1. Chapter 4 Reactions in Aqueous Solutions
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2. aqueous solutions of KMnO4
A solution is a homogenous mixture of 2 or more substances.
Solvent: the substance present in the larger amount.
Solute: the substance present in the smaller amount.
Solution
Solvent
Solute
aqueous solutions of KMnO4
Soft drink (l)
H2O
Sugar, CO2
Milk (l)
H2O
fat, protein, sugar
Air (g) N2
O2, Ar, CH4
Steel (s) Fe C

3. An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.
Nonelectrolyte
Strong electrolyte
Weak electrolyte

4. Why do they conduct electricity in solution?
Ionic Compounds dissociate (break apart) to form ions when dissolved in water
H2O
NaCl (s)
Na+ (aq) Cl- (aq)
Crystal Lattice
Ions can pass along (conduct) free electrons (electricity)

5. Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. +
Water is polar
High e- density near oxygen (-) - d+ d-
Low e- density near hydrogen (+)

6. Strong Electrolyte – 100% dissociation
CaCl2 (s) Ca+2 (aq) + 2Cl- (aq)
H2O
(NH4)2S (s) NH4+ (aq) + S-2 (aq)
H2O
Weak Electrolyte – do not completely dissociate
CH3COOH CH3COO- (aq) + H+ (aq)
~95%
~5%
A reversible reaction. The reaction occurs in both directions.
Acetic acid is a weak electrolyte because it only partially dissociates. 6

7. Nonelectrolytes do not conduct electricity
No cations (+) and anions (-) form in solution
C6H12O6 (s) C6H12O6 (aq)
H2O
(Molecules)

8. Precipitation Reactions
Precipitate: insoluble solid
2NaI (aq) + Pb(NO3)2 (aq)
Pb2+ + 2I PbI2 (s)
Na+ + 2I Pb NO3-
Na+
Na+
PbI2
Ions precipitate due to a stronger electrostatic force attraction than forces from H2O
Large charge (q) & Small ion size (r) lead to greater attraction and make them less soluble.

9. Examples of Insoluble Compounds
CdS
PbS
Ni(OH)2
Al(OH)3

10. Writing Precipitation Reaction Equations
Aqueous: soluble
Precipitate: insoluble solid
Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)
PbI2
Chemical equation
Pb2+ + 2NO3- + 2Na+ + 2I PbI2 (s) + 2Na+ + 2NO3-
Ionic equation
Pb2+ + 2I PbI2 (s)
Net Ionic equation
Na+ and NO3- are spectator ions (don’t react)

11. Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.
Most are single charged Smaller charges form weaker ionic bonds and are easier to separate
and NH4+ ions

12. DNews: Why The Government Puts Fluoride In Our Water
Soluble (aq) or Insoluble (s)
Classify the following ionic compounds as soluble or insoluble using the Solubility chart
silver sulfate (Ag2SO4)
calcium carbonate (CaCO3)
sodium phosphate (Na3PO4).
Barium Hydroxide (Ba(OH)2)
Calcium Fluoride (CaF2)
Ag2SO4 (s) is insoluble.
(b) This is a carbonate and Ca is a Group 2A metal. Therefore, CaCO3 (s) is insoluble.
(c) Sodium is an alkali metal (Group 1A) so Na3PO4 (aq) is soluble.
(d) Hydroxides are normally insoluble, but Barium is one of the exceptions along with all alkali metals making it soluble (aq).
(e) Neither Ca+2 nor F-1 are found listed as soluble, so it assumed the compound is insoluble (s)
DNews: Why The Government Puts Fluoride In Our Water

13. Writing Net Ionic Equations
Write the balanced molecular equation with phase states Swap the anions of reactants to get products. (re-criss-cross)
Al2(SO4)3(aq) + 6NaOH(aq)
*Do not carry the subscripts over (except for polyatomic ions)
Write the ionic equation showing the soluble product dissociated into ions, but the insoluble product left together.
2Al+3 + 3SO Na+ + 6OH- → 2Al(OH)3(s) + 6Na+ + 3SO4-2
3. Cancel the spectator ions on both sides of the ionic equation.
6Na+ + 3SO4-2
4. Check that charges and number of atoms are balanced
2Al+3 (aq) + 6OH-(aq) → 2Al(OH)3(s)
Al+3 (aq) + 3OH-(aq) → Al(OH)3(s)
→ 2Al(OH)3(s) + 3Na2SO4(aq)

