1. Solution Chemistry & Reactions
Unit 3
Solution Chemistry & Reactions

2. Molarity
Molarity – the number of moles of solute in one liter of solvent to make a solution
Usually the number of moles of a solid dissolved in water
The solid can be any solid capable of any amount of dissolving and/or dissociation
The solvent can be any pure liquid
Molarity is often referred to as a concentration

3. Molarity Molarity is abbreviated using M M = moles solute
Liters solvent
Molarity can also be indicated by [1.0 M]
Brackets indicate a concentration

4. Calculate…
You have grams of barium chloride and you dissolve it into 250 mL of water. Calculate the molarity of the solution.
First, turn your grams into moles
Second, turn your mL into L
Third, plug into the molarity equation
Solve for M

5. Calculate
You want to create 500 mL of a 0.4M solution of Sodium hydroxide. Calculate the mass of Sodium hydroxide required to create this solution.
First, turn your mL into L
Second, plug into the equation and solve for the number of moles necessary
Third, convert your moles into grams using a conversion factor with the molar mass of NaOH

6. Solution Stoichiometry
Reactions can occur between two solutions – but it is still the moles of each molecule (or ion) that are reacting.
It is often necessary to find the number of moles in a solution prior to using a balanced equation to solve a stoichiometry problem.

7. Solution Stoichiometry
Calculate the molarity of a hydrochloric acid solution necessary to neutralize (react completely) with 250 mL of a 0.75M solution of Calcium hydroxide if 300 mL of HCl are used.

8. Solution Stoichiometry
2HCl + Ca(OH)2 → 2H2O + CaCl2
0.75M = X moles Ca(OH) 2
0.250 L
0.19 moles Ca(OH) 2
0.19 mol Ca(OH) 2 x 2 mol HCl = 0.38 mol HCl
mol Ca(OH)2
0.38 mol HCl = 1.27 M HCl
0.300 L

9. Serial Dilutions
Solutions often come in very high concentrations that we do not use in actual laboratory experiments. It is a cheap way to obtain a lot of chemicals.
We can dilute the solution by adding water to any concentration and volume we choose
The dilution formula ONLY works if there is NO CHEMICAL REACTION
If there IS a reaction, you must use stoichiometry

10. Dilutions In a dilution M1V1 = M2V2
Where M1 is the first molarity, V1 is the first volume, M2 is the second molarity and V2 is the second volume

11. Dilutions
Hydrochloric acid is sold in a concentration of 12 M. You want 1.5 L of a 1.0 M solution for a lab. How much 12M acid must you measure out to dilute to the proper concentration?
M1 = 12 M, V1 = ?
M2 = 1.0 M, V2 = 1.5 L
(12 M)(V1) = (1.0 M)(1.5 L)
V1 = L or 125 mL

12. Oxidation and Reduction (Redox)
Electrochemistry deals with nonspontaneous Redox reactions
Electrons are transferred
Spontaneous redox rxns can transfer energy
Electrons (electricity)
Heat
Non-spontaneous redox rxns can be made to happen with electricity

13. Electrochemistry Terminology
Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na+ + e-
Reduction – A process in which an element attains a more negative oxidation state Cl2 + 2e-  2Cl-
LEO says GIR
Loss of electrons = oxidation
Gain of electrons = reduction

14. Electrochemistry Terminology
Oxidizing agent
The substance that is reduced is the oxidizing agent
Reducing agent
The substance that is oxidized is the reducing agent

15. Balancing a Redox reaction in Acidic Solutions
Write separate equations for the oxidation and reduction half-reactions.
For each half reaction
Balance all the elements except H and O
Balance O by using H2O
Balance H by using H+
Balance the charge using e-
If necessary, multiply one or both balanced half reactions to equalize the e- transferred.
Add the half reactions canceling identical species.

16. Examples – Acid Solution
MnO4-(aq) + Fe2+(aq)  Fe3+(aq) + Mn2+(aq)
5e- + 8H+(aq) + MnO4-(aq)  Mn2+(aq) + 4H2O(l)
5Fe2+(aq)  5Fe3+(aq) + 5e-
5Fe2+(aq) + 8H+(aq) + MnO4-(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
H+(aq) + Cr2O72-(aq) + C2H5OH(aq)  Cr3+(aq) + CO2(g) + H2O(l)
4H2O(l) + C2H5OH(aq)  2CO2(g) + H2O(l) + 12H+(aq) + 12e-
12e- + 28H+(aq) + 2Cr2O72-(aq)  4Cr3+(aq) + 14H2O(l)
16H+(aq) + C2H5OH(aq) + 2Cr2O72-(aq)  2CO2(g) + 4Cr3+(aq) + 11H2O(l)

17. Reactions with Calculations
Zinc metal reacts with Copper (I) ions to form Zinc ions and Copper metal. Calculate how much copper metal is formed if you start with 34.5 grams of Zinc. Assume everything reacts completely.

18. Reactions with Calculations
Permanganate ions react with Lead (II) ions to make both Manganese (II) and Lead (IV) ions. How much permanganate must you start with if you make 2.56 grams of manganese (II) ions in the process of the reaction?

19. Reactions with Calculations
75 mL of a 2.5 M solution of Sodium hydroxide reacts completely with an unknown concentration of sulfuric acid. If there are 25 mL of sulfuric acid, how much sodium hydroxide is necessary to neutralize all of the acid? (Neutralization means react completely)