1. Business Exam 09-22-2010 At 7:00 PM arrive early Covers chapters 1-3
16 MC questions, 4 Fill ins, and 2 work out
Time 1hr 30 min
Review during Wednesday class

2. Solution Stoichiometry

3. Solution Stoichiometry

4. Solution Concentration
Relative amounts of solute and solvent.
solute – substance dissolved.
solvent – substance doing the dissolving.
There are several concentration units.
Most important to chemists: Molarity

5. Brackets [ ] represent “molarity of ”
moles solute
liters of solution
mol L
Molarity = =
V of solution not solvent.
M ≡ mol/L
Shorthand: [NaOH] =1.00 M
Brackets [ ] represent “molarity of ”

6. Molarity
Calculate the molarity of sodium sulfate in a solution that contains 36.0 g of Na2SO4 in mL of solution.
nNa2SO4 =
36.0 g
142.0 g/mol
= mol
[Na2SO4 ] =
mol L
Unit change!
mL → L
[Na2SO4 ] = mol/L = M

7. Solution Concentration: Molarity
What is the concentration of a solution made by dissolving 23.5 g NiCl2 into a volume of 250 mL?

8. Molarity
6.37 g of Al(NO3)3 are dissolved to make a 250. mL aqueous solution. Calculate (a) [Al(NO3)3] (b) [Al3+] and [NO3-].
(a) Al(NO3)3 molar mass = (14.00) + 9(16.00) g
mol
= 213.0
nAl(NO3)3 = = x 10-2 mol
6.37 g
213.0 g/mol
[Al(NO3)3] = = M
2.991 x 10-2 mol
0.250 L

9. Solution Preparation Solutions are prepared either by:
Diluting a more concentrated solution.
or…
Dissolving a measured amount of solute and diluting to a fixed volume.

10. Preparing Solutions: Direct Addition
How many grams of NiCl2 would you use to prepare 100 mL of a M solution?

11. Solution Preparation by Dilution
MconcVconc = MdilVdil
Example
Commercial concentrated sulfuric acid is 17.8 M. If 75.0 mL of this acid is diluted to 1.00 L, what is the final concentration of the acid?
Mconc = 17.8 M Vconc = 75.0 mL
Mdil = ? Vdil = mL
Mdil = =
MconcVconc
Vdil
17.8 M x 75.0 mL
1000. mL
= 1.34 M

12. Preparing Solutions: Dilution from a concentrated (stock) solution
How many mL of a 2.60 M NiCl2 solution would you use to prepare 100 mL of a M solution?

13. Solution Preparation from Pure Solute
Prepare a M solution of potassium permanganate in a mL volumetric flask.
Mass of KMnO4 required
nKMnO4 = [KMnO4] x V
= M x L (M ≡ mol/L)
= mol
mKMnO4 = mol x g/mol
= g

14. Solution Preparation from Pure Solute
Weigh exactly g of pure KMnO4
Transfer it to a volumetric flask.
Rinse all the solid from the weighing dish into the flask.
Fill the flask ≈ ⅓ full.
Swirl to dissolve the solid.
Fill the flask to the mark on the neck.
Shake to thoroughly mix.

15. Aqueous Solution Titrations
Buret = volumetric glassware used for titrations.
Titrant: Base of known concentration
Slowly add standard solution.
End point: indicator changes color.
Determine Vtitrant added.
Unknown acid + phenolphthalein (colorless in acid)…
…turns pink in base

16. Titrations: This week’s lab
Part 1. “Standardizing” a solution of base:
23.8 mL of NaOH solution is used to neutralize g H2C2O4-2H2O. What is the concentration of the NaOH solution?

17. Titrations: This week’s lab
Part 2. Determining the molar mass of an acid:
35.2 mL of the same NaOH solution is used to neutralize g of an unknown diprotic acid. What is the molar mass of the acid?

19. Molarity & Reactions in Aqueous Solution
Grams of A B
Moles of
Liters of
B solution
A solution
Use molar mass of A
Use molar mass of B
Use solution molarity of A
Use solution molarity of B
Use mole ratio
nA = [ A ] x V
[product] = nproduct / (total volume).

20. Molarity & Reactions in Aqueous Solution
What volume, in mL, of M H2SO4 is required to neutralize 25.0 mL of M NaOH?
H2SO4(aq) NaOH(aq) → Na2SO4(aq) + 2 H2O(ℓ)
nNaOH = L
0.234 mol L
= x 10-3 mol
nH2SO4 = x 10-3 mol NaOH
1 H2SO4
2 NaOH
= x 10-3 mol

21. Molarity & Reactions in Aqueous Solution
H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(ℓ)
mol mol
V? mL
Vacid needed =
mol H2SO4
[H2SO4]
1 L
mol
Vacid = x 10-3 mol
Vacid = L
= 33.4 mL

22. Molarity & Reactions in Aqueous Solution
A g mixture of oxalic acid, H2C2O4 and NaCl was neutralized by mL of 0.550M NaOH. What was the weight % of oxalic acid in the mixture?
H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(ℓ)
nNaOH = L x mol/L = mol
1 H2C2O4 ≡ 2 NaOH
1 H2C2O4
2 NaOH
nacid = mol NaOH
= x10-3 mol

23. Molarity & Reactions in Aqueous Solution
A g H2C2O4 / NaCl mixture … Wt % of oxalic acid in the mixture?
Mass of acid consumed, macid
= x10-3 mol x (90.04 g/mol acid)
= g
Weight % = x 100%
macid
sample mass
Weight % = x 100% g
4.554 g
= 16.08%

24. Molarity & Reactions in Aqueous Solution
25.0 mL of M FeCl3 and 50.0 mL of M NaOH are mixed. Which reactant is limiting? How many moles of Fe(OH)3 will form?
FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s)
nFeCl3 = L x mol/L = mol
nNaOH = L x mol/L = mol

25. Molarity & Reactions in Aqueous Solution
FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s)
mol mol nFe(OH)3 ?
mol FeCl = mol Fe(OH)3
1 Fe(OH)3
1 FeCl3
mol NaOH = mol Fe(OH)3
1 Fe(OH)3
3 NaOH
FeCl3 is limiting; mol Fe(OH)3 produced.

26. Aqueous Solution Titrations
Titration = volume-based method used to determine an unknown concentration.
A standard solution (known concentration) is added to a solution of unknown concentration.
Monitor the volume added.
Add until equivalence is reached – stoichiometrically equal moles of reactants added.
An indicator monitors the end point.
Often used to determine acid or base concentrations.