1. Why does a raw egg swell or shrink when placed in different solutions?
Chemistry I – Chapters 15 & 16
Chemistry I HD – Chapter 15
ICP – Chapter 22
Why does a raw egg swell or shrink when placed in different solutions?
To play the movies and simulations included, view the presentation in Slide Show Mode.

2. Some Definitions
A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase.
One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

3. Parts of a Solution
SOLUTE – the part of a solution that is being dissolved (usually the lesser amount)
SOLVENT – the part of a solution that dissolves the solute (usually the greater amount)
Solute + Solvent = Solution
Solute
Solvent
Example
solid
Alloys (brass, steel)
liquid
Salt water
gas
Air bubbles in ice cubes
“suicides” (mixed drinks)
Soft drinks
Air

4. Definitions Solutions can be classified as saturated or unsaturated.
A saturated solution contains the maximum quantity of solute that dissolves at that temperature.
An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature

5. Example: Saturated and Unsaturated Fats
Saturated fats are called saturated because all of the bonds between the carbon atoms in a fat are single bonds. Thus, all the bonds on the carbon are occupied or “saturated” with hydrogen. These are stable and hard to decompose. The body can only use these for energy, and so the excess is stored. Thus, these should be avoided in diets. These are usually obtained from sheep and cattle fats. Butter and coconut oil are mostly saturated fats.
Unsaturated fats have at least one double bond between carbon atoms; monounsaturated means there is one double bond, polysaturated means there are more than one double bond. Thus, there are some bonds that can be broken, chemically changed, and used for a variety of purposes. These are REQUIRED to carry out many functions in the body. Fish oils (fats) are usually unsaturated. Game animals (chicken, deer) are usually less saturated, but not as much as fish. Olive and canola oil are monounsaturated.

6. Definitions
SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved
Supersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways:
Warm the solvent so that it will dissolve more, then cool the solution
Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

7. Supersaturated Sodium Acetate
One application of a supersaturated solution is the sodium acetate “heat pack.”

8. C. Solubility Solubility Curve
shows the dependence of solubility on temperature
C. Johannesson

9. C. Solubility Solubility Curve
shows the dependence of solubility on temperature
The line represents- saturation
Below the line- unsaturated
Above the line- Supersaturated
C. Johannesson

10. Some questions to answer
What mass of solute will dissolve in 100ml of water at the following temperatures. Which is most soluble
KNO3 at 70C
NaCl at 100C
NH4CL at 90C

11. C. Solubility Gases are more soluble at... low temperatures &
Solids are more soluble at...
high temperatures.
Gases are more soluble at...
low temperatures &
high pressures (Henry’s Law).
EX: nitrogen narcosis, the “bends,” soda
C. Johannesson

12. Henry’s Law
At a given temperature the solubility of a gasin a liquid (S) is directly proportional to the pressure of the gas above the liquid (P).
In other words if the pressure increases the solubility increases
S1 = S2
P P2

13. IONIC COMPOUNDS Compounds in Aqueous Solution
Many reactions involve ionic compounds, especially reactions in water — aqueous solutions.
K+(aq) + MnO4-(aq)
KMnO4 in water
To play the movies and simulations included, view the presentation in Slide Show Mode.

14. Aqueous Solutions
How do we know ions are present in aqueous solutions?
The solutions conduct electricity!
They are called ELECTROLYTES
HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

15. Aqueous Solutions
Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes.
Examples include:
sugar
ethanol
ethylene glycol

17. It’s Time to Play Everyone’s Favorite Game Show… Electrolyte or Nonelectrolyte!

18. Electrolytes in the Body
Carry messages to and from the brain as electrical signals
Maintain cellular function with the correct concentrations electrolytes
Make your own
50-70 g sugar
One liter of warm water
Pinch of salt
200ml of sugar free fruit squash
Mix, cool and drink

19. Concentration of Solute
The amount of solute in a solution is given by its concentration.
Molarity ( M ) =
moles solute
liters of solution

20. 1. 0 L of water was used to make 1. 0 L of solution
1.0 L of water was used to make 1.0 L of solution. Notice the water left over.

