1. Solutions
Edward Wen

2. Outline Composition of solution: solvent vs. solute
Factors affecting Solubility
Concentration of solution: Definition & Calculation
Mass%
Molarity*
Solution Stoichiometry

3. Solution Homogeneous mixtures
composition may vary from one sample to another
appears to be one substance, though contains multiple materials
most Homogeneous materials are actually Solutions
Gas state: common air
Liquid: Gasoline (dozens of compounds), Soda water (sugar or asparatame, CO2, citric acid, fructose)
Solid: Alloy such as brass

4. Solutions: Solute + Solvent
Solute: the dissolved substance.
Example: Sugar and CO2 gas in Soda
seems to “disappear”
“takes on the state” of the solvent
Solvent: the substance solute dissolves in.
Example: Water in Soda
does not appear to change state
Aqueous solutions: solutions in which the solvent is water.

5. Common Types of Solution
Solution Phase
Solute Phase
Solvent Phase
Example
Gaseous solutions
gas
air (mostly N2 & O2)
Liquid solutions
liquid
solid
soda (CO2 in H2O)
vodka (C2H5OH in H2O)
seawater (NaCl in H2O)
Solid solutions
brass (Zn in Cu)
Alloys: solutions that contain Metal solutes and a Metal solvent, such as Nickel (5 cents of 1$), Brass, Stainless steel

6. How Soluble? Solubility
Soluble: when one substance (solute) dissolves in another (solvent)  Homogeneous
Salt and Sugar are soluble in water: Saline and Soda
Acetic acid (HC2H3O2) in water: Vinegar
Oxygen gas in Nitrogen gas: Air
Insoluble: when one substance does not dissolve in another  Heterogeneous
Oil is insoluble in water: Italian salad dressing

7. Will It Dissolve? Like Dissolves Like Chemist’s Rule of Thumb –
a chemical will dissolve in a solvent if it has a similar structure to the solvent
when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other

8. Classifying Solvents: Polar vs. Nonpolar
Structural
Feature
Water, H2O
polar
O-H
Ethyl Alcohol, C2H5OH
Acetone, C3H6O
C=O
Toluene, C7H8
nonpolar
C-C & C-H
Hexane, C6H14

9. Solubility in Water, A Polar Solvent?
Ionic compound (Yes): Ions are attracted to polar Water. Salt NaCl dissolve in water
Polar molecules (Yes): attracted to polar solvents
table Sugar, Alcohol, glucose
Nonpolar molecules are NOT attracted to Water
b-carotene, (C40H56), is not water soluble; it dissolves in fatty (nonpolar) tissues
Those molecules with both polar and nonpolar structures: depends on structural features in the molecule

10. Salt Dissolved in Water

11. Solubility
Definition: the maximum amount of solute that can be dissolved in a given amount of solvent
Example, at room temperature, 100. g water can dissolve 200 g sugar. Solubility = 200 g sugar/100g water.
Usually a limit to the solubility of one substance in another
Exceptions: gases are always soluble in each other
two liquids that are mutually soluble are said to be miscible
alcohol and water are miscible

12. Descriptions of Solubility
Saturated solution has the maximum amount of solute that will dissolve in that solvent at that temperature. If more solute is added, it _______ dissolve.
Unsaturated solution is holding ______ solute than it is capable of. It can dissolve ______ solute.
Supersaturated solution is holding ____ solute than it is capable of at that temperature
Unstable
If more solute is added, ___________________.
Will not; less; more; more; more solute will separate from the solution

13. Adding Solute to various Solutions
___saturated
___saturated
_____saturated

14. Supersaturated Solution is unstable
A supersaturated solution has more dissolved solute than the solvent can hold, not stable. When disturbed, all the solute above the saturation level will separate.
Online demo: A grain of sodium acetate crystal is placed in the middle of supersatured sodium acetate solution.

15. Solubility of Solid Increases as Temperature increase
Application: To make pure crystals, make saturated solution at high temp
cooling leads to supersaturated solution, extra crystals forms at lower temp.

