1. Natural Approach to Chemistry Chapter 9 Water & Solutions
9.1 Solutes, Solvents, and Water
9.2 Concentration and Solubility
9.3 Properties of Solutions

2. 9.1 Assignments
290/1-11 (complete sentences, please),36,40-42

3. Special properties of water
Cohesive nature
Ability to moderate temperature
Expands upon freezing
Versatile solvent

4. Cohesive nature
There is a strong attraction among water molecules due to hydrogen bonding
Substances are generally denser in the solid phase than in the liquid phase.
Water is different
In ice, hydrogen bonds force water molecules to align in a crystal structure where molecules are farther apart than they are in a liquid.

5. Moderates temperature
If ice did not float, ponds would freeze from the bottom up killing everything inside.
Water boils at 100oC because hydrogen bonding keeps the molecules together and they cannot separate easily
Called the universal solvent because it dissolves both ionic and covalent compounds.

6. Water as a solvent Not chemically bonded
hydration: the process of molecules with any charge separation to collect water molecules around them.

7. Tap water contains dissolved salts and minerals.
(Tap water will conduct electricity)
Distilled water and deionized water have been
processed to remove dissolved salts and minerals.
Deionization is a specific filtration process to remove all ions.
(Won’t conduct electricity)
Distillation boils water to steam which is then condensed back to liquid water
(Won’t conduct electricity)

8. Phases and Chemical Reactions
Solids – chem rxn occur, but very slowly
Gases – occur and VERY rapidly
Low density and high mobility of molecules
Fire needs oxygen to support burning

9. Chemical reactions in Liquids – occur easily because of high density and mobility.

10. Reactions in liquids
Life involves many complex chemical reactions that only occur in aqueous solutions!
A step in the Krebs cycle – this is how energy is extracted from glucose

11. Not everything dissolves in water. Why not?
In general,
“like” dissolves “like”
Polar solvents dissolve polar solutes
Nonpolar solvents dissolve nonpolar solutes

12. 9.2 Assignments
290/76-80

13. 9.2 Concentration & Solubility
concentration: the amount of each solute compared to the total solution.

14. molality, m = moles solute kg solvent
There are several ways to express concentration
molality, m = moles solute
kg solvent

15. Suppose you dissolve 10. 0 g of sugar in 90. 0 g of water
Suppose you dissolve 10.0 g of sugar in 90.0 g of water. What is the mass percent concentration of sugar in the solution?
Asked: The mass percent concentration
Given: 10 g of solute (sugar) and 90 g of solvent (water)
Relationships:
Solve:

16. Asked: Molarity of solution Given: Volume of solution = 100.0 mL,
Calculate the molarity of a salt solution made by adding 6.0 g of NaCl to 100 mL of distilled water.
Asked: Molarity of solution
Given: Volume of solution = mL,
mass of solute (NaCl) = 6.0 g
Relationships: M = moles L
Formula mass NaCl = = g/mole
1,000 mL = 1.0 L, therefore 100 ml = 0.10 L
Moles NaCl = 6.0g NaCl x 1 mole NaCl = moles NaCl
58.44 g NaCl
Answer: M = moles = M solution of NaCl
0.100 L

17. Calculate the molality, m of a solution containing 350
Calculate the molality, m of a solution containing g of NaCl in g of water. (Density of water is 1g/L.)
Find the molar mass of NaCl:
= g/mol
Mass to mole conversion:
350.9 g NaCl mole = mole NaCl
58.44g
m = moles solute = mole NaCl = m
kg solvent kg

18. What happens when you add 10 g of sugar to 100 mL of water?
H2O
Water molecules dissolve sugar molecules
Conc. (%) = 10 g/110 g

19. What happens when you add 10 g of sugar to
100 mL of water?
But when two sugar molecules find each other,
they will become “undissolved” (solid) again…
… then, they become redissolved in water again.

20. This is an aqueous equilibrium!
What happens when you add 10 g of sugar to 100 mL of water?
Equilibrium
This is an aqueous equilibrium!

21. dissolving
Equilibrium
“undissolving”
saturation: situation that occurs when the amount of dissolved solute in a solution gets high enough that the rate of “undissolving” matches the rate of dissolving.

22. Temperature has an effect on solubility
Temperature and solubility
20oC
30oC
210 g
sugar
210 g
sugar
All the sugar is dis-
solved
Undis-
solved sugar
100 mL
H2O
100 mL
H2O
Temperature has an effect on solubility

23. solubility: the amount of a solute that will dissolve in a particular solvent at a particular temperature and pressure.

24. Temperature and solubility
Temperature does not have the same effect on the solubility of all solutes

25. Temperature affects:
- the solubility of solutes how much
- the rate of solubility how fast

26. Slow (cold) molecules are not as effective as fast (hot) molecules
Dissolving is a collision process
Slow (cold) molecules are not as effective as fast (hot) molecules
Salt dissolves faster in hot water

27. - with an increase in temperature
The rate of solubility increases:
- with an increase in temperature
- with an increase in surface area of the solute
At higher temperatures:
- solid solutes (like salt and sugar) are more soluble
- gases are less soluble

28. Seltzer water is a supersaturated solution of CO2 in water
This solution is unstable, and the gas “undissolves” rapidly (bubbles escaping)
supersaturation: term used to describe when a solution contains more dissolved solute than it can hold.

