Prepared by PhD Halina Falfushynska

If one looks at the boiling points of the hydrides of many elements, water, ammonia, and hydrogen fluoride have uniquely
High boiling points. For example water is projected to have a
Boiling point of only about -80 o C in stead of 100 o C!

Polar water molecules interact with the positive and negative ions of a salt, assisting with the dissolving process.

The ethanol molecule contains a polar
O-H bond similar to those in the water molecule. (b) The polar water molecule interacts strongly with the polar O-H bond in ethanol.

The Role of Water as a Solvent: The solubility of
Ionic Compounds
Electrical conductivity The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution.
Electrolyte A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called Strong
Electrolytes.
NaCl
(s)
+ H
2
O
(l)
Na +
(aq)
+ Cl -
(aq)
When Sodium Chloride dissolves into water the ions become solvated , and are surrounded by water molecules. These ions are called “ aqueous ” and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

Electrical Conductivity of Ionic Solutions

Electrical
Conductivity

HCL (aq) is completely ionized.

An aqueous solution of sodium hydroxide.

Acetic acid
(HC
2
H
3
O
2
) exists in water mostly as undissociated molecules.

The reaction of
NH
3 in water.

Comparison of a Concentrated and Dilute Solution

M = =
Liters of Solution L solute = material dissolved into the solvent
In air , Nitrogen is the solvent and oxygen, carbon dioxide, etc.
are the solutes.
In sea water , Water is the solvent, and salt, magnesium chloride, etc.
are the solutes.
In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

LIKE EXAMPLE
Calculate the Molarity of a solution prepared by bubbling 3.68g of
Gaseous ammonia into 75.7 ml of solution.
Solution:
Calculate the number of moles of ammonia:
3.68g NH
3
1 mol NH
3
17.03g
3
Change the volume of the solution into liters:
1 L
75.7 ml X = 0.0757 L
1000 mL
Finally, we divide the number of moles of solute by the volume of the solution:
0.216 mol NH
Molarity = = ____________ M NH
0.0757 L 3

• Quantitative concentration term
– Molarity is the ratio of moles solute per liter of solution
– Symbols: M or [ ]
– Different forms of molarity equation
M
 mol
L
L
 mol
M mol

M

L
19
Copyrig ht
McGra

Calculate the molarity of a solution prepared by dissolving 45.00 grams of
KI into a total volume of 500.0 mL.
Copyright McGraw-Hill 2009 20

Calculate the molarity of a solution prepared by dissolving 45.00 grams of
KI into a total volume of 500.0 mL.
45 .
00 g KI
500.0
mL

1 mol KI
166.0
g KI

1000 mL
1 L

0 .
5422 M
Copyright McGraw-Hill 2009 21

How many milliliters of 3.50 M NaOH can be prepared from 75.00 grams of the solid?
Copyright McGraw-Hill 2009 22

How many milliliters of 3.50 M NaOH can be prepared from 75.00 grams of the solid?
75 .
00 g NaOH

1 mol NaOH
40.00
g NaOH

1 L
3.50
mol NaOH

1000 mL
1 L

536 mL
Copyright McGraw-Hill 2009 23

• Dilution
– Process of preparing a less concentrated solution from a more concentrated one.
24
Copyrig ht
McGra

For the next experiment the class will need 250. mL of 0.10 M CuCl
2
. There is a bottle of 2.0 M CuCl
2
. Describe how to prepare this solution. How much of the
2.0 M solution do we need?
25
Copyrig ht
McGra

• Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to 300.0 ml or 0.300 l !
• What is the Molarity of the salt and each of the ions?
• Na
3
PO
4 (s)
+ H
2
O
(solvent)
= 3 Na +
(aq)
+ PO
4
-3
(aq)

