1. Lecture 4. Solutions and their coligative properties
Prepared by PhD Halina Falfushynska

2. A space-filling model of the water molecule.

4. 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 -80oC in stead of 100oC!

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

7. 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.

8. 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) + H2O(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

9. Electrical Conductivity of Ionic Solutions

10. Electrical Conductivity

11. HCL (aq) is completely ionized.

13. An aqueous solution of sodium hydroxide.

14. Acetic acid (HC2H3O2) exists in water mostly as undissociated molecules.

15. The reaction of NH3 in water.

16. Comparison of a Concentrated and Dilute Solution

17. Molarity (Concentration of Solutions)= M
Moles of Solute Moles
Liters of Solution L
M = =
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.
In brass , Copper is the solvent (90%), and Zinc is the solute(10%)

18. 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:
1 mol NH3
17.03g
3.68g NH3
X = mol NH3
Change the volume of the solution into liters:
1 L
1000 mL
75.7 ml X = L
Finally, we divide the number of moles of solute by the volume
of the solution:
0.216 mol NH3 L
Molarity = = ____________ M NH3

19. Quantitative concentration term
Molarity is the ratio of moles solute per liter of solution
Symbols: M or [ ]
Different forms of molarity equation
Copyright McGraw-Hill 2009

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

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

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

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

24. Dilution
Process of preparing a less concentrated solution from a more concentrated one.
Copyright McGraw-Hill 2009

25. For the next experiment the class will
need 250. mL of M CuCl2. There is
a bottle of 2.0 M CuCl2. Describe how to
prepare this solution. How much of the
2.0 M solution do we need?
Copyright McGraw-Hill 2009

26. Preparing a Solution - I
Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to ml or l !
What is the Molarity of the salt and each of the ions?
Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4-3(aq)

27. Preparing a Solution - II
Mol wt of Na3PO4 = g / mol
3.95 g / g/mol = mol Na3PO4
dissolve and dilute to ml
M = mol Na3PO4 / l = M
Na3PO4
for PO4-3 ions = ______________ M
for Na+ ions = 3 x M = ___________ M

28. 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:
1 g NaCl
1000 mg NaCl
1 mol NaCl
58.45g NaCl
2.5 mg NaCl x x = 4.28 x 10-5 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
L solution
V x = x 10-5 mol NaCl
Solving for Volume gives:
4.28 x 10-5 mol NaCl
0.14 mol NaCl
L solution
V = = ______________________ L
Or _________ ml of Blood!

29. Steps involved in the preparation of a standard solution.

30. Like Example 4.4 (P 98)
A Chemist must prepare a 1.00 L of a M solution of Ammonium
Carbonate, what mass of (NH4)2CO3 must be weighed out to prepare
this solution?
First, determine the moles of Ammonium Carbonate required:
0.375 M (NH4)2CO3
L solution
1.00 L x = M (NH4)2CO3
This amount can be converted to grams by using the molar mass:
94.07 g (NH4)2CO3
mol (NH4)2CO3
0.375 M (NH4)2CO3 x = g (NH4)2CO3
Or, to make 1.00L of solution, one must weigh out 35.3 g of
(NH4)2CO3, put this into a 1.00 L volumetric flask, and add water
to the mark on the flask.

31. Make a Solution of Potassium Permanganate
Potassium Permanganate is KMnO4 and has a molecular mass
of g / mole
Problem: Prepare a solution by dissolving 1.58 grams of KMnO4
into sufficient water to make ml of solution.
1 mole KMnO4
g KMnO4
1.58 g KMnO4 x = moles KMnO4
moles KMnO4
0.250 liters
Molarity = = ______________ M
Molarity of K+ ion = [K+] ion = [MnO4-] ion = _____________ M

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

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

34. Dilution of Solutions Take 25.00 ml of the 0.0400 M KMnO4
Dilute the ml to l - What is the resulting Molarity of the diluted solution?
# moles = Vol x M
l x M = Moles
Mol / 1.00 l = _______________ M

36. Reactant solutions: (a) Ba(NO3)3(aq)

37. Reactant solutions: (b) K2CrO4(aq).

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

39. Reaction of K2CrO4 (aq) and Ba(NO3)2 (aq).

40. Precipitation of silver chloride by mixing solutions of silver nitrate and potassium chloride.

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

42. Precipitation of
silver chromate by
adding potassium
chromate to a
solution of silver
nitrate.
K2CrO4 (aq) + 2 AgNO3 (aq)
Ag2CrO4 (s) + 2 KNO3 (aq)

