Review of Chemistry 11

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

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

Dissociation equations Ionic Ca(NO 3 ) 2(s) Ca NO 3 -

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

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

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

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

Molarity Calculations

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

MgCl g36.33 g MgCl mL H 2 O g MgCl g g? g

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

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

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

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

5. If the [SO 4 2- ] = 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 M M M

6. If the [Cl - ] = 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 M0.133 M 3.00 L mol 1 L x x g 1 mol = 53.4 g

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

mL of M H 2 SO 4 reacts with mL of M NaOH. Calculate the concentration of the excess acid or base. H 2 SO 4 Na 2 SO 4 HOH + NaOH L x mol 1 L I C E Lx mol 1 L mol mol mol mol mol mol [ H 2 SO 4 ]= mol L = M mL mL Total Volume

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

mL of M NaOH, 10.0 mL M KOH, and 20.0 mL of M H 2 SO 4 are poured into the same beaker. What is the resulting concentration of the excess acid or base? L x mole NaOH = mol L L x mole KOH = mol L = mol Total Base L x mole H 2 SO 4 = mol Total Acid L 2XOH+H 2 SO 4 →X 2 SO 4 +2HOH I mol mol C mol mol E mol Total Volume= 25.0 mL mL mL = 55.0 mL MolarityBases= mol= M L