1. Aqueous Reaction Concentration Dissociation and Ionization Double displacement Reactions: Precipitation gas evolving acid base neutralization Ionic Equation

2. Salt Dissolved in Water

3. Electrolytes and Nonelectrolytes
Materials that dissolve in water to form a solution that will conduct electricity are called e__________s
Materials that dissolve in water to form a solution that will not conduct electricity are called n___________s
Tro: Chemistry: A Molecular Approach, 2/e

4. Molecular View of Electrolytes and Nonelectrolytes
To conduct electricity, a material must have charged particles that are able to flow
Electrolyte solutions all contain ions dissolved in the water
ionic compounds are electrolytes because they dissociate into their ions when they dissolve
Nonelectrolyte solutions contain whole molecules dissolved in the water
generally, molecular compounds do not ionize when they dissolve in water
the notable exception being molecular acids
Tro: Chemistry: A Molecular Approach, 2/e

5. Salt vs. Sugar Dissolved in Water
ionic compounds dissociate into ions when they dissolve
molecular compounds do not dissociate when they dissolve
Tro: Chemistry: A Molecular Approach, 2/e

6. Acids and Ionization
Acids are molecular compounds that ionize when they dissolve in water
the molecules are pulled apart by their attraction for the water
when acids ionize, they form H+ cations and also anions
The percentage of ionization varies
Acids that ionize virtually 100% are called strong acids
HCl(aq)  H+(aq) + Cl−(aq)
H2SO4(aq)  2 H+(aq) + SO42−(aq)
Acids that only ionize a small percentage are called weak acids
HF(aq)  H+(aq) + F−(aq)
Tro: Chemistry: A Molecular Approach, 2/e

7. Strong and Weak Electrolytes
Strong electrolytes are materials that dissolve completely as ions
Soluble ionic compounds and Strong acids
their solutions conduct electricity well
Weak electrolytes are materials that dissolve mostly as molecules, but partially as ions
weak acids and weak bases (next semester 
their solutions conduct electricity, but not well
Tro: Chemistry: A Molecular Approach, 2/e

8. Dissociation of Ionic Compound
Dissociation: When ionic compounds dissolve in water, the anions and cations are separated from each other - however not all ionic compounds are soluble in water!
When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion 8

9. Dissociation of Ionic Compounds
Potassium iodide  potassium cations and iodide anions: KI(aq)  K+ (aq) + I-(aq)
Copper(II) sulfate  copper(II) cations and sulfate anions: CuSO4(aq)  Cu2+(aq) + SO42-(aq) K+ I- K I
SO4
SO42- Cu
Cu2+ 9

10. Polyatomic Ionic Compounds
Ammonium sulfate  ammonium cations and sulfate anions
(NH4)2SO4(aq)  2 NH4+(aq) + SO42-(aq)
SO4
SO42-
NH4+
NH4+
NH4+
NH4+ 10

11. Practice: How Ionic Compounds Dissociate?
Ammonium phosphate
Cobalt(III) sulfate
Zinc bromide
Sulfuric acid (strong acid)

12. 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.

13. 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

14. 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 14

15. Calculations involving Molarity
Molarity = mole  Volume (L)
Solve for mole:
Mole = ___________________
Solve for volume of solution in liters:
Volume (L) = ______________ 15

16. 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 16

17. 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

18. 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

19. 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

20. 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

21. 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-

22. Example – Find the molarity of sulfate ion from 0
Example – Find the molarity of sulfate ion from M aluminum sulfate solution
0.300 M 22

23. 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

24. Example—What Volume of 12. 0 M NH3 Is Needed to Make 5. 00 L of 1
Example—What Volume of 12.0 M NH3 Is Needed to Make 5.00 L of 1.50 M NH3 Solution?
Given: M1 = _____; V2 = ____ L, M2 = ____ M
Find: V1
Equation: M1 x V1 = M2 x V2
0.625 L

25. 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

26. Classifying Reactions
A. based on the process that happens (we observes)
precipitation, neutralization, formation of a gas, or transfer of electrons

27. Classifying Reactions
B. by what the atoms and/or ions do (at molecular/atomic level)
Type of Reaction
General Equation
Synthesis
A + B  AB
Decomposition
AB  A + B
Displacement
A + BC  AC + B
Double Displacement
AB + CD  AD + CB

