# 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.

## Presentation on theme: "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."— Presentation transcript:

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

Solution Stoichiometry

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

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

Molarity Calculate the molarity of sodium sulfate in a solution that contains 36.0 g of Na 2 SO 4 in 750.0 mL of solution. n Na2SO4 = 36.0 g 142.0 g/mol = 0.2534 mol [Na 2 SO 4 ] = 0.2534 mol 0.7500 L [Na 2 SO 4 ] = 0.338 mol/L = 0.338 M Unit change! mL → L

Solution Concentration: Molarity What is the concentration of a solution made by dissolving 23.5 g NiCl 2 into a volume of 250 mL?

Molarity (a) Al(NO 3 ) 3 molar mass = 26.98 + 3(14.00) + 9(16.00) 6.37 g of Al(NO 3 ) 3 are dissolved to make a 250. mL aqueous solution. Calculate (a) [Al(NO 3 ) 3 ] (b) [Al 3+ ] and [NO 3 - ]. n Al(NO3)3 = = 2.991 x 10 -2 mol 6.37 g 213.0 g/mol g mol = 213.0 [Al(NO 3 ) 3 ] = = 0.120 M 2.991 x 10 -2 mol 0.250 L

Solution Preparation Solutions 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 Addition How many grams of NiCl 2 would you use to prepare 100 mL of a 0.300 M solution?

Solution Preparation by Dilution M conc V conc = M dil V dil 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? M conc = 17.8 M V conc = 75.0 mL 1000. mL M dil = ?V dil = 1000. mL M dil = = M conc V conc V dil 17.8 M x 75.0 mL 1000. mL = 1.34 M

Preparing Solutions: Dilution from a concentrated (stock) solution How many mL of a 2.60 M NiCl 2 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 KMnO 4 required n KMnO4 = [KMnO 4 ] x V = 0.5000 M x 0.2500 L (M ≡ mol/L) = 0.1250 mol m KMnO4 = 0.1250 mol x 158.03 g/mol = 19.75 g Solution Preparation from Pure Solute

exactly Weigh exactly 19.75 g of pure KMnO 4 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

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

Titrations: This week’s lab Part 1. “Standardizing” a solution of base: 23.8 mL of NaOH solution is used to neutralize 1.020 g H 2 C 2 O 4 -2H 2 O. What is the concentration of the NaOH solution?

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 1.265 g of an unknown diprotic acid. What is the molar mass of the acid?

n A = [ A ] x V [product] = n product / (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 solution Liters 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 H 2 SO 4 is required to neutralize 25.0 mL of 0.234 M NaOH? H 2 SO 4 (aq) + 2 NaOH(aq) → Na 2 SO 4 (aq) + 2 H 2 O(ℓ) n NaOH = 0.0250 L 0.234 mol L = 5.850 x 10 -3 mol Molarity & Reactions in Aqueous Solution n H2SO4 = 5.850 x 10 -3 mol NaOH 1 H 2 SO 4 2 NaOH = 2.925 x 10 -3 mol

1 L 0.0875 mol V acid = 2.925 x 10 -3 mol V acid needed = mol H 2 SO 4 [H 2 SO 4 ] Molarity & Reactions in Aqueous Solution 0.002925 mol 0.00585 mol V? 25.0 mL H 2 SO 4 (aq) + 2 NaOH(aq) → Na 2 SO 4 (aq) + 2 H 2 O(ℓ) V acid = 0.0334 L = 33.4 mL

A 4.554 g mixture of oxalic acid, H 2 C 2 O 4 and NaCl was neutralized by 29.58 mL of 0.550M NaOH. What was the weight % of oxalic acid in the mixture? H 2 C 2 O 4 (aq) + 2 NaOH(aq) → Na 2 C 2 O 4 (aq) + 2 H 2 O(ℓ) n NaOH = 0.02958 L x 0.550 mol/L = 0.01627 mol n acid = 0.01627 mol NaOH = 8.135 x10 -3 mol 1 H 2 C 2 O 4 2 NaOH Molarity & Reactions in Aqueous Solution 1 H 2 C 2 O 4 ≡ 2 NaOH

Mass of acid consumed, m acid = 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 H 2 C 2 O 4 / NaCl mixture … Wt % of oxalic acid in the mixture? Weight % = x 100% m acid sample mass Weight % = x 100% 0.7324 g 4.554 g

FeCl 3 (aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH) 3 (s) n NaOH = 0.0500 L x 0.453 mol/L = 0.02265 mol 25.0 mL of 0.234 M FeCl 3 and 50.0 mL of 0.453 M NaOH are mixed. Which reactant is limiting? How many moles of Fe(OH) 3 will form? n FeCl3 = 0.0250 L x 0.234 mol/L = 0.005850 mol Molarity & Reactions in Aqueous Solution

0.00585 mol FeCl 3 = 0.00585 mol Fe(OH) 3 1 Fe(OH) 3 1 FeCl 3 Molarity & Reactions in Aqueous Solution 0.02265 mol NaOH = 0.00755 mol Fe(OH) 3 1 Fe(OH) 3 3 NaOH FeCl 3 is limiting; 0.00585 mol Fe(OH) 3 produced. FeCl 3 (aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH) 3 (s) 0.005850 mol 0.01925 mol n Fe(OH)3 ?

Titration Titration = volume-based method used to determine an unknown concentration. standard A standard solution (known concentration) is added to a solution of unknown concentration.  Monitor the volume added. equivalence  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. Aqueous Solution Titrations

Download ppt "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."

Similar presentations