14. #3 Precipitation Reactions
Predict what happens when potassium phosphate (K3PO4) solution is mixed with calcium nitrate [Ca(NO3)2] solution. Write the net ionic equation.
*Swap anions
*Swap anions
*Swap anions
K+ + NO3-
Ca2+ +
Balance the reactants and products to get the chemical equation
Use Solubility chart to determine (s) or (aq) for each

15. #3 Precipitation Solution
Step 2: Separate the soluble (aq) compounds to their ions for the ionic equation. Leave the insoluble (s) precipitate together
*Remember: Ions have charges, Show them (K ≠ K+)
Step 3: Cancel the spectator ions (K+ and NO3-1) on each side of the equation, we obtain the net ionic equation:

16. Writing Net Ionic Equations Practice
Pb(NO3)2 and LiCl
Al2(SO4)3 and NH4OH
MgSO4 and CoBr
Li2CO3 and Ca(NO3)2
Crash Course: Precipitation Reactions:
Argyria:
Silver ingestion
Beautiful Reactions: Precipitation

17. Chemistry In Action: An Undesirable Precipitation Reaction
Ca2+ (aq) + 2HCO3-1 (aq) CaCO3 (s) + CO2 (aq) + H2O (l)
Calcium Carbonate build up
The fix?
HCl reacts to form
soluble CaCl2
CaCO3(s) + 2HCl →
CaCl2(aq) + H2O + CO2

18. Properties of Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid.
Cause color changes in plant dyes.
React with many metals to produce hydrogen gas.
2HCl (aq) + Mg (s) MgCl2 (aq) + H2 (g)
React with carbonates and bicarbonates to produce carbon dioxide gas.
2HCl (aq) + CaCO3 (s) CaCl2 (aq) + CO2 (g) + H2O (l)
Aqueous acid solutions conduct electricity.

19. Properties of Bases Have a bitter taste.
Feel slippery. Many soaps contain bases.
Cause color changes in plant dyes.
Aqueous base solutions conduct electricity.
Examples:

20. Arrhenius acid is a substance that produces H+ (H3O+) in water.
Arrhenius base is a substance that produces OH- in water.

21. Arrhenius Acid/Base definition is exclusive to aqueous solutions
Hydronium ion, hydrated proton, H3O+
Greater electron
density O H H H
Lower electron density

22. A Brønsted acid is a proton (H+) donor
A Brønsted base is a proton acceptor
“proton” - the H+ ion consists of only a single proton
Bronsted definition is not exclusive to aqueous solutions
base
acid
acid
base
A Brønsted acid must contain at least one ionizable Hydrogen!
Ionizable – able to come off

23. Monoprotic acids (1 H+) Diprotic acids (2 H+) Triprotic acids (3 H+)
HCl H+ + Cl-
Strong electrolyte, strong acid
HNO H+ + NO3-
Strong electrolyte, strong acid
CH3COOH H+ + CH3COO-
Weak electrolyte, weak acid
Diprotic acids (2 H+)
H2SO H+ + HSO4-
Strong electrolyte, strong acid
HSO H+ + SO42-
Weak electrolyte, weak acid
Triprotic acids (3 H+)
H3PO H+ + H2PO4-
Weak electrolyte, weak acid
H2PO H+ + HPO42-
Weak electrolyte, weak acid
HPO H+ + PO43-
Weak electrolyte, weak acid

24. TedEd: The strengths and weaknesses of acids and base
Strong Acid (HCl)
Weak Acid (HF)
Complete
dissociation
Partial
dissociation
TedEd: The strengths and weaknesses of acids and base

25. Generally follow the same solubility rules for other ionic compounds.
Completely soluble would indicate strong acid (unless organic acid)
*Most acids tend to be weak with the short list of strong acids shown
Organic acids possess the Carboxylic acid functional group and are weak C H OH O
Carboxylic
Acid

26. 4.3 Classify each substance as a Brønsted acid or base in water: HBr
C2H5COOH
Glucose (C6H12O6)
Brønsted acid
Brønsted base
Acid
& Base
Propanoic Acid, Organic acids will display the ionizable H+ at the end (not grouped with other H’s)
Neither, Not an ionic compound nor is a H shown on the end

27. Neutralization Reaction
*Know the products
acid + base salt + water
HCl (aq) + NaOH (aq) NaCl (aq) + H2O
H+ + Cl- + Na+ + OH Na+ + Cl- + H2O
H+ + OH H2O
Net ionic equation
If we start with equal molar amounts of acid and base
we “neutralize” the solution and end up with water

