1. 1411 Chapter 4 Aqueous Reactions and Solution Stoichiometry

2. Solutions
Solutions are defined as homogeneous mixtures of two or more pure substances.
Aqueous solution – solution in which water is the dissolving medium
The solvent is present in greatest abundance.
All other substances are solutes; they are dissolved in the solvent.
Example: NaCl dissolved in water:
NaCl = solute
water = solvent
Water can dissolve many ionic or molecular compounds – “universal solvent”

3. Dissociation
When an ionic substance dissolves in water, the solvent separates the individual ions from the crystal.
This process is called dissociation.

4. Dissociation
This occurs because water is POLAR – it has poles: a partial positive (δ+) end and a partial negative (δ-) end δ- O H H δ+ δ+ O H H H H H
Solvation – the surrounding of ions by H2O molecules to help stabilize the ions
Keeps the ions from recombining
NaCl(aq)  Na+(aq) + Cl-(aq) O H H H O
Cl- O O
Na+ O H H H H H H O O H H

5. Electrolytes
An electrolyte is a substance that dissociates into ions when dissolved in water.
May be strong or weak
Strong electrolyte - completely dissociates into ions
Solutions conduct electricity well
Ex: HCl(aq) H+(aq) Cl-(aq)
Weak electrolyte – partially dissociates into ions, but some molecules remain intact
Solutions conduct electricity poorly
Ex: CH3COOH(aq) CH3COO-(aq) H+(aq)

6. Electrolytes
A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.
Solutions do not conduct electricity

7. Identifying Strong & Weak Electrolytes & Nonelectrolytes
Strong electrolytes
Weak electrolytes
Nonelectrolytes
1) Strong acids
Weak acids
Molecular compounds, except acids and bases
2) Strong bases
Weak bases
3) Ionic compounds
These are on the list below
Any other acid or base NOT on the list below
Memorize!

8. Solution Chemistry
Pay attention to exactly what species are present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution).
If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction… the driving force!
The driving force is what makes the reaction react (Ex: formation of a precipitate, gas, or liquid)
Without a driving force, the reaction won’t occur 8

9. Precipitation Reactions
When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

10. Solubility Rules
Solubility of a substance is the amount that can be dissolved in a given quantity of solvent at a given temperature.
Insoluble – the attraction between the ions in the solid is too great to separate the ions to any significant extent; substance doesn’t dissolve
Memorize!
Left off
***All compounds of alkali metals (Group 1A) and of NH4+ ion are soluble ***

11. Solubility Rules Classify the following as soluble or insoluble:
c) CaCl2
d) SrSO4
a) CuCO3
e) (NH4)3PO4
b) Mg(OH)2
f) Fe2S3

12. Electrolytic Properties ≠ Solubility
Do not confuse extent of solubility with strong or weak electrolyte
Examples:
Acetic acid (CH3COOH or HC2H3O2): very soluble in water but weak electrolyte because most remains in form of molecule, not ions, in solution
BaCl2: not very soluble in water but strong electrolyte because what does dissolve dissociates completely

13. Metathesis (Exchange) Reactions
Metathesis reactions: ions in the reactant compounds exchange ions.
They “exchange partners”
AB + CD  AD + CB
AgNO3 (aq) + KCl (aq) 
AgCl (s) + KNO3 (aq)

14. Metathesis (Exchange) Reactions
To complete and balance a metathesis equation:
1.) Use the chemical formulas of the reactants to determine the ions that are present
2.) Write the product formulas by combining the cation of one reactant with the anion of the other. (Remember to use the charges to come up with subscripts!!)
3.) Balance the equation
Mg(NO3) Li3PO4 

15. Metathesis (Exchange) Reactions
To complete and balance a metathesis equation:
1.) Use the chemical formulas of the reactants to determine the ions that are present
2.) Write the product formulas by combining the cation of one reactant with the anion of the other. (Remember to use the charges to come up with subscripts!!)
3.) Balance the equation
Mg(NO3) Li3PO4 
Mg NO3- Li+ PO43-

16. Metathesis (Exchange) Reactions
To complete and balance a metathesis equation:
1.) Use the chemical formulas of the reactants to determine the ions that are present
2.) Write the product formulas by combining the cation of one reactant with the anion of the other. (Remember to use the charges to come up with subscripts!!)
3.) Balance the equation
Mg(NO3) Li3PO4  Mg3(PO4) LiNO3
Mg NO3- Li+ PO43-

