1. The Nature of Aqueous Solutions and Molarity and Solution Stoichiometry Chemistry 142 B James B. Callis, Instructor Autumn Quarter, 2004 Lecture #10

3. The Role of Water as a Solvent: The solubility of Ionic Compounds Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution. Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called Strong Electrolytes. NaCl(s) + H 2 O(l) Na + (aq) + Cl - (aq) When sodium chloride dissolves in water the ions become solvated, and are surrounded by water molecules. These ions are called “aqueous” ions and are free to move throughout the solution, and can conduct electricity.

4. How Water Dissolves an Ionic Substance

5. Interaction of Water and Ethanol

6. Electrical Conductivity of Ionic Solutions

7. Strong Electrolytes Produce many ions in aqueous solution and conduct electricity well. The most common strong electrolytes are soluble salts, strong acids and strong bases. Acids are substances that produce H + ion when they dissolve in water. HCl, HNO 3 and H 2 SO 4 are common strong acids HNO 3 (aq) -> H + (aq) + NO 3 - (aq) NaOH and KOH are a common stong bases: NaOH(s) -> Na + (aq) + OH - (aq) All of the above species are ionized nearly 100%

8. Weak Electrolytes Produce relatively few ions in aqueous solution The most common weak electrolytes are weak acids and weak bases. Acetic acid is a typical weak acid: HC 2 H 3 O 2 (aq) -> H + (aq) + C 2 H 3 O 2 - (aq) Ammonia is a common weak base: NH 3 (aq) + H 2 O(l) -> NH 4 + (aq) + OH - (aq) Both of these species are ionized only 1%

9. Nonelectrolytes Dissolve in water but produce no ions in solution. Nonelectrolytes do not conduct electricity because they dissolve as whole molecules, not ions. Common nonelectrolytes include ethanol and table sugar (sucrose, C 12 H 22 O 11 )

11. Molarity (Concentration of Solutions) = M M = = Moles of Solute mol Liters of Solution L solute = material dissolved into the solvent In air, nitrogen is the solvent and oxygen, carbon dioxide, etc. are the solutes. In sea water, water is the solvent, and salt, magnesium chloride, etc. are the solutes. In brass, copper is the solvent (90%), and Zinc is the solute (10%).

13. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I Problem 10-1: How many moles of each ion are in each of the following: a) 4.0 moles of sodium carbonate dissolved in water b) 46.5 g of rubidium fluoride dissolved in water c) 5.14 x 10 21 formula units of iron (III) chloride dissolved in water d) 75.0 mL of 0.56 M scandium bromide dissolved in water e) 7.8 moles of ammonium sulfate dissolved in water a) Na 2 CO 3 (s) 2 Na + (aq) + CO 3 2- (aq) H2OH2O 2 mol Na + 1 mol Na 2 CO 3

14. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II b) RbF(s) Rb + (aq) + F - (aq) H2OH2O moles of RbF = c) FeCl 3 (s) Fe 3+ (aq) + 3 Cl - (aq) H2OH2O moles of FeCl 3 = 9.32 x 10 21 formula units x moles of Cl - = moles Fe 3+ =

15. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III d) ScBr 3 (s) Sc 3+ (aq) + 3 Br - (aq) H2OH2O Converting from volume to moles: Moles of ScBr 3 = 75.0 mL x x 1 L 10 3 mL Moles of Br - = e) (NH 4 ) 2 SO 4 (s) 2 NH 4 + (aq) + SO 4 2- (aq) H2OH2O Moles of NH 4 + = 7.8 moles (NH 4 ) 2 SO 4 x = 15.6 mol NH 4 + 2 mol NH 4 + 1 mol(NH 4 ) 2 SO 4 and 7.8 mol SO 4 2- are also present.