28. Neutralization Reactions
Write molecular, ionic, and net ionic equations for each of the following acid-base reactions:
hydrobromic acid (aq) + barium hydroxide (aq)
(b) sulfuric acid (aq) + potassium hydroxide (aq)
2HBr(aq) + Ba(OH)2(aq) BaBr2(aq) + 2H2O(l)
H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)

29. Solutions Solution Molecular equation:
2HBr (aq) + Ba(OH)2(aq) → BaBr2(aq) + 2H2O(l)
Ionic equation:
2H+(aq) + 2Br−(aq) + Ba2+(aq) + 2OH−(aq)
→ Ba2+(aq) + 2Br−(aq) + 2H2O(l)
Net ionic equation:
2H+(aq) + 2OH−(aq) H2O(l) or
H+(aq) + OH−(aq) → H2O(l)
Both Ba2+ and Br− are spectator ions.

30. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)
Solution
(b) Molecular equation:
H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)
Ionic equation:
Net ionic equation:
Note that because HSO4- is a weak acid and does not ionize appreciably in water, the only spectator ion is K+.

31. Neutralization Reaction Producing a Gas
Acid + Carbonate salt + water + CO2
2HCl (aq) + Na2CO3 (aq) NaCl (aq) + H2CO3
Carbonic acid degrades quickly
2HCl (aq) + Na2CO3 (aq) NaCl (aq) + H2O + CO2
2H+ + 2Cl- + 2Na+ + CO Na+ + 2Cl- + H2O + CO2
2H+ + CO H2O + CO2
CaCO3 + 2HCl → NaCl + CO2 + H2O Antacid
*Gastric juices include HCl
TED-Ed: Chemistry of Cookies;

32. [ ] - Acid-Base Properties of Water neutral [H+] = [OH-] acidic
Water is a weak electrolyte (1 H+ in 107 H2O) O H + - [ ]
H2O (l) H+ (aq) + OH- (aq) Auto-ionization
Solution Is
neutral
[H+] = [OH-]
Adding acid increases H+
Adding base increases OH-
acidic
[H+] > [OH-]
basic
[H+] < [OH-] 32

33. pH – A Measure of Acidity
pH describes the concentration of [H+] on a logarithmic scale.
Remember: A log scale is base 10:
pH 6 is 10x more acidic than pH 7
pH 5 is 100x more acidic than pH 7
pH 4 is 103x more acidic than pH 7
pH = -log [H+]
Solution Is
neutral
[H+] = [OH-]
[H+] = 1.0 x 10-7
pH = 7
acidic
[H+] > [OH-]
[H+] > 1.0 x 10-7
pH < 7
basic
[H+] < [OH-]
[H+] < 1.0 x 10-7
pH > 7
Practice: Compare the acidity of pH 8 to pH 10.

34. pH Meter
H+ is an electrolyte
pH meters measure electrical conductance to determine concentration and then calculate pH

35. The indicator changes color as pH changes
Acid-Base reactions are colorless and can be difficult to monitor.
pH Indicators – substances that change colors at certain pH ranges and can be used for titrations
The indicator changes color as pH changes

36. Phenolphthalein
Colorless in acid
Pink in base

37. Chemistry In Action: Antacids and the Stomach pH Balance
NaHCO3 (aq) + HCl (aq)
NaCl (aq) + H2O (l) + CO2 (g)
Mg(OH)2 (s) + 2HCl (aq)
MgCl2 (aq) + 2H2O (l)
Acid-Base Reactions in Solution: Crash Course Chemistry #8

38. Reduction-Oxidation Reactions (Redox)
(electron transfer reactions)
2Mg + O MgO
2Mg Mg2+ + 4e-
Oxidation half-reaction (lose e-)
O2 + 4e O2-
Reduction half-reaction (gain e-)
2Mg + O2 + 4e Mg2+ + 2O2- + 4e-

39. Redox Reaction reactants
Its charge is
“reduced” b/c it
gained an e-

40. Neutral Zn is insoluble, but oxidized Zn+2 is soluble
Neutral metals are insoluble,
Ionized metals are soluble.

41. Write the Half-reactions of the given redox reaction.+2 -2
Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
Zn(s) Zn2+ + 2e-
Zn is oxidized, lost electrons
Note: Lost e- shown in products
Zn is the reducing agent
Cu2+ + 2e Cu(s)
Cu2+ is reduced, gained electrons
Note: Gained e- shown in reactants
Cu2+ is the oxidizing agent
Note: Sulfate (SO4-2) is neither oxidized or reduced so it is not shown
It could be said that:“Zn reduced Cu” or “Cu oxidized Zn”
Additional practice: 2Na(s) + Mg(NO3)2(aq) → Mg(s) + 2NaNO3(aq)
4Fe + 3O2 → 2Fe2O3