29. Determine the formula mass of the solute.
Preparing a solution
How to prepare a mL solution of a 1.0 M CaCl2 solution.
M, molarity = moles solute / liter solution
Determine the formula mass of the solute.
Molar mass of CaCl2

30. Preparing a solution
How to prepare a mL solution of a 1.0 M CaCl2 solution.
Determine the formula mass of the solute.
Use the formula mass of the solute to determine the grams of solute needed.
Molar mass of CaCl2:
g/mole
We need 0.5 moles CaCl2

31. Preparing a solution Molar mass of CaCl2: 110.98 g/mole
How to prepare a mL solution of a 1.0 M CaCl2 solution.
Molar mass of CaCl2:
g/mole
Determine the formula mass of the solute.
Use the formula mass of the solute to determine the grams of solute needed.
We need 0.5 moles CaCl2
We need g CaCl2

32. Preparing a solution
How to prepare a mL solution of a 1.0 M CaCl2 solution.
Determine the formula mass of the solute.
Use the formula mass of the solute to determine the grams of solute needed.
Weigh the grams of solute on the balance.

33. Preparing a solution
How to prepare a mL solution of a 1.0 M CaCl2 solution.
Determine the formula mass of the solute.
Use the formula mass of the solute to determine the grams of solute needed.
Weigh the grams of solute on the balance.
Add the solute to a volumetric flask or graduated cylinder.

34. Preparing a solution
How to prepare a mL solution of a 1.0 M CaCl2 solution.
Determine the formula mass of the solute.
Use the formula mass of the solute to determine the grams of solute needed.
Weigh the grams of solute on the balance.
Add the solute to a volumetric flask or graduated cylinder.
Fill the flask about two thirds of the way up with distilled water.
500.0 mL mark
Do not fill all the way up

35. Preparing a solution
How to prepare a mL solution of a 1.0 M CaCl2 solution.
Determine the formula mass of the solute.
Use the formula mass of the solute to determine the grams of solute needed.
Weigh the grams of solute on the balance.
Add the solute to a volumetric flask or graduated cylinder.
Fill the flask about two thirds of the way up with distilled water.
Mix the solution until the solid dissolves completely.

36. Preparing a solution
How to prepare a mL solution of a 1.0 M CaCl2 solution.
Determine the formula mass of the solute.
Use the formula mass of the solute to determine the grams of solute needed.
Weigh the grams of solute on the balance.
Add the solute to a volumetric flask or graduated cylinder.
Fill the flask about two thirds of the way up with distilled water.
Mix the solution until the solid dissolves completely.
Fill the volumetric flask or graduated cylinder up to the correct volume marker.

37. Ways to express concentration:
Molality, m = moles solute
kg solvent
A higher temperature causes higher:
- solubility of solutes how much
- rates of solubility how fast

38. 9.3 Assignments
290/84-87.

39. 9.3 Properties of Solutions
Reaction rate is generally dependent upon concentration – greater concentration means reaction occurs faster Heat of solution – energy absorbed or released when a solute dissolves in a particular solvent exothermic, loss of energy (gives off energy) or negative heat of solution (feels hot) endothermic, energy absorbed (feels cold) or positive heat of solution

40. Exothermic – energy lost
Endothermic – energy gained

41. Heat loss must equal heat gained: net change is zero
The energy inside the system is constant

42. What changes is the enthalpy
NH4NO3(s) + H2O(l) → NH4+(aq) + NO3–(aq) ∆H = kJ/mole
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole
enthalpy: the energy potential of a chemical reaction measured in joule per mole (J/mole) or kilojoules per mole (kJ/mole).

43. Enthalpy Endothermic reaction Exothermic reaction “∆” means “change”
NH4NO3(s) + H2O(l) → NH4+(aq) + NO3–(aq) ∆H = kJ/mole
Positive value
Exothermic reaction
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole
Negative value

44. = HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole
Heat released by the reaction
Heat gained by the solution =
The energy inside the system is constant

45. = ∆Hreaction = –56 kJ/mole ∆Hsolution = +56 kJ/mole
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole
Heat released by the reaction
Heat gained by the solution =
∆Hreaction
= –56 kJ/mole
∆Hsolution
= +56 kJ/mole
Opposite signs!

