Post on 09-Jan-2017
http://lawrencekok.blogspot.com
Prepared by Lawrence Kok
IB Chemistry Molarity, Concentration, Standard Solution and Serial Dilution Preparation .
Concentration and Molarity
Solute Solvent 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
).( 3dmVolMoleConc
).().(3dmVolgMassConc
÷ RMM
x RMM
3
3
111
).(
moldmConc
dmVolMoleConc
3
3
46146
).().(
gdmConc
dmVolgMassConc
5 moles 1 L/dm3 Conc Sol5M 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
↓••• ••
VMMoledmVolMMolarityMole
)()( 3
molMoleVMMole515
molMole
VMMole525.2
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
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 flask Step 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 = M2V2M1 = Initial molarity M2 = Final molarityV1 = Initial volume V2 = Final volume
M1 V1 = M2V21M 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
Stock solution 1M NaOH
vs
Prepare 0.1M NaOH
Diluting a std sol Serial Dilution
Prepare 10 x fold serial dil
9 cm39 cm39 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
step1 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)
Stock sol 1M NaOH
Stock solution 1M NaOH
vs
Diluting a std sol Serial Dilution
Prepare 10 x fold serial dil
9 cm39 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 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
step1 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 cm35 cm35 cm35 cm3
5 cm35 cm35 cm35 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
Stock sol 1M NaOH
vs
Diluting a std sol Serial Dilution
Prepare 10 x fold serial dil
9 cm39 cm39 cm39 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
step1 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 cm35 cm35 cm35 cm3
5 cm35 cm35 cm35 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
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
).( 3dmVolMoleConc
).().(3dmVolgMassConc
÷ RMM
x RMM
5 moles 1 L/dm3 Conc Sol5M 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
↓••• ••
molMoleVMMole515
molMole
VMMole525.2
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
molMoleVMMole515
molMole
VMMole1025
Amt (mole) – CHANGE
Conc – NO CHANGE
15 mole in 2 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
).( 3dmVolMoleConc
).().(3dmVolgMassConc
÷ RMM
x RMM
5 moles 1 L/dm3 Conc Sol5M or 5 mol/dm3
1 dm3
•••• •
•••••
1 dm3
water
2 dm3
••• ••
1 dm3 2 dm3
1 dm3
••••• •••••••• ••
•••••
Vol increase ↑↓
Amt (mole) ↑↓
Conc remain same
molMoleVMMole515
molMole
VMMole1025
Amt (mole) – CHANGE
Conc – NO CHANGE
10 mole in 1 dm3
(10 M)
1 dm3
••••••••••
••• •• •••••
molMoleVMMole515
MConc
dmVolMoleConc
515
).( 3
MConc
dmVolMoleConc
5.7215
).( 3
molMoleVMMole
1525.7
Amt (mole) – CHANGE
Conc – CHANGE
Cal mass of Na2CO3 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
105.05
).().(
gdmConc
dmVolgMassConc
MConc
dmVolMoleConc
0625.05.0
03125.0).( 3
molMole
dmVolgMassMole
03125.01605
).().(3
gMassVolConcMass
dmVolgMassConc
102.050
).().(3
Conc = 50 g/dm3
325.08.02.0 dm
ConcMoleVol
VolConcMole
molMoleVolConcMole
3.0150.02
4 g of Na2CO3 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
MgMassMole
r
0377.01064
).(
MConc
dmVolMoleConc
15.025.00377.0
).( 3
MVolMoleConc
VolConcMole
6.1250.04.0
molMoleVolConcMole
1.02.05.0
H2SO4 diprotic produce 2 mol H+
molMoleVolConcMole
5.0250.02
gMassRMMMoleMass
25.185.365.0
RMM HCI = 36.5
Cal molarity of KOH when 750 cm3
water added to 250cm3, 0.8M KOH
750cm3 water250cm3
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 cm3500cm3 1500cm3
4 mol+6 mol
Cal molarity of HCI produce when 200cm3 , 2M HCI added to 300 cm3,
0.5M of HCIA 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) = 6molTotal 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 molTotal vol = (200 + 300) = 500 cm3
Moles = M x V Conc = Moles = 0.55 = 1.1 M V 0.5
Prepare 250cm3, 0.1M HCI 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 M2 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
MgMassMole
r
02683.055.742
).(
MConc
dmVolMoleConc
107.0250.002683.0
).( 3
Moles NaCI = M x V = 0.1 x 0.5 = 0.05 mol Moles NaCI = Mass RMMMass = Moles x RMM = 0.05 x 58.5 = 2.92 gWeigh 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
Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/