42. 2HCl(aq) + Al(s) AlCl3(aq) + H2(g)
Acid corrosion of metals are redox reactions
2HCl(aq) + Al(s) AlCl3(aq) + H2(g)
Al(s) Al3+(aq) + 3e-
Al is oxidized, lost electrons
2H+(aq) + 2e H2(g)
H+ is reduced, gained electrons
Neutral metals are insoluble as H2O doesn’t favor interacting with them.
Acids are able to oxidize metals to give them a charge. Once charged, water is able to dissolve them.

43. Oxidation numbers (ON’s)
Describes the charge of an atom, in a compound, if electrons were completely transferred between atoms (Both ionic and molecular compounds).
Used to predict how an atom will react with other atoms.
For ionic compounds, ON’s are equal to the ion charge as determined by each atom’s group number
For molecules, ON’s change depending on what atoms they bond to and follow a set of rules.
Ex. Cl has an ON of -1 when bonded to a metal (NaCl), but not with non-metals. (Chlorate (ClO3-), Cl ON = +5)
*Note: An ON of +5 is not equal to a true ion of charge +5 43

44. Oxidation numbers (ON’s)
1. Free elements (uncombined state) and non-compounds have an oxidation number of Zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. The oxidation number is equal to the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually -2.
(In H2O2 and O22- it is –1).
*you will not be given any of these exceptions on quizzes or tests.

45. *Unless bonded to metals in binary compounds.
4. Hydrogen: ON is +1
*Unless bonded to metals in binary compounds.
In these cases –1 (LiH, NaH, etc.).
*you will not be given any of these exceptions on quizzes or tests.
5. Alkali metals are +1
Alkaline Earth Metals are +2
Fluorine is always -1. (Other Halogens can vary)
6. Sum of all ONs of each atoms in a compound/molecule (or ion) is equal to the net charge.
CaBr2
Charge = 0
SO4-2
Charge = -2 +2 -1 +6 -2
(+21) + (-12) = 0
(+61) + (-24) = -2

46. Oxidation numbers of Transition metals
Alkali metals always have an oxidation number of +1
Alkaline Earth metals are always +2
Transition metals (d group) can have several possible oxidation states that are stable
Fe+3 can form FeCl Cu+2 can form CuCO3
Fe+2 can form FeCl Cu+1 can form Cu2CO3
The oxidation state may change during a reaction.
K2Cr2O7 is orange with Cr+6
Cr2(SO4)3 is blue with Cr+3

47. Redox Chemistry in Action: Breath Analyzer+6
3CH3CH2OH + 2K2Cr2O7 + 8H2SO4
Ethanol +3
3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2O
Acetic acid 47

48. Assign oxidation numbers to all the elements in the following compounds and ion:
Li2O
HNO3
SF4
HSO4-1
Ca3(PO4)2 +1 -2
Oxygen is always -2 and Li is +1 in a compound
(+1)•2 + (-2)•1 = 0 +1 +5 -2
Oxygen is always -2 and H is +1 in a compound.
(+1)•1 + (N) •1 + (-2)•3 = 0 ; N must = +5 +4 -1
Fluorine is always -1 and the sum must equal 0
(S)•1 + (-1)•4 = 0 ; S must = +4 +1 +6 -2
Oxygen is always -2 and H is +1 in a compound.
(+1)•1 + (S) •1 + (-2)•4 = -1 ; S must = +6 +2 +5 -2
Oxygen is always -2 and Ca is +2 in a compound.
(+2)•3 + (P) •2 + (-2)•8 = 0 ; P must = +5

49. The Oxidation Numbers of Elements in their Compounds
Red # signifies most common
oxidation state in compound 49

50. Types of Oxidation-Reduction Reactions
Combination Reaction
A + B C +3 -1
2Al + 3Br AlBr3
Decomposition Reaction
C A + B +1 +5 -2 +1 -1
2KClO KCl + 3O2

51. Types of Oxidation-Reduction Reactions
Combustion Reaction
A + O B +4 -2
S + O SO2
Derailed sulfur train video +2 -2
2Mg + O MgO