• Mol wt of Na
3
PO
4
= 163.94 g / mol
• 3.95 g / 163.94 g/mol = 0.0241 mol Na
3
PO
4
• dissolve and dilute to 300.0 ml
• M = 0.0241 mol Na
3
PO
4
/ 0.300 l = 0.0803 M
Na
3
PO
4
• for PO
4
-3 ions = ______________ M
• for Na + ions = 3 x 0.0803 M = ___________ M

Like Example
An isotonic solution, one with the same ionic content as blood is about
0.14 M NaCl. Calculate the volume of blood that would contain 2.5 mg
Of NaCl?
Find the moles in 1.0 mg NaCl:
2.5 mg NaCl x x = 4.28 x 10
1000 mg NaCl
1 mol NaCl
58.45g NaCl
-5 mol
NaCl
What volume of 0.14 M NaCl that would contain the amount of NaCl
(4.28 x 10 -5 mol NaCl):
0.14 M NaCl
V x = 4.28 x 10 -5 mol NaCl
L solution
Solving for Volume gives:
-5
V = = ______________________ L
0.14 mol NaCl
Or _________ ml of Blood!
L solution

Steps involved in the preparation of a standard solution.

Like Example 4.4 (P 98)
A Chemist must prepare a 1.00 L of a 0.375 M solution of Ammonium
Carbonate, what mass of (NH
4
)
2
CO
3 this solution?
must be weighed out to prepare
First, determine the moles of Ammonium Carbonate required:
0.375 M (NH
1.00 L x = 0.375 M (NH
L solution
)
2
CO
3
4
)
2
CO
3
This amount can be converted to grams by using the molar mass:
0.375 M (NH
4
)
2
CO
3
94.07 g (NH x = 35.276 g (NH mol (NH
4
)
2
4
)
2
CO
CO
3
4
)
2
CO
3
Or, to make 1.00L of solution, one must weigh out 35.3 g of
(NH
4
)
2
CO
3
, put this into a 1.00 L volumetric flask, and add water to the mark on the flask.

Potassium Permanganate is KMnO
4 and has a molecular mass of 158.04 g / mole
Problem: Prepare a solution by dissolving 1.58 grams of KMnO
4 into sufficient water to make 250.00 ml of solution.
1.58 g KMnO
4 x = 0.0100 moles KMnO
158.04 g KMnO
4
4
Molarity = = ______________ M
0.250 liters
Molarity of K + ion = [K + ] ion = [MnO
4
] ion = _____________ M

(a) A measuring pipette
(b) A volumetric pipette.

(a) A measuring pipette (b) Water is added to the flask. (c) The resulting solution is 1 M acetic acid.

• Take 25.00 ml of the 0.0400 M KMnO
4
• Dilute the 25.00 ml to 1.000 l - What is the resulting Molarity of the diluted solution?
• # moles = Vol x M
• 0.0250 l x 0.0400 M = 0.00100 Moles
• 0.00100 Mol / 1.00 l =
_______________ M

3
3

2
4

When yellow aqueous potassium chromate is added to a colorless barium nitrate solution, yellow barium chromate precipitates.

Reaction of K
2
CrO
4
(aq) and Ba(NO
3
)
2
(aq).

Photos and molecular-level representations illustrating the reaction of KCL(aq) with
AgNO3(aq) to form AgCl(s).

Precipitation of silver chromate by adding potassium chromate to a solution of silver nitrate.
K
2
CrO
4 (aq)
+ 2 AgNO
3 (aq)
Ag
2
CrO
4 (s)
+ 2 KNO
3 (aq)

1. Most nitrate (NO
3
) salts are soluble.
2. Most salts of Na + , K + , and NH
4
+ are soluble.
3. Most chloride salts are soluble. Notable exceptions are AgCl,
PbCl
2
, and Hg
2
Cl
2
.
4. Most sulfate salts are soluble. Notable exceptions are BaSO
4
,
PbSO
4
, and CaSO
4
.
5. Most hydroxide salts are only slightly soluble. The important soluble hydroxides are NaOH, KOH, and Ca(OH)
2
(marginally soluble).
6. Most sulfide (S 2), carbonate (CO
3
2), and phosphate (PO
4
3) salts are only slightly soluble.

