IB Chemistry on Molarity, Concentration, standard solution serial dilution

Concentration and Molarity

Solute Solvent Solution

Conc in g/dm3

↓ Mass solute in grams Vol of solution in dm3

+

Grams

measured

cm3 dm3 Moles

measured

Concentration

Conc in mol/dm3

↓ Moles solute in mol Vol of solution in dm3

measured

Find conc in g/dm3 and mol/dm3

46 g NaOH in 1.00dm3

46 g

1 dm3

= mass (g) vol dm3

= 46g 1dm3

= 46g/dm3

= moles (mol) vol dm3

= 1 mol 1dm3

= 1 mol/dm3

= 1 M

Moles = mass M = 46g 46g/mol = 1 mol

Conc in g/dm3 Conc in mol/dm3

g/dm3 ↔ mol/dm3

g/dm3 ÷ M → mol/dm3 46g/dm3 ÷ 46 = 1 mol/dm3

Conversion formula

Video on conc/molarity

Molarity in M

http://www.youtube.com/watch?v=SXf9rDnVFao

Standard solution Solution of known concentration

Preparing standard solution of conc - 1 M NaOH

1 M – 1 mole NaOH in 1 L/dm3

Mass of NaOH → 1 mole NaOH x M = 1 x 46g Step 1

Step 2

Transfer from beaker to 1L volumetric flask using filter funnel Step 3

Add water until 1L mark with wash bottle

Transfer to beaker, add water to dissolve it

Step 4

Step 5

46 g

Mix till it dissolved 1M NaOH – 1 mole of NaOH in total vol of solution (1000ml)

Video standard solution preparation

Molarity = 1 mole (1M) 1 L total solution vol (solute + solvent)

Molarity = 1 mole (1M) 1 L of vol solvent

1M NaOH

http://www.youtube.com/watch?v=z5EIAo0wdLw

http://www.youtube.com/watch?v=A2YyIo8vSCA

http://www.youtube.com/watch?v=yvxDNOCTh0Y

Solute 5 moles

Solvent 1 L/dm3

Conc Solution 5M or 5 mol/dm3 + Diluted solution

2.5M or 2.5 mol/dm3

5 moles NaOH in 1 dm3

5 M or 5mol/dm3

Dilution

Adding water Diluting a standard solution

Dilution

1 NaOH

5 moles NaOH in 2 dm3

2.5 M or 2.5mol/dm3

0.5NaOH

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

1 dm3 1 dm3

1 dm3

Vol Increase ↑, Conc decrease ↓

Number mole

NO CHANGE

Moles = Molarity(M) x Vol (dm3) = M x V = 5 x 1 = 5 moles

Moles = Molarity(M) x Vol (dm3) = M x V = 2.5 x 2 = 5 moles

After Before

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)

Standard solution , molarity - 1 M NaOH 1 M – 1 mole NaOH in 1 dm3

(stock solution)

1M NaOH in a volumetric flask Step 1

Step 2

Transfer 10cm3 to a volumetric flask Step 3

Add 90 cm3 water, to a total vol of 100cm3

Pipette 10cm3 of 1M NaOH with a pipette

Step 4

Step 5 Mix to dissolve, 0.1M NaOH

Video standard solution preparation

Moles before dilution = Moles after dilution

M1 V1 = M2V2 M1 = Initial molarity M2 = Final molarity V1 = Initial volume V2 = Final volume

1M 0.1M

M1 V1 = M2V2 1M x 10cm3 = 0.1M x 100cm3 (10 + 90) 1000 1000

10cm3

90cm3

10cm3

http://www.youtube.com/watch?v=bnQJ2q36d2E

http://www.youtube.com/watch?v=XMtm4hVCGWg

http://highered.mcgraw-hill.com/sites/0073511072/student_view0/animations.html

Stock solution 1M NaOH

Vs

Prepare 0.1M NaOH

Diluting a standard solution Serial Dilution

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)

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

Prepare a 10 x fold serial dil 1M NaOH

9 cm3 9 cm3 9 cm3 9 cm3 9 cm3

Pipette 9cm3 water to tube 1, 2, 3, 4

Pipette 1 cm3 stock to tube 1

Mix well Pipette 1 cm3 from tube 1 to 2

Mix well

Pipette 1 cm3 from tube 2 to 3

Pipette 1 cm3 from tube 3 to 4

Tube

1

Tube

2

Tube

3 Tube

4

Step 1

1 cm3

Step 2 +

Step 3

Step 4

+

Mix well +

Step 5 Mix well

Mix well

+

Serial dil 10 fold 1M → 0.1M, 0.01M, 0.001M, 0.0001M

Step 1 Pipette 9cm3 water to tube 1

1 cm3

Step 2 Pipette 1 cm3 stock to tube 1 +

Dilution 1M → 0.1M

Dilution factor = 1o parts (9 part water +1 part solute) (10x, 1:10) 1 part (1 part solute)

