Review of

Chemistry 11

Ionic Compounds: Covalent Compounds:

(Begins with a metal or NH4) (Begins with a nonmetal)

Base Salt Acid Nonacid

NaOH

(Metal + OH) (Metal + Anion) (H + ?) (Nonmetal + ?)

NH4OH

Ca(OH)2

CaCl2

(NH4)2SO4

NaF

HCl

H2SO4

CH3COOH

C3H8

NO2

HOH

Ends with COOH Water is neutral- acid + base

Classify CH3OH nonacid

C5H11COOH acid

NH4Br salt

Sr(OH)2base

H2SO4acid

H2O nonacid

Dissociation equationsIonic

Ca(NO3)2(s)Ca2+ + 2NO3

-

Al2(SO4)3(s)2Al3+ + 3SO4

2-

Dissociation equationsIonic

C12H22O11(s) C12H22O11(aq)

Dissolving equationsCovalent

Pb(NO3)2(aq) +

HCl(aq)

Pb2+(aq) + 2NO3

-(aq) + 2H+

(aq) + 2Cl-(aq) PbCl2(s) + 2H+

(aq) + 2NO3-(aq)

Net Ionic Equation Cross off spectator ions

Pb2+(aq) + 2Cl-

(aq) PbCl2(s)

Complete Ionic Equation

Formula Equation All formulas are together!

Dissociate (aq) Leave (s) (l) (g)

PbCl2(s) + HNO3 (aq)2 2

Al(s) + Cu(NO3)2(aq) → Cu(s) + Al(NO3)3(aq)

2 Al(s) + 3 Cu2+(aq) + 6 NO3

-(aq) → 3 Cu(s) + 2Al3+

(aq)+ 6 NO3-( aq)

2Al(s) + 3Cu2+(aq) → 3Cu(s) + 2 Al3+

( aq)

32 3 2

Formula Equation All formulas are together!

Complete Ionic Equation Dissociate (aq) Leave (s) (l) (g)

Net Ionic Equation Cross off spectator ions

MolarityCalculations

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 MgCl2

solution?

MgCl2

36.69 g36.33 g

MgCl2

100 mL H2O

136.9 g

MgCl2 36.69 g - 36.33 g? g ? g

= 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 MgCl2

solution?36.69 – 36.33 = 0.36 g Note the loss of sig figs

2. How many grams are there in 205. mL of a 0.172 M solution of NaCl?

1 Lx 0.172 moles0.205 L x 58.5 g

1mole= 2.06 g

= 2.62 L0.500 mole

x 1L 1 mole95.3 g

125 g x

3. How many litres of 0.500 M MgCl2 solution contain 125 g MgCl2?

4. What is the concentration of each ion in the solution formed by dissolving 80.0 g of CaCl2

in 600.0 mL of water?

[ CaCl2 ] =80.0 g 111.1 g

1 mol x

0.6000 L= 1.20 M

CaCl2 Ca2+ 2 Cl-+

1.20 M 1.20 M 2.40 M

5. If the [SO42-] = 0.100 M in 20.0 mL of Ga2(SO4)3, determine the

[Ga3+] and the molarity of the solution.

Ga2(SO4)3 2 Ga3+ 3 SO42-+

0.100 M0.0667 M0.0333 M

6. If the [Cl-] = 0.400 M, calculate the number of grams of AlCl3 that would be dissolved in 3.00 L of water.

AlCl3Al3+ 3 Cl-+

0.400 M0.133 M

3.00 L 0.13333 mol

1 Lx x 133.5 g

1 mol= 53.4 g

Titration Calculation

H2C2O4 + 2KOH K2C2O4 + 2H2O

0.00650 L 0.0100 L0.100 M ? M

= 0.130 M

0.0100 L

x 2 mole KOH 1 mole H2C2O4

x 0.100 mole 1 L

0.00650 L H2C2O4

[KOH] =

7. 6.50 mL of 0.100 M H2C2O4 is required to neutralize 10.0 mL of KOH solution in a titration. Calculate the base

concentration.

8. 250.0 mL of 0.100 M H2SO4 reacts with 600.0 mL of 0.0500 M NaOH. Calculate the concentration of the excess acid or base.

H2SO4 Na2SO4 HOH+NaOH+ 2 2

0.2500 L x 0.100 mol

1 L

I

C

E

0.6000 L x 0.0500 mol

1 L

0.0250 mol 0.0300 mol

0.0150 mol 0.0300 mol

0.0100 mol 0.0000 mol

[ H2SO4 ] =0.0100 mol

0.8500 L= 0.0118 M

250.0 mL + 600.0 mL

Total Volume

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.

KCl K+ Cl_+

CaCl2 Ca2+ + 2Cl_

0.400 M 0.160 M 0.160 M

0.600 M 0.360 M 0.720 M

40.0100.0

60.0100.0

[Cl_] = 0.720 M + 0.160 M = 0.880 M

10. 25.0 mL of 0.100 M NaOH, 10.0 mL 0.200 M KOH, and 20.0 mL of 0.100 M H2SO4 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 H2SO4 = 0.00200 mol Total Acid L

2XOH + H2SO4 → X2SO4+ 2HOHI 0.00450 mol 0.00200 molC 0.00400 mol 0.00200 molE 0.00050 mol 0.00000

Total Volume = 25.0 mL + 10.0 mL + 20.0 mL = 55.0 mL

Molarity Bases = 0.00050 mol = 0.0091 M0.0550 L