17. Metathesis (Exchange) Reactions
To complete and balance a metathesis equation:
1.) Use the chemical formulas of the reactants to determine the ions that are present
2.) Write the product formulas by combining the cation of one reactant with the anion of the other. (Remember to use the charges to come up with subscripts!!)
3.) Balance the equation
3 Mg(NO3)2 + 2 Li3PO4  Mg3(PO4) LiNO3
Mg NO3- Li+ PO43-

18. Predicting precipitation
To predict whether a precipitate forms when mixing aqueous solutions of 2 strong electrolytes:
1) Write the balanced metathesis reaction
2) Use solubility rules to determine if any of the products are insoluble 18

19. Predicting precipitation: Example
Predict the products and write the balanced chemical equation for the reaction, including the phases of all reactants and products. If no reaction, write NR.
Pb(NO3)2 mixed with Na2S 19

20. AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Molecular Equation
The molecular equation lists the reactants and products in their molecular form.
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Molecular:

21. AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Ionic Equation
A more accurate representation of the species that are found in the reaction mixture is the ionic equation.
In the ionic equation, all soluble, strong electrolytes are dissociated into their ions.
Remember, strong electrolytes = strong acids, strong bases, or ionic compounds
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Molecular:
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)  AgCl (s) + K+ (aq) + NO3- (aq)
Ionic:

22. Ag+(aq) + Cl-(aq)  AgCl (s)
Net Ionic Equation
Finally, the net ionic equation shows what is actually reacting.
To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
The only things left in the equation are those things that change (i.e., react) during the course of the reaction.
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Molecular:
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)  AgCl (s) + K+ (aq) + NO3- (aq)
Ionic:
Ag+(aq) + Cl-(aq)  AgCl (s)
Net ionic:

23. Net Ionic Equation
Those ions that didn’t change during the reaction (and so were deleted from the net ionic equation) are called spectator ions.
In this example, the spectator ions are K+ and NO3-
If all ions cancel, there is no reaction (no driving force).
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq)  AgCl (s) + K+ (aq) + NO3- (aq)

24. Writing Net Ionic Equations
Write a balanced molecular equation.
Write the ionic equation by dissociating all soluble, strong electrolytes. (Remember to use coefficients to indicate how many of each ion were in the molecular equation.)
Cross out anything that remains unchanged from the left side to the right side of the equation (spectator ions).
Write the net ionic equation with the species that remain. (Should already be balanced) 24

25. Example
Write the balanced molecular, ionic, and net ionic equation for the following, including phases of each species. If no reaction, write NR.
1.) AlCl3(aq) + K3PO4(aq) 
2.) NaCl(aq) + Zn(CH3COO)2(aq) 
Left off #2 25

26. Acids
Acids – substance that ionizes in aqueous solution to form hydrogen ion (H+), or a proton
So we say acids are proton donors
Monoprotic acids yield one H+ per molecule of acid
Ex: HNO3, HCl
HCl(aq)  H+(aq) + Cl-(aq)
Diprotic acids yield two H+ ions per molecule of acid
Ex: H2SO4
H2SO4(aq)  H+(aq) + HSO4-(aq)
HSO4 -(aq) H+(aq) + SO42-(aq)

27. Bases Bases are substances that accept (react with) H+ ions
Bases produce hydroxide inos (OH-) when dissolved in water
Some include OH- in formula:
NaOH, Ca(OH)2
Some don’t include OH- in formula:
NH3 (only one you should worry about for now)

28. Acid-Base Reactions
In an acid-base reaction, the acid donates a proton (H+) to the base.

29. Neutralization Reactions
In an acid-base (neutralization) reaction, the acid donates a proton (H+) to the base.
Generally, when solutions of an acid and a base are combined, the products are a salt (ionic compound) and water.
Water is neutral on the pH scale, unlike acids or bases, hence the name neutralization reaction
The formation of liquid water is the driving force for these reactions
HBr (aq) + LiOH (aq)  LiBr (aq) + H2O (l)
HC2H3O2 (aq) + NaOH (aq)  NaC2H3O2 (aq) + H2O (l)

30. Neutralization Reactions
When a strong acid reacts with a strong base, the net ionic equation is…
HCl (aq) + NaOH (aq)  NaCl (aq) + H2O (l)
H+ (aq) + Cl- (aq) + Na+ (aq) + OH-(aq) 
Na+ (aq) + Cl- (aq) + H2O (l)
H+ (aq) + OH- (aq)  H2O (l)

31. Gas-Forming Reactions
Some metathesis reactions do not give the product expected.
In this reaction, the expected product, H2CO3, decomposes to give a gaseous product, CO2.
CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + H2CO3 (aq)
CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + CO2 (g) + H2O (l)
Any time you form carbonic acid, H2CO3, it is unstable and will decompose to give H2O(l) and CO2(g)

32. Gas-Forming Reactions
When a carbonate or bicarbonate reacts with an acid, the products are a salt, carbon dioxide, and water.
CaCO3 (s) + HCl (aq)  CaCl2 (aq) + CO2 (g) + H2O (l)
NaHCO3 (aq) + HBr (aq)  NaBr (aq) + CO2 (g) + H2O (l)
Antacids!