28. Predicting Whether a Reaction Will Occur in Aqueous Solution
“Forces” that drive a reaction
formation of a solid (aka “Precipitation”)
formation of water
formation of a gas
transfer of electrons
when chemicals (dissolved in water) are mixed and one of these 4 things can occur, the reaction will generally happen 28

29. Reactions in Aqueous Solutions
Often the chemicals we are reacting together are dissolved in water
Aqueous solutions = mixtures of a chemical dissolved in water
Dissolving the chemicals in water helps them to react together faster
Water molecules separate the chemicals into individual molecules or ions
The free floating particles collide more frequently so the reaction speeds up 29

30. Double Displacement Reactions
Exchange ionic partners: AB + CD  AD + CB
Example: Iron(III) nitrate + sodium sulfide 
Iron(III) __________ + sodium __________
Double displacement reaction
occurs when forming
Precipitation
Water (Acid-base neutralization reaction)
Gas (one of the products to make a gas)

31. Formation of Precipitation Depends on the Solubility of Ionic Compound
a compound is soluble in a liquid if it dissolves in that liquid. It almost completely dissociate into ions.
NaCl is soluble in water
a compound is insoluble if a significant amount does NOT dissolve in that liquid. It essentially does NOT dissociate into ions.
CaCO3(marble stone) is insoluble in water
though there is a very small amount dissolved, but not enough to be significant

32. How to Predict the Solubility of Ionic Compound?
Predicting whether a compound will dissolve in water is not easy
Such knowledge comes through Experimentation:
Do some Experiments to test whether a compound will dissolve in water
Then develop some rules based on those experimental results
we call this method the Empirical method 32

33. Solubility Rule. Part I Compounds that are Generally Soluble in Water
Compounds Containing the following Ions are Generally Soluble
Exceptions
(insoluble when combined with ions on the left)
Li+, Na+, K+, NH4+;
H+ (most acids in this class)
none
NO3–, C2H3O2–, ClO4–
None
Cl–, Br–, I–
Cu+, Ag+, Hg22+, Pb2+ (CISMILE)
SO42–
Ag+, Ca2+, Sr2+, Ba2+, Pb2+

34. Solubility Rule II Compounds that are Generally Insoluble
Compounds Containing the following Ions are Generally Insoluble
Exceptions
(when combined with ions on the left the compound is soluble or slightly soluble)
OH– (hydroxide, base)
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
S2– (sulfide) 34

35. How to predict the solubility of an ionic compound using the previous table?
Knowing how to use that table ALONE is generally sufficient to predict the solubility of ionic compound.
Rule of Thumb:
if the ionic compound of interest does NOT contain any of the listed ions, it is INSOLUBLE! Example: Ca3(PO4)2
if the ionic compound of interest contains BOTH the listed anion AND the cation as EXCEPTION, it is INSOLUBLE. Example: Hg2I2

36. Predict an Ionic Compound’s Solubility in Water: HgSO4
1. Check the cation: Is it Li+, Na+, K+, or NH4+?
If YES  the compound will be soluble in water
regardless of the anion!
If NO  follow the rule for the anion (Table 7.2)
2. if a rule says the compounds are mostly soluble, then the exceptions are insoluble
3. but if a rule says the compounds are mostly insoluble, then the exceptions are soluble
note: slightly soluble  insoluble

37. Practice: Determine if the Following is Soluble in Water
Mg(OH)2
FeCl3
Pb(C2H3O2)2
BaSO4
(NH4)3PO4
Generally Soluble
Exceptions
Li+, Na+, K+, NH4+; H+
none
NO3–, C2H3O2–, ClO4–
None
Cl–, Br–, I–
CISMILE
SO42–
Ag+, Ca2+, Sr2+, Ba2+, Pb2+

38. Insoluble Ionic Compounds do not Dissociate into Ions
Insoluble Ionic compounds do not dissolve in water because the Coulombic force among the ions is too strong for water molecules to break down the compound. Therefore they do not dissociate into ions
BaSO4(s)  Ba2+(aq) + SO42-(aq)

39. Precipitation Reactions
When two soluble ionic compounds in aqueous solution mix, ion exchange will take place
AB + CD  AC + DB
if one of the products is insoluble in water, it will come out of solution as a precipitate

