1. Lecture 4: Aqueous solution chemistry
Lecture 4 Topics Brown, chapter 4
1. Solutes & solvents
Electrolytes & non-electrolytes
Dissociation
2. Solution concentration & stoichiometry – 4.6
Molarity & interconversion
Dilution
Types of aqueous chemial reactions
3. Precipitation reactions
Complete ionic equations
4. Neutralization reactions
Acids & bases
Neutralization reactions
Non-hydroxide bases produce gases
Titration
Summary of complete ionic equations
5. Reduction & oxidation reactions
Oxidation numbers
Oxidation of metals by acids & salts
Activity series

2. Solution concentration & stoichiometry
Molarity (moles/L) measures concentration.
Molarity relates mass to volume.
Dilution decreases concentration.
Solution stoichiometry: volume & moles
The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws.
It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made.
This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory.
While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

3. Solution concentration: molarity (M)
Remember what a solution is?
A homogeneous mixture of some amount of solute dissolved in a large(r) volume of solvent.
Molarity (M)
a measure of solution concentration moles M = 1 mole/L L of solution
What’s the molarity of a solution made by dissolving 23.4 g of sodium sulfate in enough water to make a final volume of 125 mL?
MW of Na2SO4 ~ 142 g/mole
23.4 g 1 mole = 1.32 M
142 g L
What is the molarity of a solution made by dissolving 5.00 g glucose (C6H12O6) to make 100 mL of solution? MW ~ 180 g/mol
5.00 g mole mL = M
180 g mL L
What are the ionic concentrations of this strong electrolyte, 0.74 M MgCl2?
MgCl2 --> Mg Cl-1
(2x0.74) = 1.48 M Cl-1
p.142-6

4. Interconversion: molarity, moles & volume
The ‘conversion flower’ works well for this. Or, remember that: moles = (M)(V)
grams
moles
liters MW
M (mol/L)
What volume of 1.50 M HCl would we need to get moles HCl?
The key is starting with a single value and ‘operating’ on it with the conversion factor.
0.250 moles L = L = 167 mL
1.50 moles
How many grams of Na2SO4 are needed to make L of M?
MW ~142
0.350 L mol g = 24.9 g Na2SO L mol
How many grams of Na2SO4 are there in 15 mL of M?
O.015 L mol g = 1.1 g Na2SO L mol
p.144-6

6. Solution stoichiometry “mall map”
atoms X
molecules A
subsc
Av#
mass A
moles A
volume A MW M
Stoic
mass B
moles B
volume B MW M
Av#
atoms Y
subsc
molecules B p

7. Dilution
Dilution is what happens to solutions of molecular compounds whose volume is increased. Mix chocolate syrup int0 a cold glass of milk; syrup is diluted.
In labs, solutions that are used often are often made up in a concentrated form so that they can be quickly diluted for use. This saves time & work. Think canned OJ concentrate or margarita mix.
How do we know how much dilution will be needed to get to the desired concentration from our concentrated stock solution?
MiVi = MfVf where ‘i’ is initial (stock) & ‘f’ is final (diluted sol’n) Units of volume don’t matter but must be same on both sides.
What’s the final concentration produced by diluting 25.0 mL of 4.50 M NaCl to a volume of 500.o mL?
(4.5o M)(25.0 mL) = (Mf)(500.0 mL)  Mf = M
How many mL of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M?
(3.0 M)(Vi) = (0.10 M)(450 mL)  Vi = 15.0 mL
p.146-8

8. Examples: solution stoichiometry
How many grams of Ca(OH)2 are needed to react with 25.0 mL of M H(NO3)?
Ca(OH)2 + 2H(NO3)  Ca(NO3)2 + 2H(OH)
25.0 mL 1 L mole acid 1 mole base g Ca(OH)2
103 mL L mole acid mole base
= g Ca(OH)2
How many mL of 2.5 M Ba3(PO4)2 are needed to react with 25.0 g of MgCl2?
MgCl2 = g/mol
Ba3(PO4) MgCl2  3BaCl2 + Mg3(PO4)2
25.0 g mol mol Ba3(PO4) L mL
95.21 g mol MgCl mole L
= 35.0 mL Ba3(PO4)2 p