1. Ch 8. Solutions

2. 8.1 Mixtures
Pure substances— elements, covalent compounds, and ionic compounds.
Most matter we come into contact with is a mixture of two or more pure substances.
A heterogeneous mixture does not have a uniform composition throughout the sample.
A homogeneous mixture has a uniform composition throughout the sample.

3. 8.1 Mixtures A. Solutions
A solution is a homogeneous mixture that contains small particles. Liquid solutions are transparent.
Solutions consist of two parts:
The solute is the substance present in a lesser amount.
The solvent is the substance present in a larger amount.
An aqueous solution has water as the solvent.

4. Solution = Solvent + Solute
Solvent: a substance that dissolves another substance
Or, the substance present in greater amount
Solute: a substance which is dissolved by a solvent
Or, the substance present in lesser amount

5. Homogeneous Mixture: a mixture in which the composition is the same throughout
Only one phase present
Cannot see interface
Particle size less than 1nm
Often called “solutions”

6. Mixtures
Heterogeneous Mixture: a mixture in which the particles of each component remain separate and can be observed as individual substances
See distinct phases
See interface
Solute particle size above 200nm

7. 8.1 Mixtures B. Colloids and Suspensions
A suspension is a heterogeneous mixture that contains large particles suspended in a liquid.
Particles are so large that they do not dissolve in a liquid, but small enough to stay suspended for a while before settling out..
They can be filtered away from the liquid or separated using a centrifuge.
Examples: Muddy water, liquid medications, paint
Most things that direct you to shake or stir before using

8. 8.1 Mixtures
A colloid is a homogeneous mixture with larger particles, often having an opaque appearance.
Particles in a colloid cannot be filtered from its other components.
They do not settle out.
Their properties are intermediate between heterogeneous and homogeneous mixtures

9. 8.2 Electrolytes and Nonelectrolytes A. Classification
A substance that conducts
an electric current in water
is called an electrolyte.
A substance that does
not conduct an electric
current in water is
called a nonelectrolyte.
NaCl(aq) dissociates into
Na+(aq) and Cl−(aq)
H2O2 does not
dissociate

10. 8.2 Electrolytes and Nonelectrolytes A. Classification
A strong electrolyte dissociates completely in water to form ions.
A weak electrolyte dissociates partially in water to form some ions, leaving mostly uncharged molecules.

11. 8.3 Solubility—General Features
Solubility is the amount of solute that dissolves in a given amount of solvent.
It is usually reported in grams of solute per 100 mL of solution (g/100 mL).
A saturated solution contains the maximum number of grams of solute that can dissolve.
An unsaturated solution contains less than the maximum number of grams of solute that can dissolve.

12. 8.3 Solubility A. Basic Principles
Solubility can be summed up as “like dissolves like.”
Most ionic and polar covalent compounds are soluble in water, a polar solvent.

13. 8.3 Solubility A. Basic Principles
Small neutral molecules with O or N atoms that can hydrogen bond to water are water soluble.

14. 8.3 Solubility A. Basic Principles
Nonpolar compounds are soluble in nonpolar solvents (like dissolves like).
Octane (C8H18) dissolves in CCl4 because both are nonpolar liquids that exhibit only London dispersion forces.
octane
CCl4
octane + CCl4

15. 8.3 Solubility B. Ionic Compounds—Additional Principles

16. 8.4 Solubility-- Effects of T and P A. Temperature (T) Effects
For most ionic and molecular solids, solubility generally increases as temperature increases.
By dissolving a solid in a solvent at high temperature and allowing it to cool slowly, a saturated solution can be made.
A saturated solution contains more than the predicted maximum amount of solute at a given temperature.
The solubility of a gas decreases with increasing temperature.