40. Precipitation Reactions
The “Yellow Tornado”:
Pb(NO3)2(aq) + 2 KI(aq)  2 KNO3(aq) + PbI2(s)

41. No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq)
all ions still present (all products are soluble),
 no reaction

42. Example: Write balanced equation between aqueous solution of sodium carbonate and aqueous solution of copper(II) chloride. Predict if precipitation forms.
Write the formulas of the reactants
Determine the ions present for each reactant
Exchange the Ions
Write the formulas of the products
cross charges and reduce
Balance the Equation
Determine the solubility of each product
Add “(s)” after the insoluble products and “(aq)” after the soluble products

43. Example: Write balanced equation between aqueous solution of sodium carbonate and aqueous solution of copper(II) chloride. Predict if precipitation forms.

44. Practice: Predict whether the following reactions will form precipitate (hint: predict the solubility of each product)
BaBr2 + Na2SO4  BaSO4 + NaBr
(NH4)3PO4 + Ca(NO3)2  Ca3(PO4)2 + NH4NO3
2HNO3 + Ca(OH)2  Ca(NO3)2 + 2 H2O
2 NaOH + CuSO4  Cu(OH)2 + Na2SO4

45. Acid + Base Salt + Water
Acid-Base Reactions
Also as Neutralization reactions : the Acid and Base neutralize each other’s properties
H+ from the Acid + OH- from the base  Water
Cation from the base + Anion from the acid  salt (spectator ions)
Acid + Base Salt + Water
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)

46. Example:- Write and Balance equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
Write the formulas of the reactants
Determine the ions present for each reactant
Exchange the ions, H+ combines with OH- to make H2O(l). Tip: write water as HOH
Write the formulas of the products
Balance the Equation (may be quickly balanced by matching the numbers of H and OH to make HOH)
Determine the solubility of the salt
Write an (s) after the insoluble products and a (aq) after the soluble products. Water exists as liquid, so (l) for water.

47. Example:- Write and Balance equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide

48. Gas-Evolving Reactions
Some reactions form a gas directly
From the ion exchange
K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g)
Other reactions form a gas by the decomposition of one of the products into a gas and water
Step 1: Na2CO3(aq) + H2SO4(aq)  Na2SO4(aq) + H2CO3(aq)
Step 2: H2CO3  H2O(l) + CO2(g)
Overall reaction:
Na2CO3(aq) + H2SO4(aq)  Na2SO4(aq) + H2O(l) + CO2(g)

49. Compounds that Undergo Gas Evolving Reactions
Reactant
Type
Reacting
With
Ion Exchange
Product
Decompose?
Gas
Formed
Example
metalnS,
metal HS
acid
H2S no
K2S(aq) + 2HCl(aq) 
2KCl(aq) + H2S(g)
metalnCO3,
metal HCO3
H2CO3
yes
CO2
K2CO3(aq) + 2HCl(aq) 
2KCl(aq) + CO2(g) + H2O(l)
metalnSO3
metal HSO3
H2SO3
SO2
K2SO3(aq) + 2HCl(aq) 
2KCl(aq) + SO2(g) + H2O(l)
(NH4)nanion
base
NH4OH
NH3
KOH(aq) + NH4Cl(aq) 
KCl(aq) + NH3(g) + H2O(l)

50. Example: Write and Balance equation between an aqueous solution of sodium sulfite and an aqueous solution of nitric acid. Predict if precipitate or gas will form.
Write the formulas of the reactants
Determine the ions present when each reactant dissociates
Exchange the Ions
Write the formulas of the products
Check to see either product H2S, or if either product decomposes into gas (H2CO3, H2SO3, NH4OH).
Balance the Equation
Determine the solubility of other product
Write “(s)”, or “(aq)”, or “(l) for water, or “(g)” for each product if appropriate

51. Example: Write and Balance equation between an aqueous solution of sodium sulfite and an aqueous solution of nitric acid. Predict if precipitate or gas will form.

