Solution
Concentration
Remember: Periodic Trends - Electronegativity
• Electronegativity = a number that describes the
ability of an atom to attract electrons
– More electronegative = stronger pull on
electrons being shared
– Less electronegative = weaker pull on electrons
being shared
Trend: Electronegativity
Increasing
Increasing
Difference in Electronegativity
If the electronegativity difference is:
• less than 0.4 = bond is non-polar covalent
• is between 0.5 and 1.6 = bond is polar covalent
• is greater than 1.7 = bond is ionic
Types of Bonds
• Non-Polar Covalent = the attractive forces between
two atoms that results when electrons are equally
shared by the atoms with similar electronegativities
• Polar Covalent = a covalent bond formed between
atoms with significantly different electronegativities
resulting in unequal sharing of electrons
• Ionic = a bond formed due a large difference in
electronegativity between atoms resulting in a
complete transfer of electrons
Comparison…
Non-Polar versus Polar Covalent
Solution and Concentration
How can we express the amount of solute to solvent in
a solution?
• This is called the concentration
•a measure of the number of particles of the solute in
the solvent
•Concentration =
Dilute Solutions
Small quantity of solute
per unit volume of
solution.
Concentrated Solutions
Large quantity of solute
per unit volume of
solution.
Ways to express concentration
3 ways of expressing concentration
Percentages
 PPM or PPB concentrations
 molar concentrations and mass
concentrations
Percentage Concentration
m ass solute( g )
m ass solution( g )
•% (w/w) =
x 100
• W is weight same as mass
•% (w/v) =
m ass solute( g )
volum esolution(m L)
x 100
•% (v/v) =
volum esolute(m L)
volum esolution(m L)
x 100
Everyday Examples
• Vinegar is 5% v/v acetic acid
– 5 mL of CH3COOH in 100 mL vinegar
• Hydrogen peroxide is 3% m/v
– 3 g of H2O2 per 100 mL of solution
• Yogurt is 2% m/m
– 2 g of milk fat per 100 g yogurt
Examples
1.
% V/V = 4.1 L / 55 L = 7.5% V/V
2.
% W/V = 16 g / 50 mL = 32% W/V
3.
% W/W = 1.7 g / 35.0 g = 4.9% W/W
More practice
1. What is the % W/W of copper in an alloy when 10 g of
Cu is mixed with 250 g of Zn?
2. What is approximate % V/V if 30 mL of pure ethanol is
added to 250 mL of water?
3. What is the % W/W if 8.0 g copper is added to enough
zinc to produce 100 g of an alloy
% Concentration: % Mass Example
3.5 g of CoCl2 is
dissolved in 100mL
solution.
Assuming the
density of the
solution is 1.0 g/mL,
what is concentration
of the solution in %
mass?
%m = 3.5 g CoCl2
100g H2O
= 3.5% (m/m)
Low concentrations
• For very dilute solutions, weight/weight
(w/w) and weight/volume (w/v)concentrations are
sometimes expressed in parts per million (ppm).
– Example:
• Toxic substances found in the environment
• Chlorine in a swimming pool
• Metals in drinking water
• Parts per million also can be expressed
– as milligrams per liter (mg/L) (w/v)
– As milligram per kilogram (mg/Kg) (w/w)
PPM learning check
1. Question: A solution has a concentration of
1.25g/L. What is its concentration in ppm?
•
•
Convert the mass in grams to a mass in milligrams:
1.25g = 1.25 x 1000mg = 1250mg
Re-write the concentration in mg/L = 1250mg/L = 1250ppm
2. Question: 150mL of an aqueous sodium chloride
solution contains 0.0045g NaCl. Calculate the
concentration of NaCl in parts per million (ppm).
ppm = mass solute (mg) ÷ volume solution (L)
mass NaCl = 0.0045g = 0.0045 x 1000mg = 4.5mg
volume solution = 150mL = 150 ÷ 1000 = 0.150L
concentration of NaCl = 4.5mg ÷ 0.150L = 30mg/L = 30ppm
Molar concentration
• is able to compare the amount of solute (moles)
dissolved in a certain volume of solution.
• Molar concentration (mol/L) or also called Molarity (M)
= Moles of solute (n)/Volume of solution (V MUST be in litres)
• is the number of moles of solute in one litre of a solution.
We use "M" to denote molar concentration and it has the
units of"mol/L".
• C=n/V
• n=CxV
Concentration: Molarity Example
If 0.435 g of KMnO4 is dissolved in enough water to give
250. mL of solution, what is the molarity of KMnO4?
As is almost always the case,
the first step is to convert the
mass of material to moles.
MM = 158.0 g/mol(39+55+(4X16)
n= m/MM
0.435 g KMnO4 / 158.0 g/ mol = 0.00275 mol KMnO4
Now that the number of moles of substance is known, this can be
combined with the volume of solution — which must be in liters
— to give the molarity. Because 250. mL is equivalent to 0.250 L .
Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 Mol/L = M
0.250 L solution
Molarity learning check
1. How many moles of H2SO4 are there in 250mL of a 0.8M
sulphuric acid solution?
2. If 20g of NaOH is dissolved in sufficient water to produce 500
mL of solution, calculate the molar concentration in molarity.
Questions
1. What is the molarity of the solution formed by dissolving 80
g of sodium hydroxide (NaOH) in 500 mLs of water? (Ar's:
Na=23, O=16, H=1) Ans = 4 M
2. What is the molarity of the solution formed by dissolving
9.8 g of sulphuric acid (H2SO4) in 1000 cm3 of water? (Ar's:
H=1, S=32, O=16) Ans = 0.1 M
3. What mass (g) of hydrogen chloride (HCl) is needed to make
up 500cm3 of a solution of concentration 0.2mol/L? (Ar's:
H=1, Cl=35.5) Ans = 3.65g
Making Solutions…
Solution Preparation
• Standard Solution = a solution for which the
precise concentration is known
• Used in research laboratories and industrial
processes
• Used in chemical analysis and precise control
of chemical reactions
Preparing a Standard Solution
Equipment:
– Electronic balance  precise measurement of solids
– Pipets (pipettes)  precise measurement of liquids
– Volumetric flask  calibrated to contain a precise
volume at a particular temperature, used for precise
dilutions and preparation of standard solutions
Accurate Reading of a Volumetric Flask
Bend down to see the meniscus
Pipets and Bulb/Pump to transfer
small quantities of liquid
Volumetric Pipet
Serological (Blow Out) Pipet
Mohr (Graduated) Pipet
Automatic Dispensers
Micropipets – Dispense µL
(microliters)
Preparing a Solution by Dilution
• Dilution = the process of decreasing the
concentration of a solution, usually by adding
more solvent
• Stock Solution = a solution that is in stock or
on the shelf (i.e., available); usually a
concentrated solution
Preparation of a Solution of Known
Concentration by Diluting a Stock Solution
Preparation of a Solution of Known
Concentration Using a Solid Solute
Calculating the New Concentration of
the Diluted Solution
•
•
•
•
C1 x V1 = C2 x V2
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
Sample Problem
• Water is added to 0.200L of 2.40mol/L
NH3(aq) cleaning solution, until the final
volume is 1.000L. Find the molar
concentration of the final, diluted solution.