The Solubility of Ionic Compounds in Water
The solubility of Ionic Compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compound and the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called
“insoluble” compounds may be several orders of magnitude less than ones that are called
“soluble” in water, for example:
Solubility of NaCl in water at 20 o C = 365 g/L
Solubility of MgCl
2 in water at 20 o C = 542.5 g/L
Solubility of AlCl
3 in water at 20 o C = 699 g/L
Solubility of PbCl
2 in water at 20 o C = 9.9 g/L
Solubility of AgCl in water at 20 o C = 0.009 g/L
Solubility of CuCl in water at 20 o C = 0.0062 g/L

The Solubility of Covalent Compounds in Water
The covalent compounds that are very soluble in water are the ones with -OH group in them and are called “Polar” and can have strong polar (electrostatic)interactions with water. Examples are compound such as table sugar, sucrose (C
12
H
22
O
11
); beverage alcohol, ethanol
(C
2
H
5
-OH); and ethylene glycol (C
2
H
6
O
2
) in antifreeze.
H
H C O H
Methanol = Methyl Alcohol
H
Other covalent compounds that do not contain a polar center, or the
-OH group are considered “Non-Polar” , and have little or no interactions with water molecules. Examples are the hydrocarbons in
Gasoline and Oil. This leads to the obvious problems in Oil spills, where the oil will not mix with the water and forms a layer on the surface!
Octane = C
8
H
18 and / or Benzene = C
6
H
6

When a solution of Na
2
SO
4
(aq) is added to a solution of Pb(NO
3
)
2
, the white solid
PbSO
4
(s) forms.

Problem: How many moles of each ion are in each of the following: a) 4.0 moles of sodium carbonate dissolved in water b) 46.5 g of rubidium fluoride dissolved in water c) 5.14 x 10 21 formula units of iron (III) chloride dissolved in water d) 75.0 ml of 0.56M scandium bromide dissolved in water e) 7.8 moles of ammonium sulfate dissolved in water a) Na
2
CO
3 (s)
H
2
O
2 Na +
(aq)
+ CO
3
-2
(aq) moles of Na + = 4.0 moles Na
2
CO
3 x 2 mol Na +
= 8.0 moles Na +
1 mol Na
2
CO and 4.0 moles of CO
3
-2
3 are present

H
2
O b) RbF
(s)
Rb +
(aq)
+ F -
(aq) moles of RbF = 46.5 g RbF x = 0.445 moles RbF
104.47 g RbF thus, 0.445 mol Rb + and 0.445 mol F are present
H
2
O c) FeCl
3 (s)
Fe +3
(aq)
+ 3 Cl -
(aq) moles of FeCl
3
= 9.32 x 10 21 formula units x
1 mol FeCl
3
6.022 x 10 23 formula units FeCl
3
= 0.0155 mol FeCl
3 moles of Cl = 0.0155 mol FeCl
3
x = _________ mol Cl -
1 mol FeCl
3 and ____________ mol Fe +3 are also present.

H
2
O d) ScBr
3 (s)
Sc +3
(aq)
+ 3 Br -
(aq)
Converting from volume to moles:
Moles of ScBr
Moles of Br -
3
= 75.0 ml x x = 0.042 mol ScBr
10 3 ml
0.56 mol ScBr
1 L
3
3
= 0.042 mol ScBr
3
x = 0.126 mol Br -
1 mol ScBr
3
0.042 mol Sc +3 are also present
H
2
O e) (NH
4
)
2
SO
4 (s)
2 NH
4
+
(aq)
+ SO
4
- 2
(aq)
Moles of NH
4
+ = 7.8 moles (NH
4
)
2
SO
4
1 mol(NH
4
)
2
4
+
SO
4
____ mol NH
4
+ and ______ mol SO
4
- 2 are also present.

