.1 M V 0.5
- 12. Prepare 250cm3 , 0.1MHCI using conc HCI, 1.63M. What vol of conc acid must be diluted. ? cm3 1.63M 250cm3 Cal conc when 2 g KCI dissolved in 250 cm3 of sol 250cm3 How to prepare 500cm3 of 0.1M NaCI sol 0.1 M 2 M 1.2 dm3 2 g KCI 0.1M 2..92 g NaCI 500 cm3 How to prepare 1.2 dm3 , 0.4M HCI sol starting from 2 M HCI ? 0.4 M 240 cm3 Mol bef dilution = Mol aft dilution M1 V1 = M2V2 1.63 x V1 = 0.1 x 250 V1 = 0.1 x 250 1.63 V1 = 15.3 cm3 molMole M gMass Mole r 02683.0 55.74 2 ).( MConc dmVol Mole Conc 107.0 250.0 02683.0 ).( 3 Moles NaCI = M x V = 0.1 x 0.5 = 0.05 mol Moles NaCI = Mass RMM Mass = Moles x RMM = 0.05 x 58.5 = 2.92 g Weigh 2.92 g NaCI, make up to 500cm3 sol in a volumetric flask Measure 240 cm3 of 2M HCI, make up to 1.2 dm3 using volumetric flask Mol bef dilution = Mol aft dilution M1 V1 = M2V2 2 x V1 = 0.4 x 1.2 V1 = 0.4 x 1.2 2 M2 = 0.24 dm3 or 240 cm3
- 13. Acknowledgements Thanks to sourceof pictures and video used in this presentation Thanks to Creative Commons for excellent contribution on licenses http://creativecommons.org/licenses/