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

Solution Stoichiometry

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

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 ”

Molarity Calculate the molarity of sodium sulfate in a solution that contains 36.0 g of Na2SO4 in 750.0 mL of solution. nNa2SO4 = 36.0 g 142.0 g/mol = 0.2534 mol [Na2SO4 ] = 0.2534 mol 0.7500 L Unit change! mL → L [Na2SO4 ] = 0.338 mol/L = 0.338 M

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

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 = 26.98 + 3(14.00) + 9(16.00) g mol = 213.0 nAl(NO3)3 = = 2.991 x 10-2 mol 6.37 g 213.0 g/mol [Al(NO3)3] = = 0.120 M 2.991 x 10-2 mol 0.250 L

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.

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

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 = 1000. mL Mdil = = MconcVconc Vdil 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 NiCl2 solution would you use to prepare 100 mL of a 0.300 M solution?

Solution Preparation from Pure Solute Prepare a 0.5000 M solution of potassium permanganate in a 250.0 mL volumetric flask. Mass of KMnO4 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 exactly 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.

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

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 H2C2O4-2H2O. 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?

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).

Molarity & Reactions in Aqueous Solution 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 mol L = 5.850 x 10-3 mol nH2SO4 = 5.850 x 10-3 mol NaOH 1 H2SO4 2 NaOH = 2.925 x 10-3 mol

Molarity & Reactions in Aqueous Solution H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(ℓ) 0.002925 mol 0.00585 mol V? 25.0 mL Vacid needed = mol H2SO4 [H2SO4] 1 L 0.0875 mol Vacid = 2.925 x 10-3 mol Vacid = 0.0334 L = 33.4 mL

Molarity & Reactions in Aqueous Solution 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 1 H2C2O4 ≡ 2 NaOH 1 H2C2O4 2 NaOH nacid = 0.01627 mol NaOH = 8.135 x10-3 mol

Molarity & Reactions in Aqueous Solution A 4.554 g H2C2O4 / NaCl mixture … Wt % of oxalic acid in the mixture? Mass of acid consumed, macid = 8.135 x10-3 mol x (90.04 g/mol acid) = 0.7324 g Weight % = x 100% macid sample mass Weight % = x 100% 0.7324 g 4.554 g = 16.08%

Molarity & Reactions in Aqueous Solution 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? FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s) nFeCl3 = 0.0250 L x 0.234 mol/L = 0.005850 mol nNaOH = 0.0500 L x 0.453 mol/L = 0.02265 mol

Molarity & Reactions in Aqueous Solution FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq) + Fe(OH)3(s) 0.005850 mol 0.01925 mol nFe(OH)3 ? 0.00585 mol FeCl3 = 0.00585 mol Fe(OH)3 1 Fe(OH)3 1 FeCl3 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.

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.