16. Problem 10-2: Preparing a Solution - 1 Problem 10-2: Prepare a solution of Sodium Phosphate by dissolving 3.95g of Sodium Phosphate into water and diluting it to 300.0 ml or 0.300 L. What is the Molarity of the salt and each of the ions? Na 3 PO 4 (s) + H 2 O(l) = 3 Na + (aq) + PO 4 3- (aq)

17. Problem 10-2: Preparing a Solution - 2 Molar mass of Na 3 PO 4 = g / mol mol Na 3 PO 4 = dissolve and dilute to 300.0 mL = volume of solution M in Na 3 PO 4 (aq) = M in PO 4 3- ions = M in Na + ions =

18. Problem 10-3: Dilution of Solutions Take 25.00 mL of the 0.0400 M KMnO 4 Dilute the 25.00 mL to 1.000 L. - What is the resulting molarity (M) of the diluted solution? # moles KMnO 4 = Vol 1 x C 1 = C 2 = final M KMnO 4 = # moles / Vol 2 = Note: V 1 x C 1 = moles solute = V 2 x C 2

19. Problem 10-4: Calculating Amounts of Reactants and Products for a Reaction in Solution Al(OH) 3 (s) + 3 HCl (aq) 3 H 2 O (l) + AlCl 3 (aq) Mass (g) of Al(OH) 3 Moles of Al(OH) 3 Moles of HCl Volume (L) of HCl ÷M (g/mol) x molar ratio ÷ M (mol/L) Problem: Given 10.0 g Al(OH) 3 (s), what volume of 1.50 M HCl(aq) is required to neutralize the base? 10.0 g Al(OH) 3

20. Problem 10-5: Solving Limiting Reactant Problems in Solution - Precipitation Problem: Lead has been used as a glaze for pottery for years, and can be a problem if not fired properly in an oven, and is leachable from the pottery. Vinegar is used in leaching tests, followed by lead precipitated as a sulfide. If 257.8 ml of a 0.0468 M solution of lead nitrate is added to 156.00 ml of a 0.095 M solution of sodium sulfide, what mass of solid lead sulfide will be formed? Plan: This is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass. Solution: Write the formulas for each ionic compound using the names of ions and their charges in Tables 2.3-2.5. Write the balanced equation: Pb(NO 3 ) 2 (aq) + Na 2 S (aq) PbS (s) + 2 NaNO 3 (aq)

21. Volume and Conc. Of Pb(NO 3 ) 2 solution Mass (g) of PbS RR of Pb(NO 3 ) 2 Volume and Conc. of Na 2 S solution RR of Na 2 S RR min x stoich coeff for PbS = Amount (mol)of PbS

22. Problem 10-5: Solving Limiting Reactant Problems in Solution – Precipitation, cont. RR Pb(NO3)2 = (V x C)/stoich coeff. = RR Na2S = (V x C) /stoich coeff. = Therefore is the Limiting Reactant! Calculation of product yield: Moles PbS = Mass of PbS = (stoich. coeff. = 1 for both.)

23. Problem 10-6: Another Limiting Reactant Problem in Solution - Precipitation Problem: When aqueous silver nitrate and sodium chromate solutions are mixed, a reaction occurs that forms solid silver chromate and a solution of sodium nitrate. If 257.8 ml of a 0.0468 M solution of silver nitrate is added to 156.00 ml of a 0.095 M solution of sodium chromate, what mass of solid silver chromate (M = 331.8 g/mol) will be formed? Plan: It is a limiting-reactant problem because the amounts of two reactants are given. After writing the balanced equation, determine the limiting reactant, then calculate the moles of product. Convert moles of product to mass of the product using the molar mass. Solution: Write the formulas for each ionic compound using the names of ions and their charges in Tables 2.3-2.5. Write the balanced equation: Tables=> Ag ion =, nitrate =, Na ion =, chromate = Therefore balanced reaction is: AgNO 3 (aq) + Na 2 CrO 4 (aq) Ag 2 CrO 4 (s) + NaNO 3 (aq)

24. Problem 10-6: Another Limiting Reactant Problem in Solution – Precipitation, cont. RR AgNO3 = V x C/reac. coeff. = RR Na2CrO4 = V x C / reac. Coeff. = __________________ is the Limiting Reactant Calculation of product yield: Mass Ag 2 CrO 4 = RR min x reac. coeff. x MM

25. Answers to Problems in Lecture #10 1.(a) 8.0 moles Na + and 4.0 moles of CO 3 2-, (b) 0.445 mol Rb + and 0.445 mol F -, (c) 0.0465 mol Cl – and 0.0155 mol Fe 3+, (d) 0.126 mol Br –, (e) 15.6 mol NH 4 + and 7.8 mol SO 4 2- 2.0.0803 M in PO 4 3- ions, 0.241 M in Na + ions 3.0.00100 M 4.256 mL 5.2.89 g PbS 6.2.00 g Ag 2 CrO 4