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Solution Stoichiometry

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. moles solute

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Solution Stoichiometry

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  1. Solution Stoichiometry

  2. Solution Stoichiometry

  3. 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

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

  5. 36.0 g 142.0 g/mol 0.2534 mol 0.7500 L nNa2SO4 = [Na2SO4 ] = Molarity Calculate the molarity of sodium sulfate in a solution that contains 36.0 g of Na2SO4 in 750.0 mL of solution. = 0.2534 mol Unit change! mL → L [Na2SO4 ] = 0.338 mol/L = 0.338 M

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

  7. g mol = 213.0 6.37 g 213.0 g/mol nAl(NO3)3= = 2.991 x 10-2 mol 2.991 x 10-2 mol 0.250 L [Al(NO3)3] = = 0.120 M 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)

  8. 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.

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

  10. MconcVconc Vdil 17.8 M x 75.0 mL 1000. mL Mdil = = 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 = 1.34 M

  11. 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?

  12. 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

  13. Solution Preparation from Pure Solute • Weighexactly 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.

  14. 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

  15. 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?

  16. 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?

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

  18. 1 H2SO4 2 NaOH nNaOH = 0.0250 L = 5.850 x 10-3 mol 0.234 mol L nH2SO4 = 5.850 x 10-3 mol NaOH 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) = 2.925 x 10-3 mol

  19. Vacid needed = mol H2SO4 [H2SO4] Vacid = 2.925 x 10-3 mol 1 L 0.0875 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= 0.0334 L = 33.4 mL

  20. 1 H2C2O4 2 NaOH 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 nacid = 0.01627 mol NaOH = 8.135 x10-3 mol

  21. macid sample mass Weight % = x 100% 0.7324 g 4.554 g Weight % = x 100% 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 = 16.08%

  22. 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? nNaOH = 0.0500 L x 0.453 mol/L = 0.02265 mol FeCl3(aq) + 3 NaOH(aq) → 3 NaCl (aq)+ Fe(OH)3(s) nFeCl3 = 0.0250 L x 0.234 mol/L = 0.005850 mol

  23. 0.02265 mol NaOH = 0.00755 mol Fe(OH)3 1 Fe(OH)3 1 FeCl3 1 Fe(OH)3 3 NaOH 0.00585 mol FeCl3 = 0.00585 mol Fe(OH)3 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 ? FeCl3 is limiting; 0.00585 mol Fe(OH)3 produced.

  24. Aqueous Solution Titrations Titration = volume-based method used to determine an unknown concentration. A standardsolution (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.

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