17. Solubility is expressed in g solute/100g solvent
Example: The solubility of Ethanol is 10g/100g 23ºC
Unsaturated Solution: solution in which the solvent can dissolve more solute
Saturated Solution: solution in which the solvent cannot dissolve any more solute

18. 8.4 Solubility—Effects of T and P B. Pressure (P) Effects
Henry’s law: The solubility of a gas in a liquid is proportional to the partial pressure of the gas above the liquid.
The higher the pressure, the higher the solubility of a gas in a solvent.
closed can of soda
higher CO2 pressure
higher CO2 solubility
open can of soda
lower CO2 pressure
lower CO2 solubility

19. 8.5 Concentration Units B. Volume/Volume Percent
For example, if a bottle of rubbing alcohol contains
70 mL of 2-propanol in 100 mL of solution, then the
(v/v)% concentration is
70 mL 2-propanol
100 mL rubbing alcohol
x 100% = 70% (v/v)

20. Tro's Introductory Chemistry, Chapter 13
Mass Percent
Parts of solute in every 100 parts solution.
If a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution.
Or 10 kg solute in every 100 kg solution.
Since masses are additive, the mass of the solution is the sum of the masses of solute and solvent.
Tro's Introductory Chemistry, Chapter 13

21. Tro's Introductory Chemistry, Chapter 13
Example 13.1—Calculate the Mass Percent of a Solution Containing 27.5 g of Ethanol in 175 mL H2O.
Given:
Find:
27.5 g ethanol, 175 mL H2O
% by mass
27.5 g ethanol, g solution
% by mass
Solution Map:
Relationships:
g sol’n
g EtOH, g H2O %
mL H2O
g H2O
1 mL H2O = 1.00 g
Solve:
Check:
The answer seems reasonable as it is less than 100%.
Tro's Introductory Chemistry, Chapter 13

22. Tro's Introductory Chemistry, Chapter 13
Practice—Calculate the Mass Percent of a Solution that Has 10.0 g of I2 Dissolved in g of Ethanol.
Tro's Introductory Chemistry, Chapter 13

23. Tro's Introductory Chemistry, Chapter 13
Practice—Milk Is 4.5% by Mass Lactose. Determine the Mass of Lactose in 175 g of Milk.
Tro's Introductory Chemistry, Chapter 13

24. Practice—Milk Is 4. 5% by Mass Lactose
Practice—Milk Is 4.5% by Mass Lactose. Determine the Mass of Lactose in 175 g of Milk, Continued.
Given: 175 g milk  175 g solution
Find: g lactose
Equivalence: 4.5 g lactose  100 g solution
Solution Map:
g solution
g Lactose
Apply Solution Map:
Check Answer:
Units are correct. Number makes sense because lactose is a
component of the mixture, therefore, its amount should be less.

25. Example—How Would You Prepare 250. 0 g of 5
Example—How Would You Prepare g of 5.00% by Mass Glucose Solution (Normal Glucose)?
Given: g solution
Find: g glucose
Equivalence: 5.00 g glucose  100 g solution
Solution Map:
g solution
g glucose
Apply Solution Map:
Answer: Dissolve 12.5 g of glucose in enough water to total g.

26. Tro's Introductory Chemistry, Chapter 13
Practice—How Would You Prepare g of 15.0% by Mass Aqueous Ethanol?
Tro's Introductory Chemistry, Chapter 13

27. 8.5 Concentration Units D. Parts Per Million
When a solution contains a very small concentration
of solute, it is often expressed in parts per million.

28. 8.6 Concentration Units—Molarity
Molarity is the number of moles of solute per liter of
solution, abbreviated as M.

29. Preparing a 1.00 M NaCl Solution
Weigh out
1 mole (58.45 g)
of NaCl and add
it to a 1.00 L
volumetric flask.
Step 1
Step 2
Add water to
dissolve the
NaCl, then
add water to
the mark.
Step 3
Swirl to mix.
Tro's Introductory Chemistry, Chapter 13

30. Tro's Introductory Chemistry, Chapter 13
Example 13.3—Calculate the Molarity of a Solution Made by Dissolving 15.5 g of NaCl in 1.50 L of Solution
Given:
Find:
15.5 g NaCl, 1.50 L solution M
Solution Map:
Relationships:
M = mol/L, 1 mol NaCl = g
g NaCl
mol NaCl
L solution M
Solve:
Check:
The unit is correct, the magnitude is reasonable.
Tro's Introductory Chemistry, Chapter 13