52. Practice: Complete and Balance the following double-displacement Reactions; Predict whether the following reactions will form precipitate, gas, or neutralization (hint: predict the solubility of each compound)
Lead(II) acetate + sodium sulfite
Calcium hydroxide + iron(III) nitrate
Nitric acid + zinc hydroxide
Ammonium carbonate + hydrochloric acid

53. Ionic Equations Molecular equations uses Chemical formulas
2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s)
Ionic equations describe the material’s structure when dissolved.
ONLY aqueous strong electrolytes are written as ions
soluble salts, strong acids, strong bases
ALL OTHER compounds, Gases, Solids (such as insoluble substances), Liquids are written in molecular form
2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq) ® 2K+(aq) + 2NO3−(aq) + Mg(OH)2(s) 53 53

54. Net Ionic Equations
Ions that are both reactants and products are called spectator ions
2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq) ® 2K+(aq) + 2NO3−(aq) + Mg(OH)2(s)
Net ionic equation: An ionic equation in which the spectator ions are removed
2 OH−(aq) + Mg2+(aq) ® Mg(OH)2(s) 54 54

55. Example: Acid-Base neutralization
NOTE: ONLY soluble salts, strong acids, and strong bases forms ions
2HNO3(aq) + Ca(OH)2(s) ® 2H2O(l) + Ca(NO3)2(aq)
Types? ________ _____ _____ ________
Ionic equation:
Spectator ions?
net ionic equation: 55

56. Example: Gas-forming ionic equation
ONLY soluble salts, strong acids, and strong bases forms ions
Na2CO3(aq) + 2HCl(aq) ® 2NaCl(aq) + CO2(g) + H2O(l)
Types? ________ ________ ________ ____ ____
Ionic equation:
Spectator ions?
net ionic equation: 56

57. Precipitation Reaction as ionic equation
ONLY __________, ___________, and ___________ forms ions!!!
K2CO3(aq) + 2 AgNO3(aq) ® 2 KNO3(aq) + Ag2CO3(s)
Types? ________ ________ _______ ___________
Ionic equation:
Spectator ions?
net ionic equation: 57

58. Practice – Write the ionic and net ionic equation
K2SO3(aq) + 2HBr(aq) ® H2O(l) + SO2(g) + 2 KBr(aq)
CO2(g) + 2KOH(aq) ® K2CO3(aq) + H2O(l)
Ba(OH)2(aq) + H2SO4(aq) ® 2H2O(l) + BaSO4(s)
2LiOH(aq) + CuSO4(aq) ® Cu(OH)2(s) + Li2SO4(aq) 58

59. More Practice – Convert and Complete, then Write the ionic and net ionic equation
chloric acid + ammonium sulfite 
silver acetate + iron(III) bromide 
zinc hydroxide + sulfuric acid 
lithium sulfite + manganese(II) chloride  59

60. Weak Electrolytes Does NOT dissociate
Weak acids (acetic acid, nitrous acid, hydrofluoric acid, etc), Weak base (ammonium hydroxide, etc), Insoluble salts largely remain undissociated in water
HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
NH4OH(aq)  NH4+(aq) + OH-(aq)
PbCl2(s) CaCO3(s) BaSO4(s)
Pure gas, liquid, or solid exists as non-dissolved states
CO2(g), H2S(g), H2O(l), Al(s), Hg(s) 60

61. Write the ionic and net ionic equation involving weak or non-electrolyte(s)
Na2CO3(aq) + 2 HC2H3O2(aq) ® 2 NaC2H3O2(aq) + CO2(g) + H2O(l)
sol. salt weak acid sol. salt nonelectrolyte
2Na+(aq) + CO32−(aq) + 2HC2H3O2(aq)
® 2Na+(aq) + 2C2H3O2−(aq) + CO2(g) + H2O(l)
NIE: CO32−(aq) + 2HC2H3O2(aq) ® 2C2H3O2−(aq) + CO2(g) + H2O(l) 61

62. Write the ionic and net ionic equation involving weak or non-electrolyte(s)
NH4OH(aq) + HC2H3O2(aq) ® NH4C2H3O2(aq) + H2O(l)
weak base weak acid sol. salt nonelectrolyte
NH4OH(aq) + HC2H3O2 (aq)
® C2H3O2−(aq) + NH4+(aq) + H2O(l)
NIE: no spectator ions, same as Complete Ionic Equations 62