33. Gas-Forming Reactions
This reaction gives the predicted product, but you had better carry it out in the hood, or you will be very unpopular!
But just as in the previous examples, a gas is formed as a product of this reaction.
Na2S (aq) + H2SO4 (aq)  Na2SO4 (aq) + H2S (g)
In these cases of gas-forming reactions, the driving force of each reaction is the formation of gas!

34. Examples
Write balanced molecular, ionic, and net ionic equations for the following pairs of reactants. Include states of matter: s, l, g, aq.
Li2CO3 (aq) + HNO3 (aq) 
KOH (aq) + SnCl2 (aq) 
(NH4)2S (aq) + HBr (aq) 
Zn(OH)2 (s) + H2SO4 (aq)  34

35. Examples
Write balanced molecular, ionic, and net ionic equations for the following pairs of reactants. Include states of matter: s, l, g, aq.
Li2CO3(aq) + 2HNO3(aq)  2LiNO3(aq) + H2O(l) + CO2(g)
2Li+(aq) + CO32-(aq) + 2H+(aq) + 2NO3-(aq)  2Li+(aq) + 2NO3-(aq) + H2O(l) + CO2(g)
CO32-(aq) + 2H+(aq)  H2O(l) + CO2(g)
b) 2KOH(aq) + SnCl2(aq)  2KCl (aq) + Sn(OH)2 (s)
2K+(aq) + 2OH-(aq) + Sn2+(aq) + 2Cl-(aq)  2K+(aq) + 2Cl-(aq) + Sn(OH)2 (s)
2OH-(aq) + Sn2+(aq)  Sn(OH)2 (s)
c) (NH4)2S(aq) + 2HBr(aq)  2NH4Br (aq) + H2S(g)
2NH4+(aq) + S2-(aq) + 2H+(aq) + 2Br-(aq)  2NH4+(aq) + 2Br-(aq) + H2S(g)
S2-(aq) + 2H+(aq)  H2S(g)
Zn(OH)2(s) + H2SO4(aq)  ZnSO4(aq) + 2H2O(l)
Zn(OH)2(s) + 2H+(aq) + SO42-(aq)  Zn2+(aq) + SO42-(aq) + 2H2O(l)
Zn(OH)2(s) + 2H+(aq)  Zn2+(aq) + H2O(l) 35

36. Oxidation-Reduction (Redox) Reactions
An oxidation occurs when an atom or ion loses electrons.
A reduction occurs when an atom or ion gains electrons.
One cannot occur without the other.
Left off
“OIL RIG”
Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)

37. Oxidation Numbers Oxidation numbers are for bookkeeping of electrons.
To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number (or oxidation state) to each element in a neutral compound or charged entity. 37

38. Rules for Assigning Oxidation Numbers
Elements in their elemental form have an oxidation number of 0.
Examples: Na, O2, H2, Cu, Ag, etc… all have oxidation numbers of 0
The oxidation number of a monatomic ion is the same as its charge.
Examples: Na+ oxidation number = +1
Cl- oxidation number = -1
O2- oxidation number = -2
Cu2+ oxdiation number = +2
Note: the remaining rules are for elements when they are part of a compound 38

39. Rules for Assigning Oxidation Numbers
3. Oxygen has an oxidation number of −2, except in the peroxide ion O22- in which it has an oxidation number of −1.
4. Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.
5. Fluorine always has an oxidation number of −1.
The other halogens have an oxidation number of −1 when they are monatomic ions; they can have positive oxidation numbers, however, most notably in oxyanions. 39

40. Rules for Assigning Oxidation Numbers
4. The sum of the oxidation numbers…
in a neutral compound is 0.
Ex: MgCl2 Mg2+ oxidation number = +2
Cl- oxidation number = -1
(+2) + 2(-1) = 0
in a polyatomic ion is the charge on the ion.
Ex: SO42- Sum of oxidation numbers is -2
We can use the sum to figure out oxidation number of S:
We know oxidation number of O is -2.
Make x = oxidation # of S and set up an algebraic equation:
x + 4(-2) = -2
x = +6 (oxidation number of S) 40

41. Examples: Oxidation Numbers
Assign oxidation numbers to each element in the following examples: H2
MoS2
ClO2- Ag
H2O2
Li3PO4
CaH2 41

