Classification of
Matter
Solutions are homogeneous mixtures

A. Concentration
• The amount of solute in a solution.
• Describing Concentration
– % by mass - medicated creams
– % by volume - rubbing alcohol
– ppm, ppb - water contaminants
– molarity - used by chemists
– molality - used by chemists

Solute
A solute is the dissolved substance in a solution.
Salt in salt water Sugar in soda drinks
Carbon dioxide in soda drinks
Solvent
A solvent is the dissolving medium in a solution.
Water in salt water Water in soda

A unit on concentration that is the ratio between moles of DISSOLVED substance and liters of solution

Molarity (M)
= moles of solute (mol) volume of solution (L)

There are many ways to represent molarity:
•
•
•
•

MAKE SURE YOUR UNITS ARE CORRECT!!!

A measurement of the concentration of a solution
Molarity (M) is equal to the moles of solute (n) per liter of solution
M = n / V = mol / L
Calculate the molarity of a solution prepared by mixing 1.5 g of
NaCl in 500.0 mL of water.
First calculate the moles of solute:
1.5 g NaCl ( 1 mole NaCl ) = 0.0257 moles of NaCl
58.45 g NaCl
Next convert mL to L : 0.500 L of solution
Last, plug the appropriate values into the correct variables in the equation:
M = n / V = 0.0257 moles / 0.500 L = 0.051 mol/L

M = n /
V
= mol /
L
How many grams of LiOH is needed to prepare 250.0 mL of a
1.25 M solution?
First calculate the moles of solute needed:
M = n /
V
, now rearrange to solve for n: n = (1.25 mol /
L
) (0.2500 L) n = MV
= 0.3125 moles of solute needed
Next calculate the molar mass of LiOH: 23.95 g/mol
Last, use deminsional analysis to solve for mass:
0.3125 moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of
LiOH

• 4:67 pg 153 a,b
Problem

C. Dilution
• Preparation of a desired solution by adding water to a concentrate.
• Moles of solute remain the same.
M
1
V
1

M
2
V
2

C. Dilution
• What volume of 15.8
M HNO
3 is required to make 250 mL of a 6.0
M solution?
GIVEN:
M
1
= 15.8
M
V
1
= ?
M
2
= 6.0
M
V
2
= 250 mL
WORK:
M
1
V
1
= M
2
V
2
( 15.8
M ) V
1
= (6.0
M )(250mL )
V
1
= 95 mL of 15.8
M HNO
3

& Dilution
Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar).
M
1
= 0.05 mol/L
V
1
= 25.0 mL
M
1
V
1
= M
2
V
2
M
V
2
2
= ?
= 50.0 + 25.0 = 75.0 mL
M
1
V
1
= M
2
V
2
= (0.05 mol/L) (25.0 mL) = 0.0167 M of
KI
75.0 mL

• 4:75 pg 153 a,b
Problem

• Substances that dissociate into ions when dissolved in water.
• A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

Electrolytes and Nonelectrolytes
Soluble ionic compounds and strong acids tend to be electrolytes.
There are only seven strong acids:
•Hydrochloric (HCl)
•Hydrobromic (HBr)
•Hydroiodic (HI)
•Nitric (HNO
3
)
•Sulfuric (H
2
SO
4
)
•Chloric (HClO
3
)
•Perchloric (HClO
4

Electrolytes and Nonelectrolytes
Molecular compounds tend to be nonelectrolytes, except for acids and bases.

• A strong electrolyte dissociates completely when dissolved in water.
• A weak electrolyte only dissociates partially when dissolved in water.

• Strong acids
• Strong bases
• Soluble ionic salts

Electrolytes?
1. Pure water
2. Tap water
3. Sugar solution
4. Sodium chloride solution
5. Hydrochloric acid solution
6. Lactic acid solution
7. Ethyl alcohol solution
8. Pure sodium chloride

ELECTROLYTES:
Answers to Electrolytes
Tap water (weak)
NaCl solution
HCl solution
Lactate solution (weak)
NONELECTROLYTES:
Pure water
Sugar solution
Ethanol solution
Pure NaCl

• 4:15 pg 150
Problem

When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

• Double replacement
• Metathesis comes from a Greek word that means “ to transpose ”
• It appears the ions in the reactant compounds exchange, or transpose, ions
Ag NO
3 ( aq )
+ K Cl
( aq )

Ag Cl
( s )
+ K NO
3 ( aq )

Need to know which ionic bonds are soluble and which form precipitates
Need to use solubility chart

1. Write a balanced molecular equation.
2. Dissociate all strong electrolytes.
3. Cross out anything that remains unchanged from the left side to the right side of the equation.
4. Write the net ionic equation with the species that remain.

Ag NO
3 (aq)
+ K Cl
(aq)

Ag Cl
(s)
+ K NO
3 (aq)
Not soluble
• Those things that didn ’ t change (and were deleted from the net ionic equation) are called spectator ions .
Ag +
(aq) + NO
3
-( aq ) + K +
(aq)
+ Cl
-( aq
AgCl ( s ) + K +
(aq) + NO
3
-( aq )
)


• Ag +
(aq) + Cl
-
( aq )

AgCl (s)

• 4.19pg 151
• 4.39 b and C
Problem

There are only seven strong acids:
• Hydrochloric (HCl)
• Hydrobromic (HBr)
• Hydroiodic (HI)
• Nitric (HNO
3
)
• Sulfuric (H
2
SO
4
)
• Chloric (HClO
3
)
• Perchloric (HClO
4
)

The strong bases are the soluble salts of hydroxide ion
(OH ):
• Alkali metals
• Calcium
• Strontium
• Barium

When a strong acid reacts with a strong base, the net ionic equation is…
HCl ( aq ) + NaOH ( aq )

NaCl ( aq ) + H
2
O ( l )
H + ( aq ) + Cl ( aq ) + Na + ( aq ) + OH ( aq )

Na + ( aq ) + Cl ( aq ) + H
2
O ( l )
H + ( aq ) + Cl -
( aq) + Na + ( aq ) + OH ( aq )

Na + ( aq ) + Cl ( aq ) + H
2
O ( l )

When a strong acid reacts with a strong base, the net ionic equation is…
H + ( aq ) + OH ( aq )

H
2
O ( l )

Observe the reaction between
Milk of Magnesia,
Mg(OH)
2
, and HCl.

• 4.81g 153
Problem