1. CHAPTER 9
Water and Solutions
9.3 Properties of Solutions

2. Which acid will dissolve the limestone fastest?
Reaction rates
Which acid will dissolve the limestone fastest?

3. Reaction rates Which acid will dissolve the limestone fastest?
Acid with
the higher concentration
Higher concentration generally means a faster reaction rate

4. Higher temperature generally means a faster reaction rate
Reaction rates
Higher temperature generally means a faster reaction rate

5. Reaction rates Faster reaction rate Higher concentration
Higher temperature
Faster reaction rate

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

7. In an exothermic process, energy is released
The heat comes from calcium chloride dissolving

8. In an endothermic process, energy is absorbed
The cooling effect comes from ammonium nitrate absorbing heat as it dissolves

9. Energy released
Energy absorbed
heat of solution: the energy absorbed or released when a solution dissolves in a particular solvent.

10. From Chapter 3.2
first law of thermodynamics: energy can neither be created nor destroyed.

11. From Chapter 3.2
first law of thermodynamics: energy can neither be created nor destroyed.
The energy inside an isolated system is constant.

12. From Chapter 3.2
first law of thermodynamics: energy can neither be created nor destroyed.
The energy inside an isolated system is constant.
The energy lost by a system must be gained by the surroundings or another system.

13. Calorimetry
A coffee cup calorimeter is an isolated system

14. Calorimetry Calori-metry
thermometer
“heat”
“measure”
Remember:
Heat and temperature are related
A coffee cup calorimeter is an isolated system

15. The energy inside the system is constant

16. 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).

17. 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

18. = The energy inside the system is constant
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

19. = ∆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!

20. If we can calculate ∆Hsolution, we can determine ∆Hreaction
qsolution = (grams of solution) x (specific heat of solution) x ΔT
qsolution = –qreaction
ΔHreaction = qreaction / moles
If we can calculate ∆Hsolution, we can determine ∆Hreaction

21. qsolution = (grams of solution) x (specific heat of solution) x ΔT
qsolution = –qreaction
ΔHreaction = qreaction / moles
Seen in Chapter 3.2

22. qsolution = (grams of solution) x (specific heat of solution) x ΔT
qsolution = –qreaction
ΔHreaction = qreaction / moles
thermometer

23. 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.

24. 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!

25. Break down the problem! Experimental setup
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

26. Break down the problem! Experimental setup What is asked
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

27. Break down the problem! Experimental setup What is asked Assumptions
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
Assumptions
Given: Isolated system: ∆Hreaction = ∆Hsolution
Density (H2O) = 1.0 g/mL

28. 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).

29. 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.

30. 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).

31. 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

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

33. What we have seen so far…
Reaction rates increase with:
increasing concentrations
increasing temperatures
Chemical reactions are accompanied by changes in enthalpy, ΔH

34. What’s next… Reaction rates increase with: increasing concentrations
increasing temperatures
Chemical reactions are accompanied by changes in enthalpy, ΔH
Solution vs. pure solvent

35. Solution vs. pure solvent
Volumes of solute and solvent do not add up to the volume of solution
20 g salt
80 mL water
87 mL solution!

36. 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!

37. The density of a solution increases as more solute is added
Solution vs. pure solvent
The density of a solution increases as more solute is added

38. 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)

39. What’s next… 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

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

41. 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.

42. 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

43. 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.

44. To calculate the freezing point of a solution:
Do not get confused with molarity, M
(moles solute / L of solution)

45. 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.

46. 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 = 1.8 m; Kf = 1.86oC/m (for water the solvent)
Relationships:

47. 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 = 1.8 m; Kf = 1.86oC/m (for water the solvent)
Relationships:
Solve:

48. 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 = 1.8 m; Kf = 1.86oC/m (for water the solvent)
Relationships:
Solve:
Answer: The freezing point is lowered by 3.35oC.

49. 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.

50. 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.

51. 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 as an example