31. Tro's Introductory Chemistry, Chapter 13
Practice—What Is the Molarity of a Solution Containing 3.4 g of NH3 (MM 17.03) in mL of Solution?
Tro's Introductory Chemistry, Chapter 13

32. Tro's Introductory Chemistry, Chapter 13
Example 13.4—How Many Liters of a M NaOH Solution Contains 1.24 mol of NaOH?
Given:
Find:
1.24 mol NaOH
volume, L
Solution Map:
Relationships:
1.00 L solution = mol NaOH
mol NaOH
L solution
Solve:
Check:
The unit is correct, the magnitude seems reasonable as the moles of NaOH > 10x the amount in 1 L.
Tro's Introductory Chemistry, Chapter 13

33. Tro's Introductory Chemistry, Chapter 13
Practice—Determine the Mass of CaCl2 (MM = ) in 1.75 L of 1.50 M Solution.
Tro's Introductory Chemistry, Chapter 13

34. If water is the universal solvent, then why does it not dissolve oil?
For a solution to form, the solvent and solute molecules must be attracted to each other
“Like Dissolves Like”
Polar solvents dissolve polar solutes
Nonpolar substances are attracted to other nonpolar substances, therefore nonpolar solvents will dissolve nonpolar solutes
Examples: I2, Hexane, Cooking Oils

36. 8.7 Dilution Dilution is the addition of solvent to decrease the
concentration of solute. The solution volume changes, but the amount of solute is constant.
M1V = M2V2
initial values
final values

37. 8.7 Dilution Sample Problem 8.11
What is the concentration of a solution formed by diluting 5.0 mL of a 3.2 M glucose solution to mL?
Step [1]
Identify the known quantities and the desired quantity.
M1 = 3.2 M
V1 = 5.0 mL V2 = 40.0 mL
known quantities
M2 = ?
desired quantity

38. 8.7 Dilution
Step [2]
Write the equation and isolate M2 on one side.

39. 8.7 Dilution Step [3] Solve the Problem.
Substitute the three known quantities into the equation and solve for M2.
M2 =
M1V1 V2 =
(3.2 M) (5.0 mL)
(40.0 mL)
mL unit
cancels
= M
2 sig. fig.
Answer

40. Tro's Introductory Chemistry, Chapter 13
Example—What Volume of 12.0 M KCl Is Needed to Make 5.00 L of 1.50 M KCl Solution?
Given:
Initial solution Final solution
Concentration M M
Volume ? L L
Find: L of initial KCl
Equation: (conc1)∙(vol1) = (conc2)∙(vol2)
Rearrange and apply equation:
Tro's Introductory Chemistry, Chapter 13

41. Making a Solution by Dilution
M1 x V1 = M2 x V2
M1 = 12.0 M V1 = ? L
M2 = 1.50 M V2 = 5.00 L
Dilute L of 12.0 M solution to 5.00 L.
Tro's Introductory Chemistry, Chapter 13

42. Example—Dilution Problems
What is the concentration of a solution made by diluting 15 mL of 5.0% sugar to 135 mL?
How would you prepare 200 mL of 0.25 M NaCl solution from a 2.0 M solution?
M1 = 5.0 % M2 = ? %
V1 = 15 mL V2 = 135 mL
(5.0%)(15 mL) = M2 x (135 mL)
M2 = 0.55%
M1 = 2.0 M M2 = 0.25 M
V1 = ? mL V2 = 200 mL
(2.0 M) x V1 = (0.25 M)(200 mL)
V1 = 25 mL
Dilute 25 mL of 2.0 M NaCl solution to 200 mL.
Tro's Introductory Chemistry, Chapter 13

43. Practice—Determine the Concentration of the Following Solutions.
Made by diluting 125 mL of 0.80 M HCl to 500 mL.
Made by adding 200 mL of water to 800 mL of 400 ppm.
Tro's Introductory Chemistry, Chapter 13