63. Practice – Write the ionic and net ionic equation
K2CO3(aq) + 2HC2H3O2(aq) ® H2O(l) + CO2(g) + 2KC2H3O2(aq)
2HI(aq) + BaSO3(s) ® BaI2(aq) + H2O(l) + SO2(g)
BaBr2(aq) + 2AgNO3(aq) ® 2AgBr(s) + Ba(NO3)2(aq)
2 NH4OH(aq) + CuSO4(aq) ® Cu(OH)2(s) + (NH4)2SO4(aq) 63

64. Beyond Double Displacement Rxns
Precipitation, Acid-Base, and Gas-evolving reactions all involved exchanging the ions in the solution
Oxidation-Reduction reactions : reactions involve transferring electrons from one atom to another –
also known as redox reactions
many of these reactions are not done by dissolving the reactants in water
Examples: Reaction between elements, Single Displacement Reaction, Combustion Reaction

65. Practice: Write and Balance Word Equations
iron(III) chloride + copper metal  iron(II) chloride + copper(II) chloride
nitrogen gas + magnesium metal  magnesium nitride
zinc phosphate + hydrochloric acid  phosphoric acid + zinc chloride
acetic acid + oxygen gas  carbon dioxide + water

66. Determine if Each of the Following is Soluble in Water
KOH Soluble, because the cation is K+
AgBr Insoluble, even though most compounds with Br- are soluble, this is an exception
CaCl2 Soluble, most compounds with Cl- are soluble
Pb(NO3)2 Soluble, because the anion is NO3-
SrSO4 Insoluble, even though most compounds with SO42- are soluble, this is an exception

67. Write the formulas of the reactants
Example: When an aqueous solution of sodium carbonate is added to an aqueous solution of copper(II) chloride, a white solid forms
Write the formulas of the reactants
Na2CO3(aq) + CuCl2(aq) 
Determine the ions present when each reactant dissociates
(Na+ + CO32-) + (Cu2+ + Cl-) 
Exchange the Ions
(Na+ + CO32-) + (Cu2+ + Cl-)  (Na+ + Cl-) + (Cu2+ + CO32-)

68. Write the formulas of the products
Example: When an aqueous solution of sodium carbonate is added to an aqueous solution of copper(II) chloride, a white solid forms
Write the formulas of the products
cross charges and reduce
Na2CO3(aq) + CuCl2(aq)  NaCl + CuCO3
Balance the Equation
Na2CO3(aq) + CuCl2(aq)  2 NaCl + CuCO3

69. Na2CO3(aq) + CuCl2(aq)  2 NaCl(aq) + CuCO3(s)
Example - When an aqueous solution of sodium carbonate is added to an aqueous solution of copper(II) chloride, a white solid forms
Determine the solubility of each product
NaCl is soluble
CuCO3 is insoluble
Write an (s) after the insoluble products and a (aq) after the soluble products
Na2CO3(aq) + CuCl2(aq)  2 NaCl(aq) + CuCO3(s)

70. (H+ + NO3-) + (Ca2+ + OH-)  (Ca2+ + NO3-) + HOH(l)
Example:- Write and Balance equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
Write the formulas of the reactants
HNO3(aq) + Ca(OH)2(aq) 
Determine the ions present when each reactant dissociates
(H+ + NO3-) + (Ca2+ + OH-) 
Exchange the ions, H+ combines with OH- to make H2O(l). Tip: write water as HOH
(H+ + NO3-) + (Ca2+ + OH-)  (Ca2+ + NO3-) + HOH(l)

71. Write the formulas of the products
Example: Write the molecular equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
Write the formulas of the products
cross charges and reduce
HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2 + HOH(l)
Balance the Equation
may be quickly balanced by matching the numbers of H and OH to make HOH
coefficient of the salt is always 1
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3) HOH(l)

72. Example Write the molecular equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
Determine the solubility of the salt
Ca(NO3)2 is soluble
Write an (s) after the insoluble products and a (aq) after the soluble products
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)

73. (Na+ + SO3-) + (H+ + NO3-)  (Na+ + NO3-) + (H+ + SO32-)
Example - When an aqueous solution of sodium sulfite is added to an aqueous solution of nitric acid, a gas evolves
Write the formulas of the reactants
Na2SO3(aq) + HNO3(aq) 
Determine the ions present when each reactant dissociates
(Na+ + SO32-) + (H+ + NO3-) 
Exchange the Ions
(Na+ + SO3-) + (H+ + NO3-)  (Na+ + NO3-) + (H+ + SO32-)

