ib chemistry serial dilution, molarity and concentration

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Prepared by Lawrence Kok

IB Chemistry Molarity, Concentration, Standard Solution and Serial Dilution Preparation .

http://lawrencekok.blogspot.com/

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

http://www.youtube.com/watch?v=NXirIFfpKdg

http://education.wichita.edu/saltymicro/ecology_interactives/serial_dilution.html

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



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


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


Stock sol 1M NaOH

Stock solution 1M NaOH

vs



9 cm39 cm39 cm39 cm3






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






- (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

Tube 2

Tube 3

Tube 4


+ mix well

Pipette 5 cm3 stock to tube 1+ mix well




+ mix well

+ mix well

+ mix well

Step 3

Step 4

Step 5

Stock sol 1M NaOH

Stock sol 1M NaOH

vs




Tube 1

Tube 2

Tube 3

Tube 4

1 cm3


1 cm3 1 cm3 1 cm3

Dilution• Start with Conc solution

(stock)• Add water to dilute it






- (2x dil), (2 fold), (1 : 2)




Tube 1

Tube 2

Tube 3

Tube 4


X 1 2

X 1 4

X 1 8

X 1 16

X 1 10

X 1 100

X 1 1000

X 1 10000



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)



+


measured

).( 3dmVolMoleConc


÷ RMM

x RMM


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)



+


measured

).( 3dmVolMoleConc


÷ RMM

x RMM


1 dm3

•••• •

•••••

1 dm3

water

2 dm3

••• ••

1 dm3 2 dm3

1 dm3

••••• •••••••• ••

•••••

Vol increase ↑↓

Amt (mole) ↑↓

Conc remain same

molMoleVMMole515

molMole

VMMole1025


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


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


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


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


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/

http://creativecommons.org/licenses/

ib chemistry serial dilution, molarity and concentration

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