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College Chemistry Chapter 4 Study Sheet
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 in water, results in a solution that doesn’t
conduct electricity.
Strong Electrolyte- 100% Dissociation
NaCl(s) ➔Na +(aq) and Cl-(aq)
Weak Electrolyte- Not Completely Dissociated
CH3COOH  CH3COO and H+(aq)
Equations for Acid- Base Reactions:
A.) Strong Acid- Strong Base
Neutralization- When solutions are mixed, H+ and OH- react with each other to form water
molecules. H+(aq) + OH- (aq)➔ H2O
B.) Weak Acid- Strong Base (2 Step Reaction)
1. Ionization of the HB molecule to H+ and B- ions. HB(aq)  H+(aq) + B- (aq)
2. Neutralization of H+ ions. H+ (aq) + B- (aq) ➔ H2O
OVERALL WRITTEN AS: HB(aq) + OH- (aq) ➔ B- (aq) + H2O
C.) Strong Acid- Weak Base (2 Step Reaction)
1. Reaction of NH3 with H2O to form NH4+ and OH- . NH3(aq) + H2O  NH4(aq)+ + OH-(aq)
2. H+ ions neutralize OH- ions. H+(aq) + OH- (aq)➔ H2O
OVERALL WRITTEN AS: H+(aq) + NH3(aq) ➔ NH4(aq)+
Precipitation Reactions: Result in the formation of an insoluble solid (precipitate) that
separates from the solution.
For precipitation to occur, the cation of one solution must combine with the anion of the
other solution to form an insoluble solid.
Ex.) AgNO3 (aq) + NaCl(aq) ➔ AgCl (s) + NaNO3 (aq)
AgCl (s) is the precipitate which is formed. (This can be found using Chem Helper.)
What will happen when the following are mixed?Will a precipitate be formed?If so,what is it?
(a) CuSO4 and NaNO3
Cu NO3 and NaSO4 are formed. Both soluble, no precipitate.
(b) Na2CO3 and CaCl2
Na2Cl2 and CaCO3 are formed. CaCO3 precipitate formed.
Writing Net Ionic Equations:
1. Write the balanced molecular equation.
2. Write the ionic equation showing strong electrolytes.
3. Determine precipitate.
4. Cancel spectator ions.
Write the ionic and net ionic equation for the following molecular equation:
Pb(NO3)2 (aq) + 2NaI+(aq) ➔Pb2 (s) + 2NaNO3
Ionic Equation: Pb2+ + 2NO3- + 2Na+ + 2I- ➔ PbI2(s) + 2NaI+ + 2NO3Net Ionic Equation: Pb2+ + 2I- ➔ PbI2(s)
Acid-Base or Neutralization Reactions: Result in the formation of water and an ionic
compound (a salt) due to the transfer of hydrogen ions (protons).
Oxidation- Reduction (Redox): Reactions: Result in the formation of new compounds due to
transfer of electrons among reactants.
Oxidation: Increase in oxidation number, loss of electrons, becomes more +
Reduction: Decrease in oxidation number, gain of electrons, becomes more –
Oxidizing Agent: Ion or molecule that accepts eReducing Agent: Ion or molecule that donates e-
Rules for Assigning Oxidation Numbers:
1. The oxidation number of any free element is 0.
2. The oxidation number of a monatomic ion is equal to the charge of the ion.
3. The oxidation number of each hydrogen atom in most compounds in 1+, except in hydrides.
4. The oxidation number of each oxygen atom in most compounds is 2 +, except in peroxides.
5. The sum of the oxidation numbers of all the atoms in a particle must equal the apparent
charge of that particle.
6. In compounds, the elements of Group IA and Group IIA and aluminum have positive
oxidation numbers numerically equal to their group number in the periodic table.
Steps to Balancing Redox Reactions Using the Ion-Electron Method:
1. Write unbalanced ionic equations for the two half-reactions.
2. Balance atoms other than H and O.
3. Balance O with H2O.
4. Balance H with H+.
5. Balance charge with appropriate number of electrons.
6. If an acidic solution, skip to step 10.
7. If the reaction occurs in a basic solution, the hydrogen ions (H+) must be neutralized by
adding an equal number of OH- ions as there is H+ ions to both sides of the equations.
8. Combine H+ ions and OH- ions to form water.
9. Subtract water where possible.
10. Rewrite the balanced half equations.
11. Multiply balanced half-reactions by appropriate coefficients so that the numbers of
electrons are equal.
12. Add the half- reactions together.
13. Cancel species that appear on both sides to get the balanced Net Ionic Equation.
14. If necessary, add spectator ions to get the balanced molecular equations
15. Check the final balance (atoms and charges).
Molarity = Moles of solute ÷ Liters of solution
Grams of solute = Molarity x Volume x Molar Mass
Dilution Formula:
M1V1 = M2V2