Precipitation Reactions: Will a Precipitate form?
If we add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate, will we get a precipitate?
KCl
(aq)
+ NH
4
NO
3 (aq)
= K +
(aq)
+ Cl -
(aq)
+ NH
4
+
(aq)
+ NO
3
-
(aq)
By exchanging cations and anions we see that we could have Potassium
Chloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium
Chloride. In looking at the solubility table it shows all possible products as soluble, so there is no net reaction!
KCl
(aq)
+ NH
4
NO
3 (aq)
= No. Reaction!
If we mix a solution of Sodium sulfate with a solution of Barium Nitrate, will we get a precipitate? From the solubility table it shows that Barium
Sulfate is insoluble, therefore we will get a precipitate!
Na
2
SO
4 (aq)
+ Ba(NO
3
)
2 (aq)
BaSO
4 (s)
+ 2 NaNO
3 (aq)

Precipitation Reactions: A solid product is formed
When ever two aqueous solutions are mixed, there is the possibility of forming an insoluble compound. Let us look at some examples to see how we can predict the result of adding two different solutions together.
Pb(NO
3
)
2 (aq)
+ NaI
(aq)
Pb +2
(aq)
+ 2 NO
3
-
(aq)
+ Na +
(aq)
+ I -
(aq)
When we add These two solutions together, the ions can combine in the way they came into the solution, or they can exchange partners. In this case we could have Lead Nitrate and Sodium Iodide, or Lead Iodide and
Sodium Nitrate formed, to determine which will happen we must look at the solubility table(P 141) to determine what could form. The table indicates that Lead Iodide will be insoluble, so a precipitate will form!
Pb(NO
3
)
2 (aq)
+ 2 NaI
(aq)
PbI
2 (s)
+ 2 NaNO
3 (aq)

a) Calcium Nitrate and Sodium Sulfate solutions are added together.
Ca(NO
3
)
2 (aq)
Molecular Equation
+ Na
2
SO
4 (aq)
CaSO
4 (s)
+2 NaNO
3 (aq)
Total Ionic Equation
Ca 2+
(aq)
+2 NO
3
-
(aq)
+ 2 Na +
(aq)
+ SO
4
-2
(aq)
CaSO
4 (s)
+ 2 Na +
(aq
+
)
2 NO
3
-
(aq)
Net Ionic Equation
Ca 2+
(aq)
+ SO
4
-2
(aq)
CaSO
4 (s)
Spectator Ions are Na + and NO
3
b) Ammonium Sulfate and Magnesium Chloride are added together.
In exchanging ions, no precipitates will be formed, so there will be no
Chemical reactions occurring!
All ions are spectator ions!

Acids A group of Covalent molecules which lose
Hydrogen ions to water molecules in solution
When gaseous hydrogen Iodide dissolves in water, the attraction of the oxygen atom of the water molecule for the hydrogen atom in HI is greater that the attraction of the of the Iodide ion for the hydrogen atom, and it is lost to the water molecule to form an Hydronium ion and an
Iodide ion in solution. We can write the Hydrogen atom in solution as either H +
(aq) or as H
3
O +
(aq) they mean the same thing in solution. The presence of a Hydrogen atom that is easily lost in solution is an “Acid” and is called an “acidic” solution. The water (H
2
O) could also be written above the arrow indicating that the solvent was water in which the HI was dissolved.
HI
(g)
+ H
2
O
(L)
H +
(aq)
+ I -
(aq)
HI
(g)
+ H
2
O
(L)
H
3
O +
(aq)
+ I -
(aq)
HI
(g)
H
2
O
H +
(aq)
+ I -
(aq)

+
Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion.
What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2.30 Liters of acid solution?
Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity.
Solution: Two moles of H + are released for every mole of acid:
H
2
SO
4 (l)
+ 2 H
2
O
(l)
2 H
3
O +
(aq)
+ SO
4
- 2
(aq)
Moles H
2
SO
4
Molarity of SO
4
- 2 x
98.09 g H
2
SO
4
1.58 mol SO
4
-2
2.30 l solution
0.687 Molar in SO
4
- 2
Molarity of H + = 2 x 0.687 mol H + = 1.37 Molar in H +
2
SO
4

