1. Review of Chemistry 11

2. Ionic Compounds:Covalent Compounds: (Begins with a metal or NH 4 ) (Begins with a nonmetal) BaseSalt AcidNonacid NaOH (Metal + OH)(Metal + Anion)(H + ?)(Nonmetal + ?) NH 4 OH Ca(OH) 2 CaCl 2 (NH 4 ) 2 SO 4 NaF HCl H 2 SO 4 CH 3 COOH C3H8C3H8 NO2NO2 HOH Ends with COOHWater is neutral- acid + base

3. Classify CH 3 OH nonacid C 5 H 11 COOHacid NH 4 Brsalt Sr(OH) 2 base H 2 SO 4 acid H2OH2O nonacid

4. Dissociation equations Ionic Ca(NO 3 ) 2(s) Ca 2+ + 2NO 3 -

5. Al 2 (SO 4 ) 3(s) 2Al 3+ + 3SO 4 2- Dissociation equations Ionic

6. C 12 H 22 O 11(s) C 12 H 22 O 11(aq) Dissolving equations Covalent

7. Pb(NO 3 ) 2 (aq) + HCl (aq) Pb 2+ (aq) + 2NO 3 - (aq) + 2H + (aq) + 2Cl - (aq)  PbCl 2 (s) + 2H + (aq) + 2NO 3 - (aq) Net Ionic EquationCross off spectator ions Pb 2+ (aq) + 2Cl - (aq)  PbCl 2(s) Complete Ionic Equation Formula EquationAll formulas are together! Dissociate (aq) Leave (s) (l) (g)  PbCl 2 (s) + HNO 3 (aq) 22

8. Al (s) + Cu(NO 3 ) 2(aq) →Cu (s) + Al(NO 3 ) 3(aq) 2 Al (s) +3 Cu 2+ (aq) + 6 NO 3 - (aq) →3 Cu (s) +2Al 3+ (aq) + 6 NO 3 - ( aq) 2Al (s) + 3Cu 2+ (aq) →3Cu (s) + 2 Al 3+ ( aq) 3232 Formula EquationAll formulas are together! Complete Ionic EquationDissociate (aq) Leave (s) (l) (g) Net Ionic EquationCross off spectator ions

9. Molarity Calculations

10. 1.A student weighs an empty beaker and determines the mass to be 36.33 g. She transfers 100.0 mL of a solution to this beaker and weighs it and finds the mass to be 136.09 g. She evaporates the water until it is dry and measures the mass of the beaker and residue to be 36.69 g. What is the molarity of the MgCl 2 solution?

11. MgCl 2 36.69 g36.33 g MgCl 2 100 mL H 2 O 136.9 g MgCl 2 36.69 g - 36.33 g? g

12. =0.038 M 0.100 L 95.3 g x 1 mole0.36 g Molarity = 1.A student weighs an empty beaker and determines the mass to be 36.33 g. She transfers 100.0 mL of a solution to this beaker and weighs it and finds the mass to be 136.09 g. She evaporates the water until it is dry and measures the mass of the beaker and residue to be 36.69 g. What is the molarity of the MgCl 2 solution? 36.69 – 36.33= 0.36 gNote the loss of sig figs

13. 2.How many grams are there in 205. mL of a 0.172 M solution of NaCl? 1 L x 0.172 moles0.205 Lx 58.5 g 1mole = 2.06 g

14. =2.62 L 0.500 mole x 1L1 mole 95.3 g 125 g x 3.How many litres of 0.500 M MgCl 2 solution contain 125 g MgCl 2 ?

15. 4. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl 2 in 600.0 mL of water? [ CaCl 2 ] = 80.0 g 111.1 g 1 mol x 0.6000 L = 1.20 M CaCl 2 Ca 2+ 2 Cl - + 1.20 M 2.40 M

16. 5. If the [SO 4 2- ] = 0.100 M in 20.0 mL of Ga 2 (SO 4 ) 3, determine the [Ga 3+ ] and the molarity of the solution. Ga 2 (SO 4 ) 3 2 Ga 3+ 3 SO 4 2- + 0.100 M0.0667 M0.0333 M

17. 6. If the [Cl - ] = 0.400 M, calculate the number of grams of AlCl 3 that would be dissolved in 3.00 L of water. AlCl 3 Al 3+ 3 Cl - + 0.400 M0.133 M 3.00 L 0.13333 mol 1 L x x 133.5 g 1 mol = 53.4 g

18. Titration Calculation H 2 C 2 O 4 + 2KOH  K 2 C 2 O 4 + 2H 2 O 0.00650 L0.0100 L 0.100 M? M =0.130 M 0.0100 L x 2 mole KOH 1 mole H 2 C 2 O 4 x 0.100 mole 1 L 0.00650 L H 2 C 2 O 4 [KOH] = 7.6.50 mL of 0.100 M H 2 C 2 O 4 is required to neutralize 10.0 mL of KOH solution in a titration. Calculate the base concentration.

19. 8. 250.0 mL of 0.100 M H 2 SO 4 reacts with 600.0 mL of 0.0500 M NaOH. Calculate the concentration of the excess acid or base. H 2 SO 4 Na 2 SO 4 HOH + NaOH+22 0.2500 L x 0.100 mol 1 L I C E 0.6000 Lx 0.0500 mol 1 L 0.0250 mol0.0300 mol 0.0150 mol0.0300 mol 0.0100 mol 0.0000 mol [ H 2 SO 4 ]= 0.0100 mol 0.8500 L =0.0118 M 250.0 mL+ 600.0 mL Total Volume

20. 9.If 40.0 mL of 0.400 M potassium chloride solution is added to 60.0 mL of 0.600 M calcium chloride, what is the resulting concentration of each ion. KClK+K+ Cl _ + CaCl 2 Ca 2+ + 2Cl _ 0.400 M0.160 M 0.600 M 0.360 M0.720 M 40.0 100.0 60.0 100.0 [Cl _ ] = 0.720 M + 0.160 M = 0.880 M

21. 10.25.0 mL of 0.100 M NaOH, 10.0 mL 0.200 M KOH, and 20.0 mL of 0.100 M H 2 SO 4 are poured into the same beaker. What is the resulting concentration of the excess acid or base? 0.0250 L x 0.100 mole NaOH = 0.00250 mol L 0.0100 L x 0.200 mole KOH = 0.00200 mol L = 0.00450 mol Total Base 0.0200 L x 0.100 mole H 2 SO 4 = 0.00200 mol Total Acid L 2XOH+H 2 SO 4 →X 2 SO 4 +2HOH I0.00450 mol0.00200 mol C0.00400 mol0.00200 mol E0.00050 mol0.00000 Total Volume= 25.0 mL + 10.0 mL + 20.0 mL = 55.0 mL MolarityBases=0.00050 mol=0.0091 M 0.0550 L