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• 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

Solution Stoichiometry

Solution Stoichiometry

Relative amounts of solute and solvent.

There are several concentration units.

Most important to chemists: MolarityMolarity

solutesolute – substance dissolved. solventsolvent – substance doing the

dissolving.

Solution Concentration

Molarity = moles soluteliters of solution

=

• V of solution notnot solvent.

molL

Brackets [ ] represent “molarity of ”

Brackets [ ] represent “molarity of ”

• Shorthand: [NaOH] =1.00 M

Molarity

M ≡ mol/L

MolarityCalculate the molarity of sodium sulfate in a solution that contains 36.0 g of Na2SO4 in 750.0 mL of solution.

nNa2SO4 = 36.0 g142.0 g/mol

= 0.2534 mol

[Na2SO4 ] =0.2534 mol0.7500 L

[Na2SO4 ] = 0.338 mol/L = 0.338 M

Unit change!mL → L

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

Molarity

(a) Al(NO3)3 molar mass = 26.98 + 3(14.00) + 9(16.00)

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

-].

nAl(NO3)3 = = 2.991 x 10-2

mol

6.37 g213.0 g/mol

gmol

= 213.0

[Al(NO3)3] = = 0.120 M2.991 x 10-2 mol0.250 L

Solution PreparationSolutions are prepared either by:

1. Diluting a more concentrated solution.

or…

2. Dissolving a measured amount of solute and diluting to a fixed volume.

Preparing Solutions: Direct AdditionHow many grams of NiCl2 would you use to prepare 100 mL of a 0.300 M solution?

Solution Preparation by DilutionMconcVconc = MdilVdil

ExampleExampleCommercial 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 = 1000. mL1000. mL

Mdil = =MconcVconc

Vdil

17.8 M x 75.0 mL1000. mL

= 1.34 M

Preparing Solutions: Dilution from a concentrated (stock) solutionHow many mL of a 2.60 M NiCl2 solution would you use to prepare 100 mL of a 0.300 M solution?

Prepare a 0.5000 M solution of potassium permanganate in a 250.0 mL volumetric flask.

Mass of KMnOMass of KMnO44 required required

nKMnO4 = [KMnO4] x V

= 0.5000 M x 0.2500 L (M ≡ mol/L)

= 0.1250 mol

mKMnO4 = 0.1250 mol x 158.03 g/mol

= 19.75 g

Solution Preparation from Pure Solute

• Weigh exactlyexactly 19.75 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.

Solution Preparation from Pure Solute

BuretBuret = volumetric glassware used for titrations.

Slowly add standard solution.

End point: indicator changes color.

Determine Vtitrant added.

Unknown acid + phenolphthalein (colorless in acid)… …turns pink

in base

Titrant: Base of known

concentration

Aqueous Solution Titrations

Titrations: This week’s labPart 1. “Standardizing” a solution of base:23.8 mL of NaOH solution is used to neutralize 1.020 g H2C2O4-2H2O. What is the concentration of the NaOH solution?

Titrations: This week’s labPart 2. Determining the molar mass of an acid:35.2 mL of the same NaOH solution is used to neutralize 1.265 g of an unknown diprotic acid. What is the molar mass of the acid?

nA = [ A ] x V

[product] = nproduct / (total volume).

Molarity & Reactions in Aqueous Solution

Grams of A

Grams of A

Grams of B

Grams of B

Moles of A

Moles of A

Moles of B

Moles of B

Liters of B solution

Liters of B solutionLiters of

A solution

Liters of 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

What volume, in mL, of 0.0875 M H2SO4 is required to neutralize 25.0 mL of 0.234 M NaOH?

H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(ℓ)

nNaOH = 0.0250 L 0.234 molL

= 5.850 x 10-3 mol

Molarity & Reactions in Aqueous Solution

nH2SO4 = 5.850 x 10-3 mol NaOH1 H2SO4

2 NaOH

= 2.925 x 10-3 mol

1 L0.0875 mol

Vacid = 2.925 x 10-3 mol

Vacid needed =mol H2SO4

[H2SO4]

Molarity & Reactions in Aqueous Solution

0.002925 mol 0.00585 mol V? 25.0 mL

H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(ℓ)

Vacid = 0.0334 L

= 33.4 mL

A 4.554 g mixture of oxalic acid, H2C2O4 and NaCl was neutralized by 29.58 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 = 0.02958 L x 0.550 mol/L = 0.01627 mol

nacid = 0.01627 mol NaOH

= 8.135 x10-3 mol

1 H2C2O4

2 NaOH

Molarity & Reactions in Aqueous Solution

1 H2C2O4 ≡ 2 NaOH

Mass of acid consumed, macid

= 8.135 x10-3 mol x (90.04 g/mol acid)

= 0.7324 g

= 16.08%

Molarity & Reactions in Aqueous Solution

A 4.554 g H2C2O4 / NaCl mixture … Wt % of oxalic acid in the mixture?

Weight % = x 100%macid

sample mass

Weight % = x 100%0.7324 g4.554 g

FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s)

nNaOH = 0.0500 L x 0.453 mol/L = 0.02265 mol

25.0 mL of 0.234 M FeCl3 and 50.0 mL of 0.453 M NaOH are mixed. Which reactant is limiting? How many moles of Fe(OH)3 will form?

nFeCl3 = 0.0250 L x 0.234 mol/L = 0.005850 mol

Molarity & Reactions in Aqueous Solution

0.00585 mol FeCl3 = 0.00585 mol Fe(OH)31 Fe(OH)3

1 FeCl3

Molarity & Reactions in Aqueous Solution

0.02265 mol NaOH = 0.00755 mol Fe(OH)3

1 Fe(OH)3

3 NaOH

FeCl3 is limiting; 0.00585 mol Fe(OH)3 produced.

FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s)0.005850 mol 0.01925 mol nFe(OH)3 ?

TitrationTitration = volume-based method used to determine an unknown concentration.

A standardstandard solution (known concentration) is added to a solution of unknown concentration.

Monitor the volume added.

Add until equivalenceequivalence is reached – stoichiometrically equal moles of reactants added.

An indicator monitors the end point.

Often used to determine acid or base concentrations.

Aqueous Solution Titrations