52. Types of Oxidation-Reduction Reactions
Displacement Reaction
A + BC AC + B +1 +2
Sr + 2H2O Sr(OH)2 + H2
Hydrogen Displacement +1 +2
2AgNO3 + Zn Ag + Zn(NO3)2
Metal Displacement -1 -1
Cl2 + 2KBr KCl + Br2
Halogen Displacement
How do we know what can displace what? Activity Series
Beautiful chemistry:
Metal displacement;

53. The Activity Series for Metals
Shows the tendency for an atom to give away its electrons
(reduce) another atom (which ones is on top reduces the other).
Hydrogen Displacement Reaction
Zn + 2HCl ZnCl2 + 2H2
Al has greater
reduction potential
Ca + 2H2O Ca(OH)2 + H2
Ag + H2O AgOH + H2
Metal Displacement Reaction
Al + FeCl AlCl3 + Fe
Fe + AlCl FeCl3 + Al

54. The Activity Series for Halogens
F2 > Cl2 > Br2 > I2
I2 can reduce F2, but F2 can’t reduce I2
Halogen Displacement Reaction -1 -1
Cl2 + 2KBr KCl + Br2
I2 + 2KBr KI + Br2 54

55. Types of Oxidation-Reduction Reactions
Disproportionation Reaction
The same element is simultaneously oxidized and reduced.
Example:
reduced +1 -1
Cl2 + 2OH ClO- + Cl- + H2O
oxidized
Elements most likely to disproportionate: each have at least 3 oxidation states

56. Redox Reactions: Crash Course Chemistry #10
Redox Reactions Problems
Classify the following redox reactions and indicate changes in the oxidation numbers of the elements:
(a)
(b)
(c)
(d)
(e) CH4(g) + O2(g) → CO2(g) + H2O(l)
Redox Reactions: Crash Course Chemistry #10

57. Redox Solutions
This is a decomposition reaction because one reactant is converted to two different products. The oxidation number of N changes from +1 to 0, while that of O changes from −2 to 0.
This is a combination reaction (two reactants form a single product). The oxidation number of Li changes from 0 to +1 while that of N changes from 0 to −3.
(c) This is a metal displacement reaction. The Ni metal replaces (reduces) the Pb2+ ion. The oxidation number of Ni increases from 0 to +2 while that of Pb decreases from +2 to 0.
(d) The oxidation number of N is +4 in NO2 and it is +3 in HNO2 and +5 in HNO3. Because the oxidation number of the same element both increases and decreases, this is a disproportionation reaction.
(e) Combination and combustion

58. Summary of Reactions Types
Precipitation: Two aqueous solutions mixing to forming a solid product.
Na2CO3(aq) + Ca(NO3)2(aq) → 2NaNO3(aq) + CaCO3(s)
Acid-Base Reactions: Transfer of H+ and/or OH- ions in H2O solution.
Neutralization: Reaction of Acid & Base to form a salt and water
2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O(l)
Acid + Metal: Oxidize to dissolve metal and produce Hydrogen gas
2HCl Mg → MgCl2 + H2
2H+(aq) + Mg(s) → Mg+2(aq) + H2(g) (net ionic equation)
Redox: Reaction involving the transfer of electrons between atoms
Combination: +3 -2
4Fe + 3O2 → 2Fe2O3
Combustion: +2 +2 +4 -2 -2 +2 -2 +4 -2
Decomposition: CaCO3 → CaO + CO2 +1 -1 +1 -1
Displacement: Na(s) + AgCl(aq) → NaCl(aq) + Ag(s)

59. Collision Theory of Chemical Reactions
The rate of a reaction increases as the number of molecular collisions increases.
Ways to increase collisions:
Increase dynamic interactions (Liquid > Gas > Solid)
Increase Concentrations (closer proximity)
Increase Temperatures (faster movements)
Any molecule in motion possesses Kinetic
energy, the faster it moves, the greater KE
High energy collisions are needed to break original bonds (Ea :Activation energy)
Ted-Ed: How to speed up chemical reactions (and get a date)

60. Disposing of excess WWII Sodium metal in 1947
2Na + 2H2O → 2NaOH + H2(g)

61. Solution Stoichiometry moles of solute M = molarity =
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution 61

62. Measures of Concentration
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
Molarity (M)
M =
moles of solute
liters of solution
Includes solute volume
Excludes solute mass
Because density (volume) can change with temperature it is helpful to express solvent by mass when sample undergoes temperature changes
Molality (m)
m =
moles of solute
mass of solvent (kg)
No volumetric measurements needed; all mass

