Topic 6: SOLUTION STOICHIOMETRY
Topic 6: SOLUTION STOICHIOMETRY
A Solution is an homogeneous mixture
gas / gas
CO2 / air
liquid / liquid
ethanol / water
solid / liquid
salt / water
gas / liquid
HCl / water
solid / solid
gold / silver
Definition of a SOLUTION:
Mixture where SOLUTES are dissolved in a
SOLVENT
(or in a mixture of solvents)
Solution
Solute
Solvent
1
6: How does a covalently bonded substance
dissolve?
6: How does an Ionic Solid dissolve?
Dissolution of ions into water
Dissolution of covently bonded molecules into water
3
6: Expressing Concentrations of Solutions
•
•
Methanol in water
4
6: Expressing Concentrations of Solutions
MOLARITY (mol.L-1 or M)
MOLARITY
•
2
Most useful (and common) expression of concentration
Units are Molarity (M) or mol.L-1
Example:
0.5 moles of KMnO4 in 250 mL of solution
Molarity = MOLES of solute per LITRE of solution
Molarity (mol/L) =
Molarity (mol/L) =
moles of solute
=
0.5 moles / 0.25 L
Volume of solution in Litres
=
2 M KMnO4 solution
moles of solute
Volume of solution in Litres
5
6
1
Square
bracketscontinued
[NaCl] imply
6: Molarity
“Concentration of NaCl”
EXAMPLE 2
6: Molarity continued
EXAMPLE 1
To begin making Coca-Cola, place 64 g of sucrose
(C12H22O11) in a beaker and make up the final volume to
600 mL with distilled water.
What is the concentration of sucrose?
•
50.0 g of NaCl are added to a solution and the final
volume is 175 mL. What is the concentration of NaCl?
Moles of NaCl = 50.0 g / (22.99 + 35.45) g.mol-1
= 0.8556 moles
-1
Moles
of sucrose
= 64
/ 342.34
g.mol
For
Molarity
(mol / L),
wegrequire
moles
and Litres.
0.1869and
moles
We currently have= grams
mL.
Volume in L = 0.6 L
Molarity
Volume in L
= 0.175 L
Molarity
= 0.8556 moles / 0.175 L
= 0.1869 moles / 0.6 L
[NaCl] = 4.89 moles.L-1
= 0.3 moles.L-1
(3 sig fig.)
(1 sig fig.)
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6: Molarity continued
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6: Molarity continued
EXAMPLE 3
EXAMPLE 3
•
How many grams of Na2SO4 are needed to make
400.0 mL of 0.25 M solution?
•
How many grams of Na2SO4 are needed to make
400 mL of 0.25 M solution?
mol.L-1
no.of moles = Volume (L) × Molarity
Convert moles to mass using the molar mass of Na2SO4
(mol.L-1)
Mass of Na2SO4
no. of moles = 0.4000 L × 0.25 mol.L-1
= 0.10 moles × 142.04 g.mol-1
= 14.2 g
= 0.10 moles Na2SO4
= 14 g
(2 sig fig.)
9
6: Molarity continued
6: Molarity continued: Gatorade
Ingredients
Water
Sucrose
Glucose
food acids;
330 – citric acid
331 – sodium citrates
EXAMPLE 4
•
In the 0.25 M solution of Na2SO4 from Example 3,
what is the concentration of sodium ions and what is the
concentration of sulfate ions?
Na2SO4 (aq) → 2 Na+ (aq) + SO42- (aq)
Interpret: 1 mole of Na2SO4 releases 2 moles of sodium ions
and 1 mole of sulfate ions.
flavour
monopotassiumphosphate (KH2PO4)
∴ A 0.25 M solution of Na2SO4 would have:
Sodium – 47 mg
Potassium – 22.5 mg
Sugars – 6 g
(sucrose – 5.5g)
colour;
129 – Allura red
133 – Brillant blue
[Na ] = 0.5 M
[SO42-] = 0.25 M
Information given:
(per 100 mL)
sodium chloride (NaCl)
+
and
10
11
12
2
6: Molarity continued: Gatorade
Ingredients
Water
Sucrose
Glucose
food acids;
330 – citric acid
331 – sodium citrates
flavour
monopotassiumphosphate (KH2PO4)
6: Molarity continued: Gatorade
Information given:
(per 100 mL)
Information given:
(per 100 mL)
Sodium – 47 mg
Potassium – 22.5 mg
Sugars – 6 g
(sucrose – 5.5g)
Sodium – 47 mg
Potassium – 22.5 mg
Sugars – 6 g
(sucrose – 5.5g)
What is the concentration
of sodium ions?
47 × 10-3 g / 22.99 g.mol-1
= 2.04 × 10-3 mol
2.04 × 10-3 mol / 0.1 L
[Na+] = 0.0204 mol.L-1
sodium chloride (NaCl)
colour;
129 – Allura red
133 – Brilliant blue
13
6: Molarity continued: Gatorade
14
6: Molarity continued: Gatorade
Ingredients
Information given:
(per 100 mL)
Water
Sucrose
Glucose
food acids;
330 – citric acid
331 – sodium citrates
flavour
monopotassiumphosphate (KH2PO4)
sodium chloride (NaCl)
What is the concentration
of potassium ions?
Sodium – 47 mg
Potassium – 22.5 mg
Sugars – 6 g
(sucrose – 5.5g)
22.5 × 10-3 g / 39.10 g.mol-1
= 5.75 × 10-4 mol
5.75 × 10-4 mol / 0.1 L
[K+] = 5.75 × 10-3 mol.L-1
In 600mL
33 g
3g
~2.88 g
~1.86 g
?
0.492 g
~ 0.45 g
colour;
129 – Allura red
133 – Brilliant blue
15
6: Molarity continued
6: Dilution
EXAMPLE 5
Kikkoman Soy Sauce has a sodium concentration of
6620 mg of sodium per 100.0 mL. Assuming all the sodium has
come from sodium chloride, what is the [NaCl] in mol.L-1?
NaCl (aq) →
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Na+ (aq) + Cl- (aq)
If you place 25.00 mL of a 2.00 M solution of NaCl into a
250.0mL flask and make it up to the mark, what is the
concentration of the new solution?
For any dilution,
#moles before dilution = #moles after dilution
No. of moles of sodium ions = 6.620 g / 22.99 g.mol-1
= 0.28795 moles Na+
Molarityconc. × Volumeconc. = Molaritydilute. × Volumedilute.
From ratio above, # moles Na+ = # moles NaCl
∴ [NaCl] = 0.28795 moles / 0.1000 L
= 2.88 M
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18
3
6: Dilution
If you place 25.00 mL of a 2.00 M solution of NaCl into a
250.0mL flask and make it up to the mark, what is the
concentration of the new solution?
Molarityconc. × Volumeconc. = Molaritydilute. × Volumedilute.
No. of moles of NaCl in 25.00 mL = 0.02500 L × 2.00 mol.L-1
= 0.0500 moles NaCl
This 0.0500 moles of NaCl is pipetted into a flask of volume
250.0 mL
∴ NEW [NaCl] = 0.0500 moles / 0.2500 L
= 0.200 M
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