46. When a student mixes 40. 0 mL of 1. 0 M NaOH and 40. 0 mL of 1
When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0 M HCl in a coffee cup calorimeter, the final temperature of the mixture rises from 22.0oC to 27oC. Calculate the enthalpy change for the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH). Assume that the coffee cup calorimeter loses negligible heat, that the density of the solution is that of pure water (1.0 g/mL), and that the specific heat of the solution is the same as that of pure water.
Break down the problem!
Experimental setup
Given: 40.0 mL of NaOH (1.0 M)
mL of HCl (1.0 M)
NaOH + HCl  NaCl + H2O
Tinitial = 22.0oC and Tfinal = 27oC

47. When a student mixes 40. 0 mL of 1. 0 M NaOH and 40. 0 mL of 1
When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0 M HCl in a coffee cup calorimeter, the final temperature of the mixture rises from 22.0oC to 27oC. Calculate the enthalpy change for the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH). Assume that the coffee cup calorimeter loses negligible heat, that the density of the solution is that of pure water (1.0 g/mL), and that the specific heat of the solution is the same as that of pure water.
Break down the problem!
Experimental setup
What is asked
Asked: Amount of heat change (DH) for NaOH and HCl reaction

48. Given: Isolated system: ∆Hreaction = ∆Hsolution
When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0 M HCl in a coffee cup calorimeter, the final temperature of the mixture rises from 22.0oC to 27oC. Calculate the enthalpy change for the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH). Assume that the coffee cup calorimeter loses negligible heat, that the density of the solution is that of pure water (1.0 g/mL), and that the specific heat of the solution is the same as that of pure water.
Given: Isolated system: ∆Hreaction = ∆Hsolution
Density (H2O) = 1.0 g/mL
Break down the problem!
Experimental setup
What is asked
Assumptions

49. Relationships:
Solve: First note that the temperature increased, so the reaction released energy to the solution. This means the reaction is exothermic and will have a negative DH. Total volume of solution is 40.0 mL mL = 80.0 mL Total mass of solution is 80.0 g using the densitywater (1.0 g/mL).

50. Relationships:
Solve: First note that the temperature increased, so the reaction released energy to the solution. This means the reaction is exothermic and will have a negative DH. Total volume of solution is 40.0 mL mL = 80.0 mL Total mass of solution is 80.0 g using the densitywater (1.0 g/mL).
The positive sign indicates heat is absorbed.
We reverse the sign as heat gained by the solution is lost by the reaction. Therefore qrxn = –1.67 kJ.

51. Solve: qrxn = –1.67 kJ To find heat on a per mole basis, we use the molarity times the volume in liters to calculate moles; 1.0 M x L = moles of both reactants (where NaOH and HCl are in equimolar amounts).

52. Solve: qrxn = –1.67 kJ To find heat on a per mole basis, we use the molarity times the volume in liters to calculate moles; 1.0 M x L = moles of both reactants (where NaOH and HCl are in equimolar amounts). Lastly, Since the temperature increased, heat was released form the reaction, making DH negative.
Answer: DH = –41.8 kJ/mole

53. What we have seen so far…
Reaction rates increase with:
increasing concentrations
increasing temperatures

54. Solution vs. pure solvent
Salt dissociates into ions, which fit in between water molecules
Volumes of solute and solvent do not add up to the volume of solution
20 g salt
80 mL water
87 mL solution!

55. Why does ice melt when salt is sprinkled on it?

56. Freezing point depression
Why does ice melt when salt is sprinkled on it?
Pure water freezes at 0oC, but a water and salt solution freezes at a lower temperature.

57. The freezing point is lowered in the presence of salt
Freezing point depression
more
less
Pure solvent
Solid formation is not hindered
Order
Entropy
Solution
Solute particles “get in the way” of solid formation
less
more
The freezing point is lowered in the presence of salt

58. Freezing point depression is a colligative property
more
less
Pure solvent
Solid formation is not hindered
Freezing point depression is a colligative property
Order
Entropy
Solution
Solute particles “get in the way” of solid formation
less
more
colligative property: physical property of a solution that depends only on the number of dissolved solute particles not on the type (or nature) of the particle itself.

59. D Tf = Kf x m To calculate the freezing point of a solution:
depression constant
D Tf = Kf x m
Change in freezing molality
Point, oC
Do not get confused with molarity, M
(moles solute / L of solution)

60. Calculate the freezing point of a 1
Calculate the freezing point of a 1.8 m aqueous solution of antifreeze that contains ethylene glycol (C2H6O2) as the solute.
Asked: The freezing point of a 1.8 m solution of ethylene glycol
Given: molality, m = 1.8 m; Kf = 1.86oC/m (Kf , freezing point depression for water, the solvent)
Relationships:
Solve:
Answer: The freezing point is lowered by 3.35oC.

61. Electrolyte solutions
Aqueous solutions containing dissolved ions are able to conduct electricity
1 mole of solute → 2 moles of ions
1 mole of solute → 3 moles of ions
electrolyte: solute capable of conducting electricity when dissolved in an aqueous solution.

62. Electrolyte solutions
Aqueous solutions containing dissolved ions are able to conduct electricity
1 mole of solute → 2 moles of ions
1 mole of solute → 3 moles of ions
The greater the number of particles in solution, the greater the effects.
colligative property: physical property of a solution that depends only on the number of dissolved solute particles not on the type (or nature) of the particle itself.

63. Reaction rates increase with:
increasing concentrations
increasing temperatures
Chemical reactions are accompanied by changes in enthalpy, ΔH
Solution vs. pure solvent
density (solution) > density (pure solvent)
colligative properties: freezing point depression is an example