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
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
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
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
Stock solution 1M NaOH
Serial Dilution
Prepare 10x fold serial dil 1M NaOH
9 cm3 9 cm3 9 cm3 9 cm3 5 cm3
Pipette 9cm3 water to tube 1, 2, 3, 4
Pipette 1cm3 stock to tube 1
Mix well Pipette 1cm3 from tube 1 to 2
Mix well
Pipette 1cm3 from tube 2 to 3
Pipette 1cm3 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
Pipette 5cm3 water to tube 1, 2, 3, 4
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
Pipette 5cm3 stock to tube 1
Pipette 5cm3 from tube 1 to 2
Pipette 5cm3 from tube 2 to 3
Pipette 5cm3 from tube 3 to 4
+
+
+
Mix well
Mix well
Mix well
Serial dil 2 fold 1M → 0.5M, 0.25M, 0.125M, 0.0625M
Prepare 2x fold serial dil 1M NaOH
Dilution factor = 1o parts (9 part water +1 part solute) (10x, 1:10) 1 part (1 part solute)
Animation serial dilution Video serial dilution
1 cm3 1 cm3 1 cm3
5 cm3 5 cm3 5 cm3
Stock solution 1M NaOH
Serial Dilution
Prepare 10x fold serial dil 1M NaOH
9 cm3 9 cm3 9 cm3 9 cm3 5 cm3
Pipette 9cm3 water to tube 1, 2, 3, 4
Pipette 1cm3 stock to tube 1
Mix well Pipette 1cm3 from tube 1 to 2
Mix well
Pipette 1cm3 from tube 2 to 3
Pipette 1cm3 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
Pipette 5cm3 water to tube 1, 2, 3, 4
5 cm3
+
Dilution factor = 1o parts (5 part water +5 part solute) (2x, 1:2) 5 part (5 part solute)
5 cm3 5 cm3 5 cm3
Tube
1
Tube
2
Tube
3
Tube
4
Pipette 5cm3 stock to tube 1
Pipette 5cm3 from tube 1 to 2
Pipette 5cm3 from tube 2 to 3
Pipette 5cm3 from tube 3 to 4
+
+
+
Mix well
Mix well
Mix well
Serial dil 2 fold 1M → 0.5M, 0.25M, 0.125M, 0.0625M
Prepare 2x fold serial dil 1M NaOH
Dilution factor = 1o parts (9 part water +1 part solute) (10x, 1:10) 1 part (1 part solute)
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
Moles bef dilution = Moles aft dilution
M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol
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
Moles bef dilution = Moles aft dilution
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
Moles bef dilution = Moles aft dilution
M1 V1 = M2V2 M1 = Ini molarity M2= Final molarity V1 = Initial vol V2 = Final vol
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
Moles bef dilution = Moles aft dilution
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.
IB Questions on Conc and 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?
IB Questions on Conc and Molarity
Cal molarity of KOH when 750cm3 water added to 250cm3, 0.8M KOH
9
Answer
Moles bef dilution = Moles aft dilution
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
Moles bef dilution = Moles aft dilution
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
IB Questions on Conc and Molarity
Prepare 250cm3, 0.1M HCI using concentrated HCI, 1.63M. What vol of concentrated acid must be diluted.
13
Answer
Moles bef dilution = Moles aft dilution
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
Moles bef dilution = Moles aft dilution
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