1. Solutions, Acids & Bases
Chapters 16 & 19

2. Matter Pure substances Mixtures Elements Compounds – chemically bonded
Ex. aluminum, gold, etc.
Compounds – chemically bonded
Ex. water, carbon dioxide
Mixtures
Homogeneous - mixed uniformly (solutions)
Ex. sweet tea
Heterogeneous – not mixed uniformly
Ex. chocolate chip cookies, salads

3. Solutions (Homogeneous Mixtures)
Solute: becomes dispersed in the solvent
Solvent: dissolves the solute
Aqueous Solutions: substance dissolved in water. (water = solvent)

4. Dissolving Ionic Compounds
1) Dissociation: The NaCl splits up into ions.
NaCl (s)  Na+ (aq) + Cl- (aq)

5. Dissolving Ionic Compounds
2) Hydration: Water surrounds the ions.

6. Dissolving Ionic Compounds
The Na+ is MORE attracted to the negative side of water (oxygen side) than it is to Cl-.
The Cl- is MORE attracted to the positive side of water (hydrogen side than it is to Na+.
**It’s all based on
FORCES OF
ATTRACTION!!

7. Insoluble Compounds
A compound is INSOLUBLE if the forces of attraction within the compound are GREATER than the forces of attraction to the solvent.

8. “Like Dissolves Like”
Dissolving is a physical process in which particles of a solute are held apart by particles of the solvent.
RULE: Like Dissolves Like
Nonpolar solvents (ex. oil) dissolve nonpolar compounds.
Polar solvents (ex. water) dissolve polar compounds and most ionic compounds (ex. NaCl).

9. Polarity Review To be NONPOLAR, a molecule must fit BOTH conditions:
No extra electron pairs on the central atom.
The same element all the way around the outside.

10. Electrolytes
Electrolyte: a compound that conducts an electric current when it is in an aqueous state.
Mobile ions are required for the conduction of electric current.
Ex. Ionic Compounds (salts), acids, and bases.
Nonelectrolytes: cannot conduct electricity
Ex. sugars, alcohols, and nonpolar covalent compounds

11. Solubility
Solubility: The amount of solute that dissolves in a given quantity of a solvent at a specified temperature and pressure.
Concentration of Solute in a Solution
Concentrated: Relatively more solute in the solution.
Dilute: Relatively less solute in the solution.
Example: A 0.02 M solution is more dilute than a 2 M solution.

12. Saturation
Unsaturated Solution: the amount of solute dissolved is less than the maximum that could be dissolved.
Ex. Earth’s oceans = unsaturated salt solution
Saturated Solution: the solution holds the maximum amount of solute.
Supersaturated Solution: contains more solute than the usual maximum amount and is unstable (may release solute suddenly).

13. Solubility Curves
A solution has 132 g of NaNO3 in 100 g of water at 75oC. Is the solution saturated, unsaturated, or supersaturated?
What about 95 g of KNO3 in 100g water at 50oC?

14. Solubility Curves
3) How many grams of potassium chloride would dissolve in 100 g of water at 90oC? 4) How much would dissolve in 200 g of water?

15. Factors Affecting Solubility
1) Temperature
Solids dissolving in liquids  solubility increases with increasing temperature
Gases dissolving in liquids  solubility decreases with increasing temperature
Carbonated soda

16. Factors Affecting Solubility
2) Pressure
Solids and Liquids  little effect
Gases  solubility increases under increased pressure. - ex. carbonated beverages

17. Calculating the Concentration of Solutions
Concentration is measured in molarity.
Molarity (M) = moles of solute
liters of solution
3M NaCl = “three molar solution of sodium chloride”
grams (use p.t.)
mL (1000mL = 1L)

18. Molarity Problems
1) What volume would have to be used to make a 3 molar solution with 5 moles of NaCl?
M = mol L
3M = 5 mol  (x L)(3M) = 5 mol  x L = 5mol
x L M
Volume = 1.67 L

19. More molarity problems…
2) If 20 grams of Mg(OH)2 was dissolved in 500 mL of water, what is the concentration?
M = mol L
20 g Mg(OH) mol Mg(OH)2 = mol Mg(OH)2
58.3 g Mg(OH)2
M = 0.34 mol = 0.68 M
0.5 L
500 mL L = L
1000 mL

21. Making Dilutions
When you dilute a solution, you increase the amount of solvent (the number of solute particles stays the same).
M1 x V1 = M2 x V2
M = molarity V = volume (in L!)

