Solutions
Remember:
 Solute is something dissolved in something else.
 Solvent is the substance that dissolves the solute.
 Solute goes in solvent!
 Solutions are different from pure liquids.
Liquid water must always contain twice as many hydrogens as oxygen. (constant composition)
Solutions can have different amounts of solvent and solute. (variable composition)
Dissolving a solute in a solvent is a physical change, not a chemical one!
Types of Solutions





Solid in solid
Liquid in liquid
Solid in liquid
Gas in liquid
Gas in gas
Examples
1.
2.
3.
Tea
Air
Salt Water
Important Definitions:
Aqueous solution: a solution where water is the solvent. Denoted by (aq) in a chemical
equation.
Misciple liquids: liquids that can be combined in any proportion.
Ex. Water and ethanol
Immiscible liquids: liquids that do not mix well.
Ex. Oil and water
Alloys: solutions of metals
Solubility
 Ability for a solvent to dissolve a solute
 Solute is soluble if: more than 1 g can be dissolved in 100 mL of solvent.
 Solute is insoluble if: less than 1 g can be dissolved in 100 mL of solvent.
Depends on forces of attraction between particles
Polar solvents dissolve polar solutes
Non-polar solvents dissolve non-polar solutes.
Example:
Water does not dissolve vegetable oil.
Kerosene or gasoline does dissolve vegetable oil.
Water does dissolve salt.
Kerosene does not dissolve salt.
Intermolecular Forces:
 Affect solubility
 Dipole-dipole attractions
 Ion-dipole attractions: ions are attracted to molecules with strong dipoles (polar
molecules)
Ex:
Water molecules will surround an Na+ ion so that the negative oxygen ends are pointed towards
the positive ion.
Saturated solutions: solutions where no more solute can be dissolved, and excess solute is
present.
Unsaturated solutions: solutions where more solute can be dissolved and there is no excess
solute.
Factors that Affect Solubility
 Temperature
 Agitating a mixture
 Size of particle
 Pressure
Dissolving a Solute: Up Close and Personal
1.
Break intermolecular forces between particles in the solute. Remember this ALWAYS
requires energy. Break ionic bonds in an ionic solid:
NaCl  Na+ + Cl2.
Break some intermolecular forces between particles in the solvent to make room for
solute. More energy required!
3.
Attraction between solute and solvent particles. Energy is released.
Covalent Compounds
Cannot dissolve in water if are not polar.
Cannot make new hydrogen bonds or dipole-dipole bonds with water molecules.
Factors that Affect Solubility
Temperature
Liquids and Aqueous
Increase temp, increase amount of energy
More bonds can be broken
Decrease temp, decrease amount of energy
Fewer bonds can be broken
Increasing temperature increases solubility
Gases
Already have high kinetic energy
When dissolved in liquid lose energy
If heated gas will leave solution.
Therefore increasing temperature decreases solubility
Think of warm pop vs. cold pop. Cold pop always tastes fizzier.
Size of particle
The smaller the particle the more soluble it is (can fit easily between particles in solvent)
Larger particles must make hydrogen bonds with water.
If not, low solubility
Pressure (gas in liquid)
Increase pressure, increase solubility of gases.
Think: opening a pop bottle, bubbles form when pressure is decreased.
Calculating Concentration
Concentration = mass of solute (g)
Volume of solution (mL)
Mass/volume percent: concentration x 100
Example Calculation:
A pharmacist adds 2.00 mL of distilled water to 4.00 g of a powdered drug. The final volume of
the solution is 3.00 mL. What is the concentration of the drug in g/100 mL? What is the percent
(m/v) of the solution?
Solution:
x = 4.00 g
100 mL 3.00 mL
= 1.33 g/mL x 100mL
= 133g (in 100 mL)
Careful: the concentration in g/mL is 1.33!
133% (m/v)
Example Calculation 2:
Trisodium phosphate is a cleaning product. The recommended concentration is 1.7% (m/v).
What mass of TSP is needed to make a 2.0 L solution?
1.7 % = 1.7/100
1.7 g = x
100mL 2000mL
.x = 1.7g x 200mL
100mL
.x = 34 g
Therefore, 34 g of TSP are needed to make 2.0 L of cleaning solution.
Mass/Mass Concentrations
Mass/mass = Mass of solute (in g)_
Mass of solution (in g)
To find mass/mass percent, multiply mass/mass concentration by 100
Vol/Vol Concentrations
Volume/volume = Volume of solute (in mL)
Volume of solution (in mL)
v/v percent = v/v x 100
Parts Per Million
Ppm means 1 part solute per 1 million parts solvent.
Ppm = Mass of solute
Mass of solution
x 106
OR
Mass of solute
Mass of solution
=
xg
10 g of solution
6
Example Calculation
There is a deadly fungus. Any shipment of peanuts that contains more than 25 ppb (parts per
billion) is dangerous and is rejected. A company receives 20 tons of peanuts fo make peanut
butter, what is the maximum mass (in g) of fungus that is allowed given that 1 ton = 1 000 000
g
25 ppb allowed
Mass is 20 tons (t) =
.ppb = Mass of fungus
x 109
Mass of peanuts
X g solute
=
6
20 x 10 g solution
25 g solute
1 x 10 9g solution
X = 0.5 g
The maximum mass allowed is 0.5 g.
Molarity
More often we will refer to the molarity (M) of a solution.
That is moles per litre.
Molar Concentration (M) = moles of solute
Volume of solution (L)
Notice here we use L and NOT mL!
This is more useful for stoichiometry problems involving aqueous solutions.
We can’t do stoichiometry without moles!