An Acid is a substance that produces H + (H
3
O + ) ions when dissolved in water, and is a proton donor
A Base is a substance that produces OH ions when dissolved in water.
the OH ions react with the H + ions to produce water, H
2
O, and are therefore proton acceptors.
Acids and Bases are electrolytes, and their strength is categorized in terms of their degree of dissociation in water to make hydronium or hydroxide ions. Strong acids and bases dissociate completely, and are strong electrolytes. Weak acids and bases dissociate weakly and are weak electrolytes.
The generalized reaction between an Acid and a Base is:
HX
(aq)
+ MOH
(aq)
MX
(aq)
+ H
2
O
(L)
Acid + Base = Salt + Water

Selected Acids and Bases
Acids Bases
Strong Strong
Hydrochloric, HCl Sodium hydroxide, NaOH
Hydrobromic, HBr Potassium hydroxide, KOH
Hydroiodoic, HI Calcium hydroxide, Ca(OH)
2
Nitric acid, HNO
3
Strontium hydroxide, Sr(OH)
2
Sulfuric acid, H
2
SO
4
Perchloric acid, HClO
4
Barium hydroxide, Ba(OH)
2
Weak Weak
Hydrofluoric, HF Ammonia, NH
3
Phosphoric acid, H
3
PO
4
Acetic acid, CH
3
COOH
(or HC
2
H
3
O
2
)

Problem: Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following Chemical reactions: a) Calcium Hydroxide(aq) and Hydroiodoic acid(aq) b) Lithium Hydroxide(aq) and Nitric acid(aq) c) Barium Hydroxide(aq) and Sulfuric acid(aq)
Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts.
Solution: a) Ca (OH)
2 (aq)
+ 2 H I
(aq)
CaI
2 (aq)
+ 2 H
2
O
(l)
Ca 2+
(aq)
+ 2 OH -
(aq)
+ 2 H +
(aq)
+ 2 I -
(aq)
Ca 2+
(aq)
+ 2 I -
(aq)
+ 2 H
2
O
(l)
2 OH -
(aq)
+ 2 H +
(aq)
2 H
2
O
(l)

b) Li OH
(aq)
+ H NO
3 (aq)
LiNO
3 (aq)
+ H
2
O
(l)
Li +
(aq)
+ OH -
(aq)
+ H +
(aq)
+ NO
3
-
(aq)
Li +
(aq)
+ NO
3
-
(aq)
+ H
2
O
(l)
OH -
(aq)
+ H +
(aq)
H
2
O
(l) c) Ba (OH)
2 (aq)
+ H
2
SO
4 (aq)
Ba 2+
(aq)
+ 2 OH -
(aq)
+ 2 H +
(aq)
+ SO
4
2-
(aq)
Ba 2+
(aq)
+ 2 OH -
(aq)
+ 2 H +
(aq)
+ SO
4
2-
(aq)
BaSO
4 (s)
+ 2 H
2
O
(l)
BaSO
4 (s)
+ 2 H
2
O
(l)
BaSO
4 (s)
+ 2 H
2
O
(l)

G E N E R A L P R O P E R T I E S O F S O L U T I O N S
1. A solution is a homogeneous mixture of two or more components.
2. It has variable composition.
3. The dissolved solute is molecular or ionic in size.
4. A solution may be either colored or colorless nut is generally transparent.
5. The solute remains uniformly distributed throughout the solution and will not settle out through time.
6. The solute can be separated from the solvent by physical methods.