22. Dilutions problems
A chemistry teacher needs 6 liters of 2 M HCl for a lab. In the stock room, she found a bottle of 5 M HCl. How many milliliters would she use to dilute it? M1V1 = M2V2 (5M)(?L) = (2M)(6L) Volume = 2.4 L  2400 mL

24. Colligative Properties
Colligative Properties – depend only on HOW MUCH you have…not WHAT you have.
Vapor-Pressure Lowering
Osmotic Pressure Elevation
Freezing-Point Depression
Ex. Salt on the icy roads
Boiling-Point Elevation
(For aqueous solutions, BP will be higher than 100oC.)
Ex. Adding salt to cook spaghetti

25. Acids & Bases

26. Acids Sour taste Examples: Release H+
HCl (stomach acid)
Citric acid (lemon juice)
Release H+
(Note: H+ and H3O+ are interchangeable
H+ + H2O  H3O+ )
Electrolytes – because they produce charged particles

27. Acids (cont’d) Acid/Base Indicators: Turns litmus paper red.
Phenolphthalein is clear.

28. Acids continued… Ex. HNO3, HCl, HC2H3O2 Ex. H2SO4, H2SO3 Ex. H3PO4
H+ refers to a proton.
Monoprotic: 1 H+ to give away
Ex. HNO3, HCl, HC2H3O2
Diprotic: 2 H+
Ex. H2SO4, H2SO3
Triprotic: 3 H+
Ex. H3PO4

29. Bases Bitter taste, slippery feel Examples: NaOH, NH3 (ammonia), soap
Most release OH- (hydroxide ions) when dissolved in water.
Electrolytes – because they produce charged particles
Acid/Base Indicators:
Turns litmus paper blue.
Phenolphthalein turns bright pink.

30. 2 Definitions of Acids & Bases
1) Arrhenius Theory
Acids: produce H+ ions
Bases: produce OH- ions
Neutralization Reaction:
Acid + Base  Salt + Water

31. Neutralization Reaction
ACID + BASE  SALT + WATER
Salt: ionic compound formed from the negative part of the acid and the positive part of the base.
Examples:
  2HCl + Mg(OH)2 MgCl2 + 2H2O
H2CO3 + 2NaOH  Na2CO3 + 2H2O
What type of reaction is this? (Synthesis, decomposition, single replacement, double replacement, combustion)

32. Complete and Balance the Neutralization Reactions
HCl + NaOH 
HC2H3O2 + Ca(OH)2 
HBr + Al(OH)3 
NaCl + H2O
Ca(C2H3O2)2 + 2H2O
AlBr3 + 3H2O

33. Definitions (cont’d) 2) Brønsted-Lowry Theory
Acids: proton (H+) donors – ex. HCl
Bases: proton (H+) acceptors – ex. NH3
Acid + Base  Conjugate Base + Conjugate Acid
Ex. HCl + NH3  Cl- + NH4+
Acid Base CB CA

34. H+ When you add an H+ to a compound, the charge goes up.
Ex. HSO4- + H+  H2SO4
NH3 + H+  NH4+
When you take away an H+, the charge goes down.
Ex. HCl  H+ + Cl-
H3PO4  H+ + H2PO4-
H2PO4-  H+ + HPO42-

35. Conjugate Acids and Bases
Show the direction of H+ transfer.
Label: Acid, Base, Conjugate Base, Conjugate Acid
NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)
base acid conj. acid conj. base
acid base conj. acid conj. base

36. More Examples Show the direction of H+ transfer.
Label: Acid, Base, Conjugate Base, Conjugate Acid
H2SO4 + OH- HSO H2O
HSO41- + H2O SO H3O+
acid base conj. base conj. acid
acid base conj. base conj. acid

37. pH Scale
pH Scale: logarithmic scale in which [H3O+] is expressed as a number from 0 to 14.
0-6.9 (acid) (neutral) – 14 (base)
The numbers change by a factor of 10.
[OH-] = [H3O+] when pH = 7.

