Solutions and
Concentration (p123 -136)
Lesson 2
Dec 7th – 8th
Homework from Monday
P 276 figure 6- predict solubility,
 p 269 # 8,
 p 271 # 4, 5, 6,
 p 277 # 4,5,
 p 279 # 9-12,
 p280 # 2-4

Concentration of solutions

The concentration of a solution is the
ratio of the quantity of solute to the
quantity of solution.
Whenever possible, reactions are carried out
with all of the reactants in the same fluid
phase.
 Liquids with liquids and gases with gases.

Types of solutions

Dilute solution - ratio of solute to solvent is
very small.
◦ Example - a few crystals of sugar in a glass of water.

Concentrated solution the ratio of solute
to solvent is large.
◦ Example - Maple syrup is a concentrated solution of
sugar in water.

Saturated solution - no more solute
can be dissolved in the solvent at a
particular temperature.

Unsaturated solution - the ratio of
solute to solvent is lower than that of the
corresponding saturated solution. If more
solute is added to an unsaturated
solution, at least some of it should
dissolve


Supersaturated solution is an unstable
system in which the ratio of dissolved solute
to solvent is higher than that of a saturated
solution. A supersaturated solution can be
made by gently cooling a hot saturated
solution.
At a lower temperature the dissolved solute
can be made to precipitate out when a seed
crystal is added. The process is
called precipitation and the substance that
forms is the precipitate.
Supersaturated solution

The amount of solute needed to make a
saturated solution in a given quantity of
solvent at a specific temperature is called
the solubility of the solution.
Concentrations
There are 3 basic ways to express
concentration
1. Percent concentration
2. Very low concentrations
3. Molar Concentrations

Percent Concentration

Can be expressed in V/V (volume by
volume), W/W (weight by weight) or W/V
(weight by volume)
Volume by volume (V/V)
concentration – 2 liquids Example 1

What is the % V/V if 30 mL of pure
ethanol is added to 250 mL of water?
vsolute = 30 mL
vsolution = 250 mL + 30 mL =
280 mL
csolution = ?
csolution = vsolute x 100%
vSolution
csolution = 30 mL X 100%
280 mL
csolution = 10.7 % V/V
Therefore, 30 ml of ethanol
dissolved in 250 ml of water is
a 10.7% solution.
Weight by volume (W/V) concentration –
liquid and solid Example 2

3 grams of H2O2 topical antibiotic solution is
dissolved into 50mL of solution. What is the
w/v concentration?
mH2O2 = 3 g
vsolution = 50 mL
csolution = ?
csolution = msolute x 100%
vSolution
csolution = 3 g x 100%
50 ml
csolution = 6 % solution
Therefore, the solution has a
mass by volume concentration
of 6%.
Weight by weight (W/W)
concentration – Example 3

What is the % W/W of copper in an alloy
when 10 g of Cu is mixed with 250 g of Zn?
mCu= 10 g = solute
mZn = 250 g + solute = 260
g
Concentration = ?
c = msolute x 100%
msolution
Concetration = 10 g Cu x 100%
260 g
csolution = 3.8 % solution
Therefore the alloy has a percent
concentration of 3.8 % copper.

Page 284 # 2-8
VERY LOW
CONCENTRATIONS

Very low concentrations are expressed in
parts per million
Example 4

Dissolved O2 in water shows a
concentration of 333 mL of water At SATP
and 6.5 mg of O2. What is the concentration
in ppm?
mO2 = 6.5 mg
look at chart above and see that
ppm = 1mg/1L
vH2O = 333mL = 0.333L
cO2 = ?
cO2 = m / v
cO2 = 6.5mg / 0.333 L
= 19.5 ppm
Therefore, there are 19.5 parts
of O2 for every million of
water.

Example 5
The maximum acceptable concentration of
fluoride ions in municipal water supplies is
1.5 ppm. What is the maximum mass of
fluoride ions you would get from a 0.250 L
glass of water?
cF = 1.5 mg/L
look at chart above and see that
ppm = 1mg/1L
ratio = 1.5 mg / 1 L =
concentration
Therefore the maximum
vH2O = 0.25 L
amount of fluoride ions in 250
mF = ?
ml of water is 0.375 mg.
c = m / v rearrange to
m=cxv
questions

Page 287 # 11-17
MOLAR
CONCENTRATIONS AND
MOLARITY

Molar concentration or Molarity, is a way
of specifying the amount of solute in one
litre of solvent (mol /L). The abbreviated
form is M or c. In some cases molar
concentration is indicated by using square
brackets. [NH3(aq)] = 0.5 mol/L
Example 6

In a quantitative analysis, a stiochiometric
calculation produced 0.15 mol of NaOH
in a 0.250 mL solution. Calculate the
molar concentration of sodium hydroxide.
nNaOH = 0.15 mol
v = 0.250 ml
c =?
c = n/v
= 0.6 mol/L
Therefore, the molar
concentration of sodium
hydroxide is 0.6 M.
Example 7

A student requires 0.250 moles of NaCl
for an experiment. The only thing
available to them is a bottle with a
solution labelled "0.400 M NaCl." What
volume of the solution should be
used? Give the answer in millilitres.
n = 0.250 mol
c = 0.400 mol/L = 0.400 mol / 1000 mL
v=?
v=n/c
= 0.625 L
Therefore, the student would
need 625 ml of the 0.400mol /L
solution.
Example 8
How many grams of NaOH would be
needed to produce 500 mL of solution with
a molarity of 1 M.
v = 500 mL = 0.5 L
Now we must find the
c = 1 mol/L
mass
n=?
m=nxM

n = 0.5 mol
m = 20 g.
Example 9

A student is heating water to cook pasta,
they add 25 g of sodium chloride to 1.5 L of
water. What is the molar concentration of
the salt water?
mNa = 25 g
v = 1.5 L
Molar concentration = ?
M = 0.28 mol/L
Therefore the molar
concentration of the salt
water is 0.28 mol/L
Questions
Page 290 # 19 – 22
 Page 290 # 2 - 8