1 cm3 1 cm3 1 cm3


Serial Dilution

Prepare 10x fold serial dil 1M NaOH

9 cm3 9 cm3 9 cm3 9 cm3 5 cm3


Pipette 1cm3 stock to tube 1

Mix well Pipette 1cm3 from tube 1 to 2

Mix well

Pipette 1cm3 from tube 2 to 3


Tube

1

Tube

2

Tube

3

Tube

4

Step 1

1 cm3

Step 2 +

Step 3

Step 4

+

Mix well +

Step 5 Mix well

Mix well

+



5 cm3

+

Dilution factor = 1o parts (5 part water +5 part solute) (2x, 1:2) 5 parts (5 part solute)

5 cm3 5 cm3 5 cm3

Tube

1

Tube

2

Tube

3

Tube

4





+

+

+

Mix well

Mix well

Mix well




Animation serial dilution Video serial dilution

1 cm3 1 cm3 1 cm3

5 cm3 5 cm3 5 cm3

https://nationalvetcontent.edu.au/alfresco/d/d/workspace/SpacesStore/520c7516-1f08-49a7-b6be-7eaea672d339/412/_TEST501/act3201SerialDilution.htm

http://hlsweb.dmu.ac.uk/ahs/elearning/RITA/Serial4/Serial4.html

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

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


Serial Dilution


9 cm3 9 cm3 9 cm3 9 cm3 5 cm3



Mix well Pipette 1cm3 from tube 1 to 2

Mix well



Tube

1

Tube

2

Tube

3

Tube

4

Step 1

1 cm3

Step 2 +

Step 3

Step 4

+

Mix well +

Step 5 Mix well

Mix well

+



5 cm3

+


5 cm3 5 cm3 5 cm3

Tube

1

Tube

2

Tube

3

Tube

4





+

+

+

Mix well

Mix well

Mix well




X 1

2 X 1

10

X 1

100

X 1

1000

X 1

10000

X 1

4

X 1

8

X 1

16

Dilution factor

Dilution factor

1 cm3 1 cm3 1 cm3 5 cm3 5 cm3

5 cm3

Concept Map

Moles = Mass Molar mass

Moles = M(Molarity) x V(Vol)

Moles in solution = M x V M = Molarity -M or mol/dm3 V = Vol in dm3

Moles in solution = M x V 1000 M = Molarity- M or mol/dm3 V = Vol in cm3

Moles bef dilution = Moles aft dilution

M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol

1 mole

gas

Mole = Vol gas (stp) 22.4

mole

changes

mole

NO change

concentration

change

Moles = M x V

= 5M x 1 dm3

= 5 mol



concentration

change

Moles vs Concentration/Molarity

1 dm3

1 dm3

1 dm3

1 dm3 2 dm3

5 moles

Moles = M x V

= 5M x 1 dm3

= 5 mol

Mole = M x V

= 2.5M x 2dm3

= 5 mol

Mole = M x V

= 5M x 2dm3

= 10 mol


M1 V1 = M2V2 5 x 1 = M2 x 2

M2 = 5/2 M2 = 2.5M

mole

NO change

Conc = 5M Conc = 2.5M

1 dm3

1 dm3

1 dm3 2 dm3

Mole = M x V

= 5M x 1dm3

= 5 mol

Molarity = 5M

mole

change

Mole = M x V

Conc = Mole V

= 5mol 1dm3

Conc = 5M

Mole = M x V

Conc = Mole V

= 10mol 2dm3

Conc = 5M

Concentration

NO change

Molarity = 5M

Moles = M x V

= 5M x 1 dm3

= 5 mol



concentration

change

Moles vs Concentration/Molarity

1 dm3

1 dm3

1 dm3

1 dm3 2 dm3

5 moles

Moles = M x V

= 5M x 1 dm3

= 5 mol

Mole = M x V

= 2.5M x 2dm3

= 5 mol

Mole = M x V

= 7.5M x 2dm3

= 15 mol


M1 V1 = M2V2 5 x 1 = M2 x 2

M2 = 5/2 M2 = 2.5M

mole

NO change

Conc = 5M Conc = 2.5M

1 dm3

1 dm3

1 dm3 2 dm3

Mole = M x V

= 5M x 1 dm3

= 5 mol

Molarity = 5M

mole

change

Mole = M x V

Conc = Mole V

= 5mol 1dm3

= 5M

Mole = M x V

Conc = Mole V = 15mol 2dm3

= 7.5M

Concentration

change

Molarity = 10M

Total = 5 + 10 = 15 mol moles

Cal mass of Na2CO3 require to dissolve in water to prepare 200cm3

solution, containing 50 g/dm3

Answer: Vol = 200cm3 → 0.2dm3

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

Answer: Mass copper(II) sulphate = 5.00g Vol solution = 500cm3 → 500/1000 → 0.5dm3