43. Simple Rules for Solubility of Salts in Water
Most nitrate (NO3-) salts are soluble.
Most salts of Na+, K+, and NH4+ are soluble.
Most chloride salts are soluble. Notable exceptions are AgCl,
PbCl2, and Hg2Cl2.
Most sulfate salts are soluble. Notable exceptions are BaSO4,
PbSO4, and CaSO4.
Most hydroxide salts are only slightly soluble. The important
soluble hydroxides are NaOH, KOH, and Ca(OH)2
(marginally soluble).
Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-)
salts are only slightly soluble.

44. 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 20oC = 365 g/L
Solubility of MgCl2 in water at 20oC = g/L
Solubility of AlCl3 in water at 20oC = 699 g/L
Solubility of PbCl2 in water at 20oC = 9.9 g/L
Solubility of AgCl in water at 20oC = g/L
Solubility of CuCl in water at 20oC = g/L

45. 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 (C12H22O11); beverage alcohol, ethanol
(C2H5-OH); and ethylene glycol (C2H6O2) in antifreeze. H
Methanol = Methyl Alcohol H C
O H 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 = C8H and / or Benzene = C6H6

46. When a solution of Na2SO4 (aq) is added to a solution of Pb(NO3)2, the white solid PbSO4(s) forms.

47. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I
Problem: How many moles of each ion are in each of the following:
a) 4.0 moles of sodium carbonate dissolved in water
b) g of rubidium fluoride dissolved in water
c) x 1021 formula units of iron (III) chloride dissolved in water
d) ml of 0.56M scandium bromide dissolved in water
e) moles of ammonium sulfate dissolved in water
a) Na2CO3 (s) Na+(aq) + CO3-2(aq)
moles of Na+ = 4.0 moles Na2CO3 x
= 8.0 moles Na+ and 4.0 moles of CO3-2 are present
H2O
2 mol Na+
1 mol Na2CO3

48. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II
H2O
b) RbF(s) Rb+(aq) + F -(aq)
1 mol RbF
g RbF
moles of RbF = 46.5 g RbF x = moles RbF
thus, mol Rb+ and mol F - are present
H2O
c) FeCl3 (s) Fe+3(aq) + 3 Cl -(aq)
moles of FeCl3 = 9.32 x 1021 formula units
1 mol FeCl3
6.022 x 1023 formula units FeCl3 x
= mol FeCl3
3 mol Cl -
1 mol FeCl3
moles of Cl - = mol FeCl3 x = _________ mol Cl -
and ____________ mol Fe+3 are also present.

49. Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - III
H2O
d) ScBr3 (s) Sc+3(aq) + 3 Br -(aq)
Converting from volume to moles:
1 L
103 ml
0.56 mol ScBr3
1 L
Moles of ScBr3 = 75.0 ml x x = mol ScBr3
3 mol Br -
1 mol ScBr3
Moles of Br - = mol ScBr3 x = mol Br -
0.042 mol Sc+3 are also present
H2O
e) (NH4)2SO4 (s) NH4+(aq) + SO4- 2(aq)
2 mol NH4+
1 mol(NH4)2SO4
Moles of NH4+ = 7.8 moles (NH4)2SO4 x = ____ mol NH4+
and ______ mol SO4- 2 are also present.

50. 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) + NH4NO3 (aq) = K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(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) + NH4NO3 (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!
Na2SO4 (aq) + Ba(NO3)2 (aq) BaSO4 (s) + 2 NaNO3 (aq)

51. 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(NO3)2 (aq) + NaI(aq) Pb+2(aq) + 2 NO3-(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(NO3)2 (aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq)

52. Predicting Whether a Precipitation Reaction Occurs; Writing Equations:
a) Calcium Nitrate and Sodium Sulfate solutions are added together.
Molecular Equation
Ca(NO3)2 (aq) + Na2SO4 (aq) CaSO4 (s) +2 NaNO3 (aq)
Total Ionic Equation
Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq) CaSO4 (s) + 2 Na+(aq+) 2 NO3-(aq)
Net Ionic Equation
Ca2+(aq) + SO4-2(aq) CaSO4 (s)
Spectator Ions are Na+ and NO3-
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!