21. Steps to make a solution from a SOLID
Step 1: Weigh out the amount of solid needed
Step 2: place the weighed solid into a VOLUMETRIC FLASK
Step 3: while mixing add solvent to bring the level up to
Mark on the flask neck. 

22. PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the Molarity.
Step 1: Calculate moles of NiCl2•6H2O
Step 2: Calculate Molarity
[NiCl2•6 H2O ] = M

23. moles = M•V USING MOLARITY What mass of oxalic acid, H2C2O4, is
required to make 250. mL of a M
solution?
moles = M•V
Step 1: Change mL to L.
250 mL * 1L/1000mL = L
Step 2: Calculate.
Moles = ( mol/L) (0.250 L) = moles
Step 3: Convert moles to grams.
( mol)(90.00 g/mol) = g

24. Learning Check
How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?
1) 12 g
2) 48 g
3) 300 g

25. Solution M = moles of solute Liters of solution M * V = moles
3.0 mol/L * L = 1.2 mol NaOH
1.2 mole NaOH x g NaOH mole NaOH
= 48 g NaOH

26. Molarity from Solubility curve
Determine the Molarity of a saturated NaCl solution at 25C
Go back to solubility curve

27. Concentration Units
An IDEAL SOLUTION is one where the properties depend only on the concentration of solute.
Need conc. units to tell us the number of solute particles per solvent particle.
The unit “molarity” does not do this!

28. Two Other Concentration Units
MOLALITY, m
m of solution =
mol solute
kilograms solvent
% by mass
grams solute X100
grams solution
% by mass =

29. Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate molality and % by mass of ethylene glycol.

30. Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate m & % of ethylene glycol (by mass).
Calculate molality
Calculate weight %

31. Learning Check
A solution contains 15 g Na2CO3 and 235 g of H2O? What is the mass % of the solution?
1) 15% Na2CO3
2) 6.4% Na2CO3
3) 6.0% Na2CO3

32. Solution mass solute = 15 g Na2CO3
mass solution = 15 g g = 250 g
%(by mass) = 15 g Na2CO3 x g solution
= 6.0% Na2CO3 solution

33. 250 g NaCl soln x 10.0 g NaCl = 25 g NaCl 100 g NaCl soln
Using mass %
How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution?
250 g NaCl soln x g NaCl = 25 g NaCl g NaCl soln

34. Try this molality problem
25.0 g of NaCl is dissolved in mL of water. Find the molality (m) of the resulting solution.
m = mol solute / kg solvent
25 g NaCl mol NaCl
58.5 g NaCl
= mol NaCl
Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg
0.427 mol NaCl
5 kg water
= m salt water

35. Mole Fraction (✗)
N is the number of moles of each substance solute and solvent
✔ all mole fractions add up to = 1

36. Problems If I add 1.65 L of water to 112 grams of sodium nitrate
a) What is the molality of NaNO3 in this solution?
b) What is the percent by mass of sodium nitrate in this solution?
c) What is the mole fraction of water in this solution?

37. If I add 1.65 L of water to 112 grams of sodium acetate…
a) What is the molality of NaC2H3O2 in this solution?
0.82 m 
b) What is the percent by mass of sodium acetate in this solution?
6.36%
c) What is the mole fraction of water in this solution?
0.985 

38. Colligative Properties
On adding a solute to a solvent, the properties of the solvent are modified.
Vapor pressure decreases
Melting point decreases
Boiling point increases
Osmosis is possible (osmotic pressure)
These changes are called COLLIGATIVE PROPERTIES.
They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

39. Change in Freezing Point
Ethylene glycol/water
solution
Pure water
The freezing point of a solution is LOWER than that of the pure solvent

40. Change in Freezing Point
Common Applications of Freezing Point Depression
Ethylene glycol – deadly to small animals
Propylene glycol

41. Change in Freezing Point
Common Applications of Freezing Point Depression
Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision?
sand, SiO2
Rock salt, NaCl
Ice Melt, CaCl2