63. moles = Liters x Molarity moles M = liters moles L = Molarity (grams)
1,000 mL = 1 L
1,000 g = 1 kg
(mL)
(grams)
L =
moles
Molarity

64. Molarity/Molality Problem
Calculate the Molarity and Molality of a H2SO4 solution containing 24.4 g of sulfuric acid in 198 g of water at 70°C to produce a 204 mL solution.
M =
moles
L solution
m =
moles
kg solvent
24.4 g H2SO4
1 mol H2SO4
= mol H2SO4
98.1 g H2SO4
M =
0.249 mol
0.204 L
m =
0.249 mol
0.198 kg
= 1.22 Molar H2SO4
= 1.26 molal H2SO4

65. K2CO3(s) → 2K+(aq) + CO3-2(aq)
Ion Molarity Problem
What is the Molar concentration of the Potassium ion [K+] when 9.85 g of K2CO3 are dissolved in a 250 mL solution?
9.85 g K2CO3
1 mol K2CO3
2 mol K+
138.2 g K2CO3
1 mol K2CO3
K2CO3(s) → 2K+(aq) + CO3-2(aq)
= mol K+
M =
0.143 mol
0.250 L
= 0.57 M K+
Practice: Molarity of [NO3-1] in 2.5 grams Mg(NO3)2 in 1.4 L solution.

66. Preparing a Solution of Known Concentration from solids
Volumetric Flask
Mix till dissolved
Bring to
desired volume

67. (doubles from subscript)
Molarity Problem
What is the Molar concentration of the Sodium ion [Na+] when 23.4 g of NaCl and 34.1 g of Na2O are dissolved in 0.60 L H2O?
23.4 g NaCl 1 mol
= 0.40 mol NaCl
= 0.40 mol Na+
58.5 g NaCl
34.1 g Na2O 1 mol Na2O
2 mol Na+
= 1.08 mol Na+
62.8 g Na2O
1 mol Na2O
(doubles from subscript)
0.40 mol Na mol Na+ = 1.48 mol Na+
M =
1.48 mol
0.60 L
= 2.5 M Na+

68. Practice: Grams of C3H8O3 in 250 mL solution for 0.75 M ?
Molarity Problem
How many grams of potassium dichromate (K2Cr2O7) are required to prepare a 250-mL solution whose concentration is 2.16 M?
M =
mol L
mol = ML
2.16 M 0.250 L = mol K2Cr2O7
0.540 mol
294.2 g K2Cr2O7
= 159 g K2Cr2O7
1 mol K2Cr2O7
Practice: Grams of C3H8O3 in 250 mL solution for 0.75 M ?

69. Number of moles does not change
Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution (stock).
Dilution
Add Solvent
Moles of solute
before dilution (1)
after dilution (2) =
M1V1
M2V2 =
Number of moles does not change 69

70. *A description required to answer the question
Dilution Practice
Describe how you would prepare 250 mL of a 1.75 M H2SO4 solution, starting with an 8.6 M stock solution of H2SO4.
Keep in mind that in dilution, the concentration of the solution decreases but the number of moles of the solute remains the same.
Stock
New soln
M1 = M2 =1.75
V1 = ? V2 = 250 mL
M1V1 = M2V2
(8.6 M V1) = (1.75 M 250 mL)
V1 = 50.9 mL
Dilute 50.9 mL of the 8.6 M H2SO4 stock with water to give a final volume of 250 mL.
*A description required to answer the question

71. Dilution Problem
Describe how you would prepare 300 mL of a 0.4 M C6H12O6 solution, starting with an 2.5 M stock solution of C6H12O6.
M1V1 = M2V2
(2.5 M V1) = (0.4 M 300 mL)
V1 = 48 mL
Thus, we must dilute 48 mL of the 2.5 M solution with water to give a final volume of 300 mL

72. Crash Course: Water and Solutions for Dirty Laundry
More Practice
You have 250 mL of a 3.0 M Ba(OH)2 solution. What is the concentration if we add 150 mL of water to the solution?
M1 = 3.0 M
V1 = 250 mL
M2 = ?
V2 = 250 mL mL = 400 mL
(3.0 M 250 mL) = (M2 400 mL)
M2 = 1.9 M Ba(OH)2
Crash Course: Water and Solutions for Dirty Laundry

73. Bell Ringer
1) How many grams of solid NaNO3 are needed to produce 125 mL of a 0.85 M NaNO3 solution?
2) What is the Molar concentration of the Sodium ion [Na+] when 2.8 g of Na3PO4 and 4.5 g of Na2CO3 are dissolved in 85 mL?
3) How would you prepare 250 mL of a 0.65 M H2SO4 solution from a stock of 6.5 M H2SO4 solution?
4) You have 50 mL of 6.0 M NaF and 450 mL water are added, what is the new Molarity?