16. Solubility of Gases Decreases at higher Temperature
Observation 1: Warm soda pop fizzes more than cold soda pop
Cause: Solubility of CO2 in water is ________ (higher/lower) at high temperature.
Observation 2: When water is heated up, gas bubbles appear even before boiling occurs.
Cause: Solubility of air in water decreases as temperature increases.

17. Solubility of Gas depends on Pressure
higher pressure = higher solubility
CO2 is dissolved under Pressure into bottled/canned soda

18. Under which conditions does oxygen gas have the best solubility?
high temperature and high partial pressure (PO2)
high temperature and low partial pressure
low temperature and low partial pressure
low temperature and high partial pressure

19. Solution Concentration Descriptions
Diluted solutions have low solute concentrations. Soda drink from the soda fountain.
Concentrated solutions have high solute concentrations. Syrup in the storage tank in the soda fountain.

20. Concentrations – Quantitative Descriptions of Solutions
Solutions have variable composition. Salt vs. Water in Seawater
To describe a solution accurately, you need to describe the components and their relative amounts
Concentration = amount of solute in a given amount of solution
Seawater: Salt concentration 3.4%
Dead Sea: Salt concentration 30%
Vinegar: Acetic acid concentration 5%

21. Mass Percent (%) mass of solute (gram) in every 100 gram of solution
if a solution is 0.9% by mass, then there are 0.9 grams of solute in every 100 grams of solution
Mass of solution
= Mass of solute + Mass of solvent
Mass% = _________________ x _____

22. Information Given: 27.5 g C2H6O; 175 mL H2O Find: % by mass
Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C2H6O) and 175 mL of H2O (assume the density of H2O is 1.00 g/mL)
Information
Given: 27.5 g C2H6O; 175 mL H2O
Find: % by mass
Mass of Solvent = 175 g
Mass of Solution = g
Mass of Solute = 27.5 g
Mass% = 13.6

23. Using Concentrations as Conversion Factors
concentrations show the relationship between the amount of solute and the amount of solvent
12% by mass sugar(aq) means 12 g sugar  100 g solution
The concentration can then be used to convert the amount of solute into the amount of solution, or visa versa

24. Example: A soft drink contains 11. 5% sucrose (C12H22O11) by mass
Example: A soft drink contains 11.5% sucrose (C12H22O11) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL)
Information
Given: 85.2 g C12H22O11
Find: mL sol’n
V = 741 mL

25. Preparing a Solution
What we need to know: Amount of solution & Concentration of solution
Calculate the mass of solute needed
start with amount of solution
use concentration as a conversion factor
5% by mass solute Þ 5 g solute  100 g solution

26. Preparing a Solution by Mass%
Example - How would you prepare g of 5.00% by mass glucose solution (normal glucose)?
12.5 g of glucose
237.5 g of water

27. Solution Concentration: Molarity
Definition: Moles of solute per 1 liter of solution
Purpose: describing how many molecules of solute in each liter of solution
Unit: mole/L, abbreviated as “M”.
If a sugar solution concentration is 2.0 M , 1 liter of solution contains 2.0 moles of sugar.
molarity =
moles of solute
liters of solution

28. Why Molarity?
Many reagents used in chemistry, even many biology labs, are in the form of solution.
Molarity concentration of solution is particularly important and useful because:
Easy to use: To obtain given amount (mole) of reagent, just calculate the volume of solution to be used:
Easy to prepare a solution to a given molarity 28

29. Calculations involving Molarity
Molarity = mole  Volume (L)
Solve for mole:
Mole = ___________________
Solve for volume of solution in liters:
Volume (L) = ______________ 29

30. Example – How to prepare 500 mL of 0.020 M NaCl solution?
Part A. Calculate the mass NaCl needed:
0.010 mol NaCl
0.58 g NaCl 30

31. Example: How to prepare 500 mL of 0. 020 M NaCl solution. Part B
Example: How to prepare 500 mL of M NaCl solution? Part B. Preparation
Weigh out
0.58 g NaCl and
Add it to a 500mL
Volumetric flask.
Step 1
Step 2
Add water to
dissolve the
NaCl, then
add water to
the mark.
Step 3
Put the lid on,
Invert the flask
to homogenize
the solution