42. Change in Boiling Point
Common Applications of Boiling Point Elevation

43. Boiling point elevation
The solute gets in the way of the molecules escaping to become vapor. .

44. Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
i = van’t Hoff factor = number of particles produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -)
Compound Theoretical Value of i
glycol 1
NaCl 2
CaCl2 3
Ca3(PO4)2 5

45. Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
m = molality
K = molal freezing point/boiling point constant
Substance Kf
benzene
5.12
camphor
40.
carbon tetrachloride
30.
ethyl ether
1.79
water
1.86
Substance Kb
benzene
2.53
camphor
5.95
carbon tetrachloride
5.03
ethyl ether
2.02
water
0.52

46. Change in Boiling Point
Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution?
Kb = 0.52 oC/molal for water (see Kb table).
Solution ∆TBP = Kb • m • i
1. Calculate solution molality = 4.00 m
2. ∆TBP = Kb • m • i
∆TBP = oC/molal (4.00 molal) (1)
∆TBP = oC
BP = = oC (water normally boils at 100)

47. Freezing Point Depression
Calculate the Freezing Point of a 4.00 molal glycol/water solution.
Kf = 1.86 oC/molal (See Kf table)
Solution
∆TFP = Kf • m • i
= (1.86 oC/molal)(4.00 m)(1)
∆TFP = 7.44
FP = 0 – 7.44 = oC (because water normally freezes at 0)

48. Freezing Point Depression
At what temperature will a 5.4 molal solution of NaCl freeze?
Solution
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = oC
FP = 0 – 20.1 = oC

49. Molecular mass determination
A Solution of 7.50 g of a nonvolital compound in g of water boils at °C at 760 mm Hg. What is the molecular mass of the solute. (assume i =1)
ΔTb= Kb X m X i  m = ΔT Kb
The Kb for water is °C/molal
m = 0.78°C
0.512 °C/molal
m =1.5

50. Now calculate moles of solute in solution
1.5 m X 22,6 g of water X 1kg/1000g =0.034 mol solute
Finally use the number of moles of solute and its mass to determine the molecular mass of the solute.
MM of solute = mass of solute/moles of solute
=7.50 g/0.0344mol= 2.2X102 g/mol

51. Problem
The freezing point for water is lowered to C when 3.90 grames of a molecular solute is dissolved in 475 g of water. Calculate the molar mass of the solute.

52. osmosis

53. Setup for titrating an acid with a base

54. Titration 1. Add solution from the buret.
2. Reagent (base) reacts with compound (acid) in solution in the flask.
Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base)
This is called NEUTRALIZATION.

55. LAB PROBLEM #1: Standardize a solution of NaOH — i. e
LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.
35.62 mL of NaOH is neutralized with 25.2 mL of M HCl by titration to an equivalence point. What is the concentration of the NaOH?

56. M x V = moles Ma Va = Mb Vb Ma Va = Mb Vb
35.62 mL of NaOH is neutralized with 25.2 mL of M HCl by titration to an equivalence point. What is the concentration of the NaOH?
Ma Va = Mb Vb
Ma Va
= Mb Vb
( M) (25.2 mL)
= M
(35.62 mL)
M x V = moles
Since the moles of the acid and base are equal and do not change:
M x V (acid) = M x V (base)

57. Preparing Solutions
Weigh out a solid solute and dissolve in a given quantity of solvent.
Dilute a concentrated solution to give one that is less concentrated.

58. Safety: “Do as you oughtta, add the acid to the watta!”

59. PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
Add water to the 3.0 M solution to lower its concentration to 0.50 M
Dilute the solution!

60. PROBLEM: You have 50. 0 mL of 3. 0 M NaOH and you want 0. 50 M NaOH
PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?
Conclusion:
add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

61. Preparing Solutions by Dilution
A shortcut
M1 • V1 = M2 • V2

62. You try this dilution problem
You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400 mL of 0.10 M HCl. How much of the acid and how much water will you need?
M1 V1 = M2 V2
M1 = 12.1 M
V1 = ??? L
M2 = 0.10 M
V2 = 400 mL  L
V1 = L (or 3.1 mL HCl)
Then add enough water so that the total volume is 400 mL.
It should be ABOUT mL (400 – 3.1), but it will be off slightly due to the density of the HCl not being 1.00 g/mL