74. Titrations
In a titration, a solution of accurately known concentration is added gradually to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Standard solution – solution with known concentration to be precisely added for comparison
Equivalence (End) point – the point at which the reaction is complete
example) 1 mol H2SO4  2 mol NaOH
(obtained from balanced equation) 74

75. Titrations Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color

76. Titrations can be used in the analysis of: Acid-base reactions
Redox reactions
H2SO4 + 2NaOH H2O + Na2SO4
5Fe2+ + MnO4- + 8H+
Mn Fe3+ + 4H2O
*Can involve color changes without indicator 76

77. Titration Steps MsVs = MuVu
Write Balanced equation for stoichiometry
(mole-to-mole ratio)
2) Determine moles of Standard solution used.
(mol = M x L)
3) With stoichiometry, convert molesstandard to molesunknown
(train tracks)
4) Determine unknown Molarity using given volume (L)
(M = mol/L)
MsVs = MuVu
Alternative Equation:
Coefficient #
Coefficient #

78. Titration Problem #1 1) 1Ba(OH)2 + 2HCl → BaCl2 + 2H2O (Reacts 1:2)
It takes 32 mL of 2.0M HCl standard to neutralize a 500. mL solution of Ba(OH)2. What is the concentration of Ba(OH)2?
1) 1Ba(OH)2 + 2HCl → BaCl2 + 2H2O
(Reacts 1:2)
2) mol HCl = 2.0M x L = mol HCl
3) mol HCl 1 mol Ba(OH)2
= mol Ba(OH)2
2 mol HCl
4) M Ba(OH)2 = moles
0.500 L
= M Ba(OH)2
Alternative:
MHVH = MOHVOH
2.032 = MOH500
Coefficient # 2 1
Coefficient #
MOH = 0.064

79. Titration: Finding the Molar Mass of an Unknown
Lauric Acid is a short-chain fatty acid that is solid at room temperature and monoprotic. We dissolve grams into 500. mL of water and then titrate it with M NaOH.
If it takes 11.0 mL of NaOH to neutralize the fatty acid, what is the Molar Mass of Lauric Acid?
Molar mass has the units grams/mole. We weighed out the mass of the solid acid in grams. Titration can tell us how many moles of acid are present in the same sample.
0.022 grams
= ? Molar mass
Moles H+
Because it is monoprotic, it will react 1:1 with NaOH.

80. Large component of coconut oil
Molar Mass Titration Solution
1) Given information states it reacts 1:1
2) (0.010 M NaOH) x ( L) = 1.1 x 10-4 mol NaOH
3) 1.1 x 10-4 mol NaOH 1 mol Lauric acid
= 1.1 x 10-4 mol Lauric Acid
1 mol NaOH
0.022 grams
4) Molar Mass =
= 200 g/mol
1.1 x 10-4 moles
Large component of coconut oil
~ 3-6% of milk
C11H23COOH

81. Example: Redox Titration
A mL volume of M KMnO4 solution is needed to oxidize mL of a FeSO4 solution in an acidic medium.
What is the concentration of the FeSO4 solution in molarity?
The net ionic equation is
need to find
want to calculate
given

82. Redox Titration Solution
Solution The number of moles of KMnO4 (in mL) = M x L
From the net ionic equation we see that 5 mol Fe2+  1 mol MnO4-
( M KMnO4) x ( L) = x 10-3 mol
M =
moles of solute
liters of solution =
1.090 x 10-2 mol
0.025 L
= M FeSO4

83. Titration Bell Ringer
50 mL of Ba(OH)2 is measured out into an Erlenmeyer flask. The concentration is unknown.
It takes 8.5 mL of 2.5 M H3PO4 standard to reach the equivalence point and neutralize the unknown base.
Determine the concentration of Ba(OH)2 solution.