42. Examples: Oxidation Numbers
Assign oxidation numbers to each element in the following examples:
H2 H = 0
MoS2 S = -2 Mo = +4
ClO2- O = -2 Cl = +3
Ag Ag = 0
H2O2 O = -1 H = +1
Li3PO4 Li = +1 O = -2 P = +5
CaH2 Ca = +2 H = -1 42

43. Displacement Reactions
A + BX  AX + B
In displacement reactions, ion B oxidizes an element A.
The ion B, then, is reduced to elemental form.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

44. Displacement Reactions
A + BX  AX + B
In displacement reactions, ion B oxidizes an element A.
The ion B, then, is reduced to elemental form.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) O +1 -1 +2 -1 O
Oxidation number of Mg increases from 0 to +2
Mg is losing electrons so Mg is being oxidized!
Oxidation number of H decreases from +1 to 0
H is gaining electrons so H is being reduced!
Oxidation number of Cl remains the same
Cl is a spectator ion

45. Displacement Reactions
In this reaction,
silver ions oxidize
copper metal.
Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)

46. Displacement Reactions
The reverse reaction,
however, does not
occur.
Cu2+ (aq) + 2 Ag (s)  Cu (s) + 2 Ag+ (aq) x

47. Activity Series
Different metals vary in how easily they are oxidized (how easily they lose electrons)
Activity series – list of metals arranged in order of decreasing ease of oxidation
Allows us to predict whether a metal will be oxidized or not
Any neutral metal (or hydrogen) on the list can be oxidized by the ions of elements below it
That is why silver ion oxidized copper metal, but copper ion could not oxidize silver metal
Do NOT need to memorize activity series, but you should know how to use it if given this table. 47

48. Activity Series Examples:
Predict the outcome of the following reactions. If no reaction occurs, write NR. If a reaction occurs, write a balanced equation.
a) Fe(s) + MgCl2(aq)
b) Cr(s) + Pb(NO3)2(aq)
c) Na(s) + HCl(aq) 
Left off part c 48

49. Activity Series Examples:
Predict the outcome of the following reactions. If no reaction occurs, write NR. If a reaction occurs, write a balanced equation.
a) Fe(s) + MgCl2(aq) NR
b) 2Cr(s) + 3Pb(NO3)2(aq) 2Cr(NO3)3(aq) + 3Pb(s)
c) 2Na(s) + 2HCl(aq)  2NaCl(aq) + H2(g) 49

50. moles of solute liters of solution
Molarity
Concentration – tells how much solute is dissolved in a solvent
Molarity is one way to measure the concentration of a solution.
moles of solute
volume of solution in liters
Molarity (M) =
Ex: M MgCl2 = 2.1 mol MgCl2/L MgCl2 solution
(read as:) “2.1 molar solution of MgCl2”
Because the units of molarity are mol/L, we can use molarity as a conversion factor in stoichiometric calculations to interconvert:
moles of solute liters of solution
2.1 mol MgCl2
1 L solution
1 L solution
2.1 mol MgCl2 or

51. Using Molarities in Stoichiometric Calculations51

52. Molarity: Examples
Calculate the molarity of 142 mL of an aqueous solution containing g NaNO3.
Calculate the number of moles of Cl- ions in 32.3 mL of 0.45 M CaCl2.
Consider the reaction of aluminum chloride with silver acetate. How many milliliters of M AlCl3 would be needed to react completely with 20.0mL of M AgC2H3O2 solution?
1.10 M
0.029 mol
13.3 mL

53. Dilution: Adding water
If an original solution is diluted, the molarity of the new solution can be determined from the equation:
M1  V1 = M2  V2
where M1 and M2 are the molarity of the initial (concentrated) and final (dilute) solutions, respectively, and V1 and V2 are the volumes of the two solutions.
moles of solute before dilution = moles of solute after dilution
(because all we are doing is adding solvent to the original solution)

54. Dilution: Example
If a stock solution of 25.0 mL of 1.00 M HCl is diluted to a total of mL, what is the final concentration of the HCl solution?
M1  V1 = M2  V2
M2 = M1V1 V2
M2 = (1.00 M)( L) L
M2 = M

55. Titration
Titration is an analytical technique in which one can calculate the concentration of a solute in a solution.

56. Titration
A sample of known concentration (standard solution) is added to solution of unknown concentration.
The point at which stoichiometrically equivalent quantities are reached is the equivalence point.
For acid-base reactions, the equivalence point is observed with an indicator.
Ex: Phenolphthalein is colorless in acidic solutions but turns pink in basic solutions.

57. Titration Example
In a titration, mL of M NaOH was needed to react with mL of H2SO4 solution. What is the molarity of the acid? M 57