74. Example - When an aqueous solution of sodium sulfite is added to an aqueous solution of nitric acid, a gas evolves
Write the formulas of the products
cross charges and reduce
Na2SO3(aq) + HNO3(aq)  NaNO3 + H2SO3
Check to see either product H2S - No
Check to see of either product decomposes – Yes
H2SO3 decomposes into SO2(g) + H2O(l)
Na2SO3(aq) + HNO3(aq)  NaNO3 + SO2(g) + H2O(l)

75. Example - When an aqueous solution of sodium sulfite is added to an aqueous solution of nitric acid, a gas evolves
Balance the Equation
Na2SO3(aq) + 2 HNO3(aq)  2 NaNO3 + SO2(g) + H2O(l)
Determine the solubility of other product
NaNO3 is soluble
Write an (s) after the insoluble products and a (aq) after the soluble products
Na2SO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + SO2(g) + H2O(l)

76. Practice: Complete and Balance the following Word equations
copper(II) nitrate + sodium carbonate
silver acetate + ammonium chloride
potassium carbonate + phosphoric acid
barium hydroxide + sulfuric acid
Cu(NO3)2(aq) + Na2CO3(aq)
 2 NaNO3(aq) + CuCO3(s)
AgC2H3O2(aq) + NH4Cl(aq)
 AgCl(s) + NH4C2H3O2(aq)
3K2CO3(aq) + 2H3PO4(aq)
 2K3PO4(aq) + 3H2O(l) + 3CO2(g)
Ba(OH)2(aq) + H2SO4(aq)
 BaSO4(s) + 2 H2O(l)

77. Example–Ionic and net ionic equation
NOTE: ONLY soluble salts, strong acids, and strong bases forms ions
2HNO3(aq) + Ca(OH)2(s) ® 2 H2O(l) + Ca(NO3)2(aq)
Forms Ions? ______ _____ _____ ________
Ionic equation:
Spectator ions?
NIE: 2H+(aq) + Ca(OH)2(s) ® 2 H2O(l) + Ca2+(aq) 77

78. Practice –ionic equation
ONLY soluble salts, strong acids, and strong bases forms ions
Na2CO3(aq) + 2 HCl(aq) ® 2 NaCl(aq) + CO2(g) + H2O(l)
sol. salt sol.acid sol. salt nonelectrolyte
2Na+(aq) + CO32−(aq) + 2H+(aq) + 2Cl−(aq)
® 2Na+(aq) + 2Cl−(aq) + CO2(g) + H2O(l)
NIE: CO32−(aq) + 2 H+(aq) ® CO2(g) + H2O(l) 78

79. Practice – Write the ionic and net ionic equation
K2CO3(aq) + 2 AgNO3(aq) ® 2 KNO3(aq) + Ag2CO3(s)
sol. salt sol.salt sol. salt insoluble salt
2K+(aq) + CO32−(aq) + 2Ag+(aq) + 2NO3−(aq)
® 2K+(aq) + 2NO3−(aq) + Ag2CO3(s)
NIE: 2 Ag+(aq) + CO32−(aq) ® Ag2CO3(s) 79

80. Write the ionic and net ionic equation involving weak or non-electrolyte(s)
Na2CO3(aq) + 2 HC2H3O2(aq) ® 2 NaC2H3O2(aq) + CO2(g) + H2O(l)
sol. salt weak acid sol. salt nonelectrolyte
2Na+(aq) + CO32−(aq) + 2HC2H3O2(aq)
® 2Na+(aq) + 2C2H3O2−(aq) + CO2(g) + H2O(l)
NIE: CO32−(aq) + 2HC2H3O2(aq) ® 2C2H3O2−(aq) + CO2(g) + H2O(l) 80

81. Write the ionic and net ionic equation involving weak or non-electrolyte(s)
NH4OH(aq) + HC2H3O2(aq) ® NH4C2H3O2(aq) + H2O(l)
weak base weak acid sol. salt nonelectrolyte
NH4OH(aq) + HC2H3O2 (aq)
® C2H3O2−(aq) + NH4+(aq) + H2O(l)
NIE: no spectator ions, same as Complete Ionic Equations 81