44. Tro's Introductory Chemistry, Chapter 13
Practice—Determine the Concentration of the Following Solutions, Continued.
Made by diluting 125 mL of 0.80 M HCl to 500 mL.
Made by adding 200 mL of water to 800 mL of 400 ppm.
M1 = 0.80 M M2 = ? M
V1 = 125 mL V2 = 500 mL
(0.80 M)(125 mL) = M2 x (500 mL)
M2 = 0.20 M
M1 = 400 ppm M2 = ? ppm
V1 = 800 mL V2 = mL
(400 PPM)(800 mL) = M2 x (1000 mL)
M2 = 320 PPM
Tro's Introductory Chemistry, Chapter 13

45. Tro's Introductory Chemistry, Chapter 13
Example—To What Volume Should You Dilute L of 15.0 M NaOH to Make 3.00 M NaOH?
Sort information.
Given:
Find:
V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M
V2, L
Strategize.
Solution Map:
Relationships:
M1V1 = M2V2
V1, M1, M2 V2
Follow the solution map to Solve the problem.
Solve:
Check.
Check:
Since the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does.
Tro's Introductory Chemistry, Chapter 13

46. 8.3 Solubility A. Basic Principles
Small neutral molecules with O or N atoms that can hydrogen bond to water are water soluble.

47. 8.3 Solubility A. Basic Principles
Nonpolar compounds are soluble in nonpolar solvents (like dissolves like).
Octane (C8H18) dissolves in CCl4 because both are nonpolar liquids that exhibit only London dispersion forces.
octane
CCl4
octane + CCl4

48. 8.3 Solubility B. Ionic Compounds—Additional Principles

49. Solubility
Solubility: the amount of solute that will dissolve in a specific solvent under given conditions
Polar solutes will be more soluble in polar solvents
Non-polar solutes will be more soluble in non-polar solvents
Amphiphilic solutes will be soluble in both polar and non-polar solvents
Have hydrophobic and hydrophilic regions
Example: Phospholipids

50. Page 584
Solubility: SOAP

51. Figure 17.3

52. Page 585(middle)

53. 8.9 Osmosis and Dialysis
The membrane that surrounds living cells is a semipermeable membrane.
Semipermeable membranes allow water and small molecules to pass across, but ions and large molecules cannot.
Osmosis is the passage of a solvent, usually water, across a semipermeable membrane from a solution of low solute concentration to a solution of higher solute concentration.

54. 8.9 Osmosis and Dialysis A. Osmotic Pressure
Osmotic pressure is the pressure that prevents the flow of additional solvent into a solution on one side of a semipermeable membrane.

55. Tro's Introductory Chemistry, Chapter 13
Osmotic Pressure
Solvent flows through a semipermeable membrane to make the
solution concentration equal on both sides of the membrane.
The pressure required to stop this process is osmotic pressure.
Tro's Introductory Chemistry, Chapter 13

56. 8.9 Osmosis and Dialysis B. Focus on the Human Body—Osmosis
Two solutions with the same osmotic pressure are
said to be isotonic.
Solutions isotonic to the body:
0.92% (w/v) NaCl solution
5.0% (w/v) glucose solution
isotonic
solution

57. 8.9 Osmosis and Dialysis B. Focus on the Human Body—Osmosis
A hypotonic solution has a lower osmotic pressure
than body fluids.
The concentration of particles outside the cell is lower than the concentration of particles inside the cell.
Water diffuses into the cell, so the cell swells and eventually bursts.
For red blood cells, this swelling and rupture is called hemolysis.
hypotonic
solution

58. 8.9 Osmosis and Dialysis B. Focus on the Human Body—Osmosis
A hypertonic solution has a higher osmotic pressure
than body fluids.
The concentration of particles outside the cell is higher than the concentration of particles inside the cell.
Water diffuses out of cell, so the cell shrinks.
This process is called crenation.
hypertonic
solution

60. 8.2 Electrolytes and Nonelectrolytes B. Equivalents
An equivalent (Eq) is the number of moles of charge that a mole of ions contributes to a solution.
The number of equivalents per mole of an ion equals the charge on the ion.
In solutions that contain ions, there must be a balance between total positive and total negative charge.