Transcript Unit 4 Solubility - Prairie Spirit School Division

Unit 4 Solubility
4.2.1- DILUTIONS
4.2.2- MOLARITY
4.2.3 CONCENTRATION
4.2.4 DILUTION OF SOLUTIONS
4.2.5CONCENTRATION OF IONS
4.2.1 Dilute & Concentrated Solutions
 We are often concerned with how much solute is
dissolved in a given amount of solution. We will
begin our discussion of solution concentration with
two related and relative terms dilute and concentrated.
 A dilute solution is one in which there is a relatively
small amount of solute dissolved in the solution.
 A concentrated solution contains a relatively large
amount of solute.
4.2.2 Molarity
 There are several different ways to indicate the
amount of solute dissolved in a given volume of
solution. The most common is called molarity.
 Molarity = moles solute
litre solution
 Symbol for Molarity = M
 As an example: The concentration of a solution
might be expressed as a "0.50 M NaOH".
 You would read as "a 0.50 molar sodium hydroxide
solution", meaning that there are 0.50 moles of
NaOH dissolved per one litre of solution.
Sample Problem
 1. Antifreeze is a solution of ethylene glycol,
C2H6O2 in water. If 4.50 L of antifreeze contains 27.5
g of ethylene glycol, what is the concentration of the
solution?
Solution:
 The question asks moles solute/litre solution
 Thus we will need to calculate the molar mass of
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C2H6O2 in order to convert between mass and moles.
the molar mass of C2H6O2 to be 62.1 g·mol-1
arrange the information to end up with the desired unit:
Mol = 27.5 g × 1 mol × 1__
L
62.1 g
4.5 L
= 0.098 mol/L
 Our final answer: [C2H6O2 ] = 0.098 M
 Practice Questions 4.2.2
4.2.3 Other Measures of Concentration
 There are some other ways to express solution
concentration. Here are some alternatives:
1. Molality- Molality is defined as the number of
moles of solute dissolved in one kilogram of solvent
Note:
 Molarity uses mass rather than volume, and
 Molality uses solvent instead of solution.
2. Mass per unit volume. This unit is typically
milligrams per milliliter (mg/mL) or milligrams
per cubic centimeter (mg/cm3). A useful note is
that 1 mL = 1 cm3 and that cm3 is sometimes
referred to as a "cc" (cubic centimeter).
 Mass per unit volume is handy when discussing how
soluble a material is in water or a particular solvent.
3. Percent by Mass. Also called weight percent or
percent by weight. This is simply the mass of the
solute divided by the total mass of the solution and
multiplied by 100%:
 % mass = mass of component x 100%
mass of solution
4. Parts per million (PPM). Parts per million
works like percent by mass, but when there is only
a small amount of solute present. PPM is defined as
the mass of the component in solution divided by
the total mass of the solution multiplied by 106 (one
million):
 A solution with a concentration of 1 ppm has 1 gram
of substance for every million grams of solution. In
general, one ppm implies one mg of solute per liter of
solution.
Example
 A sample of water contains 25 ppm of lead ions,
Pb2+. Convert this concentration to molarity, mol·L-1.
Solution:
 Since the conversion involves mass, we will need to
know the molar mass of Pb2+, which is 206.2 g·mol-1
1. change g to mL
2. change mL to L
3. change g to mol
 25 ppm = 25 g Pb2+ = 25 g Pb2+= 1 mol Pb2+=
106mL H2O 103L H2O
=1.2 × 10-4mol
L
206.2 g
4.2.4 Dilutions of Standard Solutions
 Imagine we have a salt water solution with a certain
concentration (i.e a certain amount of salt dissolved
in a certain volume of solution), to dilute this
solution - we add more water
 M1 = moles1
litre1
M2 =moles2
litre2
 rearrange the equations to find moles:
 moles1 = M1× litre1
 moles2= M2 × litre2
 By adding more water, we changed the volume of the
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solution and it's concentration. But the number of
moles of solute did not change.
So, moles1 = moles2
Therefore, M1 × litre1 = M2 × litre2
Which we will usually express as:
M1V1= M2V2
Where M1 and M2 are the concentrations of the original
and diluted solutions and V1 and V2 are the volumes of
the two solutions
* note C may be used for M *
Example:
 What volume of concentrated sulfuric acid, 18.0 M,
is required to prepare 5.00 L of 0.150 M solution by
dilution with water?
Solution:
 In a dilution question there are 4 variables - M1, V1, M2 and
V2. You will know three of these values and have to calculate
the fourth.
 Write out what you know and what you are trying to find out
1. M1 = 18.0 M M2= 0.150 M V1 = ? V2 = 5.00 L
2. Set up the formula, and rearrange to solve for the unknown,
3. M1V1= M2V2
18.0 × V1= 0.150 × 5.00
V1= 0.150 × 5.00
18.0
 V1= 0.0417 L = 41.7 mL
Example:
 Describe how you would prepare 500.0 mL of a 0.100 M
standard solution of KNO3
 Solution:
 We must first determine the mass of KNO3 that will be
needed to prepare 500.0 mL (0.500 L) of a 0.100 M
solution. The molar mass of KNO3 is 101.1 g·mol-1:
 g = 101.1 g × 0.100mol× 0.500 L= 5.05 g
mol
L
 To prepare the standard solution we will carefully
measure out 5.05 g of KNO3 and dissolve it in some
water. Then we add enough water to make 500 ml
Solution:
 We must first determine the mass of KNO3 that will
be needed to prepare 500.0 mL (0.500 L) of a 0.100
M solution. The molar mass of KNO3 is 101.1 g·mol-1:
 g = 101.1 g × 0.100mol× 0.500 L= 5.05 g
mol
L
 To prepare the standard solution we measure out
5.05 g of KNO3 and dissolve it in some water. Then
we add enough water to make 500 ml
 Practice 4.2.4
4.2.5 Concentration of Ions in Solution
 The solutions often involve ionic compounds and
acids. Both of these produce electrolytic solutions,
meaning they conduct an electrical current due to
the production of ions in solution.
 It will often be important for us to be able to
determine the concentration of these ions, not just
the overall concentration of the solution.
 It will be important for you to be able to write
balanced reactions that show how these substances
break down into ions. If you are given the formula of
the compound (either ionic or acid) you will need to
be able to determine:
 which ions are produced and
 in what mole ratio.
 For example, sodium carbonate, Na2CO3 dissociates into
ions as:
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Na2CO3 (s) → 2 Na+(aq)+ CO32-(aq)
Notice that two molse of Na+ ions are produced for every
one mole of Na2CO3 . The total volume of the solution
remains unchanged.
If we have a 0.20 M solution of Na2CO3 , what will be the
concentration of our two ions, Na+ and CO32-? This can
easily be determined from the coefficients in our
balanced equation:
[Na+] = 2 × [Na2CO3] = 2 × 0.20 M = 0.40 M
[CO32-] = 1 × [Na2CO3] = 1 × 0.20 M = 0.20 M
 Practice problems 4.2.5
 Assignment 4.2.5