38. pH Scale

40. What does the pH scale mean?
 Neutral Solution: [H3O+] = [OH-]
A pH 11 solution has ____ times more hydroxide ions than a pH 10 solution.
A pH 11 solution has _____ times more hydroxide ions than a pH 9 solution.
A pH 3 solution has _____ times the number of hydronium ions compared to a pH 4 solution. 10
100
10, 100, 1/10 10

41. Strong Acids and Bases Strong Acids: completely ionize in water
Ex. HCl, HBr, HI, HNO3, H2SO4, HClO3
Strong Bases: completely dissociate into ions in water
All hydroxides bonded to an alkali or alkaline earth metal are strong.
Ex. NaOH, LiOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2

42. Weak Acids and Bases Weak Acids: only some molecules ionize in water
Ex: acetic acid (less than 0.5% of molecules ionize)
Weak Bases: do not completely dissociate into ions in water
Ex: ammonia (only 0.5% of molecules dissociate)

43. Concentrated vs. Strong
“Concentrated” – refers to the amount dissolved in solution.
“Strong” – refers to the fraction of molecules that ionize.  
For example, if you put a lot of ammonia into a little water, you will create a highly concentrated solution. However, since only 0.5% of ammonia molecules ionize in water, this basic solution will not be very strong.

44. Note  [ ] = concentration (molarity)

45. pH [H+] pOH [OH-] [H+] = 10-pH [OH-] = 10-pOH pH = -log[H+]
pH + pOH = 14
[OH-] = 10-pOH
pOH = -log[OH-]

46. Practice – Using the equations
Find the pH of the following solutions. Is the solution acidic or basic?
0.01 M HCl
0.050 M KOH
2.6 x M NH4OH
pH = -log(0.01) pH = acidic
pOH = -log(0.05) pOH = 1.3
pOH + pH = 14  pH = pH = 12.7
basic
pOH = -log(2.6 x 10-12) pOH = 11.6
pH =  pH = 2.4, acidic

47. More pH practice…
4) Find the concentration of hydrogen ions if the pH is 3.
[H3O+] = 10-pH [H3O+] = 10-3M
5) Find the concentration of hydroxide ions if the pH is 5.
pOH = 14 – pOH = 9
[OH-] = 10-pOH [OH-] = 10-9M
6) Find the [H3O+] in a solution if [OH-] = 10-6 M.
[H3O+][OH-] = [H3O+] = [H3O+] = 10-8M
10-6

48. Acid-Base Titration
Uses a neutralization reaction to determine the concentration of an acid or base.
Standard Solution: the reactant that has a known molarity
Endpoint: the point at which the unknown has been neutralized.

49. Titration Problems
Look for “titration” or “neutralization” in the problem.
Use this equation: MAVA = MBVB
nA nB
“A” means “acid”.
“B” means “base”.
M – molarity V – volume
n – moles (from the coefficient in the balanced reaction)

50. Titration Examples MAVA = MBVB nA nB MA(20.0mL) = (0.100M)(8.0 mL) 1 1
Example #1) 8.0 mL of 0.100M NaOH is used to neutralize 20.0 mL of HCl. What is the molarity of HCl?
NaOH + HCl  NaCl + H2O
MAVA = MBVB
nA nB
MA(20.0mL) = (0.100M)(8.0 mL)
= 0.04 M HCl

51. Titration Examples (cont’d)
Example #2) A 0.1M Mg(OH)2 solution was used to titrate an HBr solution of unknown concentration. At the endpoint, 21.0 mL of Mg(OH)2 solution had neutralized 10.0 mL of HBr. What is the molarity of the HBr solution?
Mg(OH)2 + 2HBr  MgBr2 + 2H2O
MA(10.0mL) = (0.1M)(21.0mL)
= 0.42 M HBr

52. Titration Practice = 0.073 M Al(OH)3
Example #3) What is the molarity of an Al(OH)3 solution if 30.0 mL of the solution is neutralized by 26.4 mL of a 0.25 M HBr solution?
Al(OH) HBr  AlBr H2O
(0.25M)(26.4mL) = MB(30.0mL)
= M Al(OH)3

53. Titration Curves
Strong Acid Weak Acid Strong Acid Weak Base Strong Base Strong Base pH pH pH 7 7 7
Reaction
Reaction
Reaction

54. Buffers
Buffer – A solution in which the pH remains relatively constant when small amounts of acid or base are added.
Titration Curve with a buffer added to the solution: pH 7
Reaction