Conc = Mass Vol = 5.00g 0.5dm3

= 10.0g/dm3

Alternative 0.5dm3 → 5.00g 1 dm3 → 5.00 x 1 0.5 = 10g/dm3

Convert g/dm3 to mol/dm3

RMM copper(II) sulphate = 160 mol/dm3 → g/dm3

RMM = 10 160 = 0.0625M

5.00g of copper(II) sulphate dissolve in water form 500cm3 solution. Cal conc in g/dm3 and mol/dm3

IB Questions on Conc and Molarity

Cal moles of NH3 in 150cm3 of 2M aqueous NH3 solution.

Answer: Moles = M x V 1000 = 2 X 150 1000 = 0.3 mol

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

1

5.00g → 0.5dm3

0.5 dm3

5.00 g

2

0.2 dm3

10g

3 4

0.8M

0.2mol

150cm3

2M

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

Convert g/dm3 to mol/dm3 RMM = 106 mol/dm3 → g/dm3

RMM = 16 106 = 0.15M

4.0g of Na2CO3 dissolved in water, making up 250cm3

Calculate its molarity.

Answer: Moles = M x V 1000 0.4 = M X 250 1000 M = 0.4 x 1000 250 = 1.6M Alternative 0.25 dm3 → 0.4 mol HNO3

1 dm3 → 0.4 x 1 mol HNO3

0.25 = 1.6M

250cm3 of HNO3 contain 0.4moles. Cal its molarity.


5

250cm 3 → 0.25dm3

0.25 dm3

0.0377mol

6

0.4mol

0.25 dm3

HCI acid has conc of 2.0M. Cal the mass of hydrogen chloride gas in 250cm3 in HCI.

7

Answer: Moles = M x V 1000 = 2.0 X 250 1000 = 0.5 mol RMM HCI= 36.5 Mass of HCI = Moles x RMM = o.5 x 36.5 = 18.25g

2.0M

0.25 dm3

Mass ?

Cal moles of hydrogen ions H+ in 200cm3 of 0.5M H2SO4 8

Answer: Moles = M x V 1000 = 0.5 X 200 1000 = 0.1 mol H2SO4 diprotic produce 2 mol H+ ions Moles H+ = 2 x 0.1 mol = 0.2 mol

H2SO4 → 2 H+ + SO42-

0.5M

0.2 dm3

Moles?


Cal molarity of KOH when 750cm3 water added to 250cm3, 0.8M KOH

9

Answer


M1 V1 = M2V2 0.8 x 250 = M2 x 1000

M2 = 0.8 x 250 1000

M2 = 0.2M

Final vol 750 + 250

= 1000cm3

750cm3

250cm3 1000cm3

Cal vol of water added to 60cm3, 2.0M of H2SO4 to produce 0.3M H2SO4

10

Answer


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

? cm3

60cm3

2M 0.3M

Cal molarity of NaOH produce when 500cm3 , 2mol NaOH added to 1500cm3, 4mol NaOH

11

Answer 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

2 mol

2000cm3 500cm3 1500cm3

4 mol

+

6 mol

Answer 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

Cal molarity of HCI produce when 200cm3 , 2M HCI added to 300cm3, 0.5M of HCI

12

A B C

+ 2 M 0.5M ? M

Mole = M x V B 1000 = 0.5 x 300 = 0.15mol 1000

Mole = M x V A 1000 = 2.0 x 200 = 0.4mol 1000

200cm3 300cm3 500cm3

0.8M


Prepare 250cm3, 0.1M HCI using concentrated HCI, 1.63M. What vol of concentrated acid must be diluted.

13

Answer


M1 V1 = M2V2 Moles aft dilution = M x V = 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

? cm3

1.63M

250cm3

Cal conc formed when 2.00g KCI dissolved in 250cm3 of solution 14

Answer Moles KCI = mass (solid) M = 2.00 = 0.02683 mol 74.55 Moles KCI = M x V (solution) M = moles = 0.02683 V 0.250 M = 0.107 mol/dm3

250cm3

How to prepare 500cm3 of 0.1M NaCI solution ? 15

Answer 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

0.1 M

Answer Moles of HCI = M x V = 0.4 x 1.2 = 0.48 mol


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

Measure 240 cm3 of 2M HCI and make up to 1.2 dm3 with water using volumetric flask.

16

2 M

1.2 dm3

2.00g

250 cm3 → 0.250dm3

0.1M

250 cm3 → 0.250dm3

2..92g NaCI

500cm3

500 cm3 → 0.5dm3

How to prepare 1.2dm3 of 0.4M HCI solution starting from 2.0M HCI ?

0.4 M

240 cm3

IB Chemistry on Molarity, Concentration, standard solution serial dilution

Education