53. 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 H3O+(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 (H2O) could also be written
above the arrow indicating that the solvent was water in which the HI
was dissolved.
HI(g) + H2O(L) H+(aq) + I -(aq)
HI(g) + H2O(L) H3O+(aq) + I -(aq)
H2O
HI(g) H+(aq) + I -(aq)

54. Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids
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:
H2SO4 (l) + 2 H2O(l) H3O+(aq) + SO4- 2(aq)
1 mole H2SO4
98.09 g H2SO4
Moles H2SO4 = = 1.58 moles H2SO4
155 g H2SO4 x
1.58 mol SO4-2
2.30 l solution
Molarity of SO4- 2 = = Molar in SO4- 2
Molarity of H+ = 2 x mol H+ = 1.37 Molar in H+

55. Acid - Base Reactions : Neutralization Rxns.
An Acid is a substance that produces H+ (H3O+) 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, H2O, 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) + H2O(L)
Acid Base = Salt Water

56. 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 Strontium hydroxide, Sr(OH)2
Sulfuric acid, H2SO Barium hydroxide, Ba(OH)2
Perchloric acid, HClO4
Weak Weak
Hydrofluoric, HF Ammonia, NH3
Phosphoric acid, H3PO4
Acetic acid, CH3COOH
(or HC2H3O2)

57. Writing Balanced Equations for Neutralization Reactions - I
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) + 2HI(aq) CaI2 (aq) + 2H2O(l)
Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq)
Ca2+(aq) + 2 I -(aq) + 2 H2O(l)
2 OH -(aq) + 2 H+(aq) H2O(l)

58. Writing Balanced Equations for Neutralization Reactions - II
b) LiOH(aq) + HNO3 (aq) LiNO3 (aq) + H2O(l)
Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq)
Li+(aq) + NO3-(aq) + H2O(l)
OH -(aq) + H+(aq) H2O(l)
c) Ba(OH)2 (aq) + H2SO4 (aq) BaSO4 (s) + 2 H2O(l)
Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq) BaSO4 (s) + 2 H2O(l)
Ba2+(aq) + 2 OH -(aq) + 2 H+(aq) + SO42-(aq) BaSO4 (s) + 2 H2O(l)

59. GENERAL PROPERTIES OF SOLUTIONS
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.

60. Colligative Properties of Solutions
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: DT=Tf-Ti or in this case DT=Tsolution-Tsolvent
Boiling point elevation
Freezing point depression
Osmosis
Vapor pressure lowering

61. 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 non-volatile solute (B),    and 
The vapor pressure lowering relative to pure solvent is  , which is proportional to the mole fraction of solute.

62. Boiling point elevation (ebullioscopy)
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, Kb is the ebullioscopic constant of the solvent (equal to 0.512°C kg/mol for water), and m is themolality of the solution

63. Freezing point depression (cryoscopy)
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 Kf 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.

64. Osmotic pressure
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.

65. MOLALITY Molality = moles of solute per kg of solvent
m = nsolute / 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 mL of pure water at 20oC?
First calculate the mass in kilograms of solvent using the density of solvent:
250.0 mL of H2O (1 g/ 1 mL) = g of H2O (1 kg / 1000 g) = kg of H2O
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 / kg = 1.8 m (molal) solution

66. Freezing Point Depression DTf = - kf m
Q. Estimate the freezing point of a 2.00 L sample of seawater (kf = 1.86 oC kg / mol), which has the following composition:
0.458 mol of Na mol of Mg mol Ca2+
0.010 mol K mol Cl mol HCO3-
0.001 mol Br 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 = moles of solute
Now calculate the molality of the solution:
m = moles of solute / kg of solvent = mol / 2.00 kg
= mol/kg
Last calculate the temperature change:
DTf = - kf m = -(1.86 oC kg/mol) ( mol/kg) = oC
The freezing point of seawater is Tsolvent - DT = 0 oC oC
= oC

67. Boiling Point Elevation DTb = kb m
Q. The boiling point of a solution containing 40.0 g of an unknown substance dissolved in g of water is oC . Calculate the molar mass of the compound.
Since the solvent is water, the change in temperature (DT) would be oC = 5.3 oC. You can also find the kb in the table in your textbook, kb = oC kg/mol.
From this data, you can calculate the molality:
m = DTb / kb = 5.3 oC / oC 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 ( 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