32. Example: Calculate the molarity of a solution made by putting 15
Example: Calculate the molarity of a solution made by putting g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution.
Information
Given: 15.5 g NaCl; 1.50 L sol’n
Find: molarity, M
CF: g = 1 mol NaCl;
0.265 mol NaCl
0.177 M NaCl

33. Example: How many liters of a 0. 114 M NaOH solution contains 1
Example: How many liters of a M NaOH solution contains 1.24 mol of NaOH?
Information
Given: 1.24 mol NaOH
Find: L solution
CF: mol = 1 L
SM: mol → L
10.9 L

34. Molarity of Ions: Dissociation of Ionic Compound
When strong electrolytes dissolve, virtually all the solute particles dissociate into ions
From the formula of the compound and the molarity of the solution
 Determine the Molarity of the dissociated Ions by simply multiplying the salt Concentration by the Number of Ions

35. Molarity & Dissociation
CaCl2(aq) = Ca2+(aq) + ____ Cl-(aq)
1 mole “molecules”
= ___ mole ions + ___ mole ions
1 M CaCl2
=____M Ca2+ ions + ____ M Cl- ions
0.25 M CaCl2
= _____ M Ca ____ M Cl-

36. Making a Solution by Dilution
Example: A student added 1.00 L water to 2.00 L 1.00 M HCl. The final volume became 3.00 L.
mole HCl before mixing = M1 x V1 = 2.00 mole
= mole HCl after mixing more water.
When mixing more solvent into a solution, the volume of final solution is greater than the original solution
The mole of solute remains the same before and after mixing more solvent
For dilution, mole solute
= M1 x V1 = M2 x V2

37. Example—What Volume of 12. 0 M KCl Is Needed to Make 5. 00 L of 1
Example—What Volume of 12.0 M KCl Is Needed to Make 5.00 L of 1.50 M KCl Solution?
Given: M1 = _____; V2 = ____ L, M2 = ____ M
Find: V1
Equation: M1 x V1 = M2 x V2
0.625 L

38. Example—What is the final concentration (molarity) if 10. 0 mL 12
Example—What is the final concentration (molarity) if 10.0 mL 12.0 M HCl is diluted to a final volume of 5.00 L?
Given: M1 = _____; V1 = ____ L, V2 = ____
Find: V1
Equation: M1 x V1 = M2 x V2 M

39. Solution Stoichiometry
Stoichiometry: Balanced chemical equation shows the mole ratio among reactants and products in a reaction.
2 H2(g) + O2(g) → 2 H2O(l)
Solution molarity  moles of solute and liters of solution
Measure the moles of a material in a reaction in solution by knowing its molarity and volume.

40. Example: How much M KI solution, in liters, is required to completely precipitate all the Pb2+ in L of M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
0.115 mol KI  1 L solution
0.225 mol Pb(NO3)2  1 L solution
2 mol KI  1 mol Pb(NO3)2
Solution Map:
L Pb(NO3)2 → mol Pb(NO3)2 →
mol KI → L KI
= L

41. Full solution for Examples

42. Example: Calculate the mass percent of a solution containing 27
Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C2H6O) and 175 mL of H2O (assume the density of H2O is 1.00 g/mL)
Information
Given: 27.5 g C2H6O; 175 mL H2O
Find: % by mass
Eq’n:
CF: 1.00 g H2O = 1 mL H2O
Design a Solution Map:
Mass Solute
&
Vol Solvent
Mass Percent
density
Mass Solution
Mass Solvent
Solution =
solute + solvent

43. Apply the Solution Maps
Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C2H6O) and 175 mL of H2O (assume the density of H2O is 1.00 g/mL)
Information
Given: 27.5 g C2H6O; 175 mL H2O
Find: % by mass
Eq’n:
CF: 1.00 g H2O = 1 mL H2O
SM: mass sol & vol solv → mass solv → mass sol’n → mass percent
Apply the Solution Maps
Mass of Solution = Mass C2H6O + Mass H2O
= 27.5 g C2H6O g H2O
= g

