IB Chemistry Serial Dilution, Molarity and Concentration

1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
IB Chemistry Molarity, Concentration, Standard Solution and Serial
Dilution Preparation .

2. 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
).( 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 1M NaOH
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
Stock solution 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
Diluting a 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 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
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 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
).( 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 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
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 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
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 of KOH 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
, 0.1M 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 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 source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/