84. Chemistry in Action: Metals from the Sea
CaCO3 (s) CaO (s) + CO2 (g)
Many metals are found
in the earth’s crust,
but it is cheaper to
“mine” from the sea
Precipitation
CaO (s) + H2O (l) Ca2+ (aq) + 2OH- (aq)
Slightly soluble
Mg2+ (aq) + 2OH- (aq) Mg(OH)2 (s)
Precipitation
Mg(OH)2 (s) + 2HCl (aq) MgCl2 (aq) + 2H2O (l)
1.3 g of Magnesium/
Kg seawater
Electrolysis of MgCl2 (redox)
Mg2+ + 2e Mg
2Cl Cl2 + 2e-
MgCl2 (aq) Mg (s) + Cl2 (g) 86

85. Gravimetric Analysis
Analytical technique based on the measurement of mass
Precipitation: the analyte is precipitated out of solution by adding another reagent to make it insoluble. Then filtered and weighed.
Volatilization: the analyte is converted to a gas and removed. The loss of mass from the starting material indicates the mass of gas. 87

86. Precipitation Gravimetric Analysis Steps
Dissolve unknown substance in water (if not already dissolved)
React unknown with precipitating reagent to form a solid precipitate
Reagent: chemical added to another substance to bring about a change
Filter, dry, and weigh precipitate.
Use chemical formula and mass of precipitate to determine amount of unknown analyte (chemical of interest in experiment) 88

87. *Alternate Method: Finding the % Mass of Ba+2 can also be used
Gravimetric Analysis Problem #1
A sample of an unknown soluble compound contains Ba+2 and is dissolved in water and treated with excess sodium phosphate.
If g of Barium phosphate precipitates out of solution, what mass of Barium in found in the unknown compound?
0.411 g Ba3(PO4)2
1 mol
3 mol Ba+2
137.3 g Ba+2
= g Ba+2
601.9 g
1 mol Ba3(PO4)2
1 mol Ba+2
*Alternate Method: Finding the % Mass of Ba+2 can also be used
3 x g Ba+2
601.9 g Ba3(PO4)2
x 0.411g Ba3(PO4)2 = g Ba+2

88. Gravimetric Analysis Problem #2
We have 250 mL of a Copper (Cu+1) solution. We add excess Na2CO3 to precipitate out 3.8 g of Cu2CO3. What is the [Cu+1] Molarity of the original solution?
Since we need to find moles of Cu+1 , it will be quicker to use train-tracks instead of % composition (either would work).
3.8 g Cu2CO3
1 mol Cu2CO3
2 mol Cu+1
= mole Cu+1
187.0 g Cu2CO3
1 mol Cu2CO3
0.041
mol Cu+
M =
0.250 L
= 0.16 M Cu+1

89. Gravimetric Analysis Problem #3
A g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3.
If g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound?
35.45 g Cl-
143.4 g AgCl
x g AgCl = grams Cl

90. More Gravimetric Review Problems
1) An unknown ionic compound contains Carbonate (CO3-2).
To precipitate the carbonate, we add excess CaCl2 and collect 25.3 grams of CaCO3 precipitate. Calculate mass of Carbonate present in the original compound.
2) 50 mL of a solution contains an unknown amount of Ni+ ions. We add excess Na3PO4 to precipitate out 5.67 grams of Ni3PO4. What is the Molarity of Ni+?
3) 340 mL of an unknown solution contains the Silver ion (Ag+). When excess Na2S is added, 15.8 grams of precipitated Ag2S are formed. What is the Molarity of Ag+ in the solution?

91. H2O reactions Exam Review Problems
Write the net ionic equation for: (NH4)2S and MgSO4
Write the labeled half reactions: 2N2O → 2N2 + O2
How many grams NH4NO3 needed for 250 mL of 3.5M?
Describe preparation of 300 mL 0.5M HCl (5M stock)
120 mL of H3PO4 is titrated with 35 mL 2.4 M standard KOH, what is concentration of H3PO4?
340 mL of an unknown solution contains the Silver ion (Ag+). When excess Na2S is added, 15.8 grams of precipitated Ag2S are formed. What is the Molarity of Ag+ in the solution?

92. Aqueous Reactions & Calculations Terms
Acid
Base
Neutralization Reaction
Redox Reaction
Precipitate
Electrolyte
Solvent
Solute
Ionic Equation
Net-Ionic Equation
Weak Electrolyte
Acid and Metals
Spectator Ions
Indicator
Equivalence point
Titration
Standard
Gravimetric Analysis
Molality
Molarity
Dilution
Analyte
Precipitating reagent
Concentration
Stock