 Depends on the concentration of the solute particles and not on the identity of the solute.
 Dissolved particles alter and interfere with the dynamic process of a solution.
 NOTE: D T=T f
-T i or in this case
D T=T solution
-T solvent

Boiling point elevation

Freezing point depression

Osmosis

Vapor pressure lowering

• For an ideal solution , the equilibrium vapor pressure is given by Raoult's law as
•
- is the vapor pressure of the pure component i
(= A, B, ...) and is the mole fraction of the component i in the solution
• For a solution with a solvent (A) and one nonvolatile solute (B), and
• The vapor pressure lowering relative to pure solvent is , which is proportional to the mole fraction of solute.

•
The boiling point of a pure solvent is increased by the addition of a non-volatile solute, and the elevation can be measured by ebullioscopy .
•
Here i is the van't Hoff factor as above, K b is the ebullioscopic constant of the solvent (equal to
0.512°C kg/mol for water), and m is the molality of the solution

• The freezing point of a pure solvent is lowered by the addition of a solute which is insoluble in the solid solvent, and the measurement of this difference is called cryoscopy .
•
Here K f is the cryoscopic constant , equal to
1.86°C kg/mol for the freezing point of water.
Again i is the van't Hoff factor and m the molality.

• The osmotic pressure of a solution is the difference in pressure between the solution and the pure liquid solvent when the two are in equilibrium across a semipermeable membrane.
• If the two phases are at the same initial pressure, there is a net transfer of solvent across the membrane into the solution known as osmosis .

•
Molality = moles of solute per kg of solvent
•
m = n solute
/ kg solvent
•
If the concentration of a solution is given in terms of molality, it is referred to as a molal solution.
Q. Calculate the molality of a solution consisting of 25 g of KCl in
250.0 mL of pure water at 20 o C?
First calculate the mass in kilograms of solvent using the density of solvent:
250.0 mL of H
2
O (1 g/ 1 mL) = 250.0 g of H
2
O (1 kg / 1000 g) = 0.2500 kg of H
2
O
Next calculate the moles of solute using the molar mass:
25 g KCl (1 mol / 54.5 g) = 0.46 moles of solute
Lastly calculate the molality:
m = n / kg = 0.46 mol / 0.2500 kg = 1.8 m (molal) solution

D
T f
= - k f m
Q. Estimate the freezing point of a 2.00 L sample of seawater (k f has the following composition:
= 1.86 o C kg / mol), which
0.052 mol of Mg 2+ 0.458 mol of Na +
0.010 mol K + 0.533 mol Cl-
0.010 mol Ca 2+
0.002 mol HCO
3
-
0.001 mol Br0.001 mol neutral species.
Since colligative properties are dependent on the NUMBER of particles and not the character of the particles, you must first add up all the moles of solute in the solution.
Total moles = 1.067 moles of solute
Now calculate the molality of the solution: m = moles of solute / kg of solvent = 1.067 mol / 2.00 kg
= 0.5335 mol/kg
Last calculate the temperature change:
D
T f
= - k f m = -(1.86 o C kg/mol) (0.5335 mol/kg) = 0.992 o C
The freezing point of seawater is T solvent
-
D
T = 0 o C - 0.992 o C
= - 0.992 o C

D
T b
= k b m
Q. The boiling point of a solution containing 40.0 g of an unknown substance dissolved in
100.0 g of water is 105.3 o C . Calculate the molar mass of the compound.
Since the solvent is water, the change in temperature (
D
T) would be 105.3 - 100.0 o C = 5.3 o C. You can also find the k kg/mol.
b in the table in your textbook, k b
= 0.512 o C
From this data, you can calculate the molality: m =
D
T b
/ k b
= 5.3 o C / 0.512 o C kg/mol = 10.4 mol/kg
Molality is also defined as the moles of solute per kg of solvent: m = n /(kg solvent), can be rearranged to be n = m (kg of solvent) n = 10.4 mol /kg (0.1000 kg) = 1.04 mol of solute
The molar mass can be calculated by using the equation, MW = m/n
MW = 40.0 g / 1.04 mol = 38.5 g/mol