44. Apply the Solution Maps - Equation
Example: Calculate the mass percent of a solution containing 27.5 g of ethanol (C2H6O) and 175 mL of H2O (assume the density of H2O is 1.00 g/mL)
Information
Given: 27.5 g C2H6O; 175 mL H2O
Find: % by mass
Eq’n:
CF: 1.00 g H2O = 1 mL H2O
SM: mass sol & vol solv → mass solv → mass sol’n → mass percent
Apply the Solution Maps - Equation
= %
= 13.6%

45. Example: A soft drink contains 11. 5% sucrose (C12H22O11) by mass
Example: A soft drink contains 11.5% sucrose (C12H22O11) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL)
Information
Given: 85.2 g C12H22O11
Find: mL sol’n
CF: 11.5 g C12H22O11  100 g sol’n
1.00 g sol’n = 1 mL sol’n
Design a Solution Map:
Mass Solute
Volume Solution
mass
percent
density
Mass Solution

46. Apply the Solution Map = 740.87 mL = 741 mL
Example: A soft drink contains 11.5% sucrose (C12H22O11) by mass. What volume of soft drink in milliliters contains 85.2 g of sucrose? (assume the density is 1.00 g/mL)
Information
Given: 85.2 g C12H22O11
Find: mL sol’n
CF: 11.5 g C12H22O11  100 g sol’n
1.00 g sol’n = 1 mL sol’n
SM: g sucrose → g sol’n → mL sol’n
Apply the Solution Map
= mL
= 741 mL

47. Preparing a Solution by Mass%
Example - How would you prepare g of 5.00% by mass glucose solution (normal glucose)?
Dissolve 12.5 g of glucose in g of water to give the solution

48. Example: Calculate the molarity of a solution made by putting 15
Example: Calculate the molarity of a solution made by putting g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution.
Information
Given: 15.5 g NaCl; 1.50 L sol’n
Find: molarity, M
CF: g = 1 mol NaCl;
Design a Solution Map:
Mass Solute
Mole Solute
Molarity
Volume Solution
L Solution
already
liters

49. Example: Calculate the molarity of a solution made by putting 15
Example: Calculate the molarity of a solution made by putting g of NaCl into a beaker and adding water to make 1.50 L of NaCl solution.
Information
Given: 15.5 g NaCl; 1.50 L sol’n
Find: molarity, M
CF: g = 1 mol NaCl;
Apply the Solution Map
= M NaCl

50. Example: How many liters of a 0. 114 M NaOH solution contains 1
Example: How many liters of a M NaOH solution contains 1.24 mol of NaOH?
Information
Given: 1.24 mol NaOH
Find: L solution
CF: mol = 1 L
SM: mol → L
Apply the Solution Map

51. Find: volume of KI solution, L
Example: How much M KI solution, in liters, is required to completely precipitate all the Pb2+ in L of M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
Write down the given quantity and its units.
Given: L Pb(NO3)2
Write down the quantity to find and/or its units.
Find: volume of KI solution, L

52. Collect needed conversion factors:
Example: How much M KI solution, in liters, is required to completely precipitate all the Pb2+ in L of M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
Information:
Given: L Pb(NO3)2
Find: L KI
Collect needed conversion factors:
0.115 M KI  mol KI  1 L solution.
0.225 M Pb(NO3)2  mol Pb(NO3)2  1 L solution.
Chemical equation  2 mol KI  1 mol Pb(NO3)2.

53. Example: How much M KI solution, in liters, is required to completely precipitate all the Pb2+ in L of M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
Information:
Given: L Pb(NO3)2
Find: L KI
Conversion Factors:
0.115 mol KI  1 L solution
0.225 mol Pb(NO3)2  1 L solution
2 mol KI  1 mol Pb(NO3)2
Design a solution map:
L Pb(NO3)2
mol Pb(NO3)2
mol KI
L KI

54. Apply the solution map:
Example: How much M KI solution, in liters, is required to completely precipitate all the Pb2+ in L of M Pb(NO3)2? 2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
Information:
Given: L Pb(NO3)2
Find: L KI
Conversion Factors:
0.115 mol KI  1 L solution
0.225 mol Pb(NO3)2  1 L solution
2 mol KI  1 mol Pb(NO3)2
Solution Map:
L Pb(NO3)2 → mol Pb(NO3)2 →
mol KI → L KI
Apply the solution map:
= L
= L