Transcript Document

Review
of
Chemistry 11
Ionic Compounds:
Covalent Compounds:
(Begins with a metal or NH4)
(Begins with a nonmetal)
Base
Salt
(Metal + OH)
(Metal + Anion)
Acid
(H + ?)
Nonacid
(Nonmetal + ?)
NaOH
CaCl2
HCl
C3H8
Ca(OH)2
(NH4)2SO4
H2SO4
NO2
NH4OH
NaF
CH3COOH
HOH
Ends with COOH
Water is neutral- acid + base
Classify
CH3OH
nonacid
C5H11COOH
acid
NH4Br
salt
Sr(OH)2
base
H2SO4
acid
H2O
nonacid
Dissociation equations
Ionic
Ca(NO3)2(s)
Ca2+
+
2NO3-
Dissociation equations
Ionic
Al2(SO4)3(s)
2Al3+
+
3SO42-
Dissolving equations
Covalent
C12H22O11(s)
C12H22O11(aq)
Formula Equation
Pb(NO3)2(aq)
All formulas are together!
+ 2 HCl(aq)
Complete Ionic Equation

PbCl2(s) +
Dissociate (aq)
2 HNO3 (aq)
Leave (s) (l) (g)
Pb2+(aq) + 2NO3-(aq) + 2H+(aq) + 2Cl-(aq)  PbCl2(s) + 2H+(aq) + 2NO3-(aq)
Net Ionic Equation
Pb2+(aq) + 2Cl-(aq)
Cross off spectator ions

PbCl2(s)
Formula Equation
2 Al(s) + 3 Cu(NO3)2(aq)
Complete Ionic Equation
All formulas are together!
→
3 Cu(s)
+ 2 Al(NO3)3(aq)
Dissociate (aq)
Leave (s) (l) (g)
2 Al(s) + 3 Cu2+(aq) + 6 NO3-(aq) → 3 Cu(s) + 2Al3+(aq)+ 6 NO3-( aq)
Net Ionic Equation
2Al(s)
+
3Cu2+(aq)
Cross off spectator ions
→
3Cu(s) +
2 Al3+( aq)
Molarity
Calculations
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
10036.33
mL Hg2O
136.9 g
MgCl2
MgCl2
?g
?g
36.69 g - 36.33 g
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
0.36 g
Molarity =
x
1 mole
95.3 g
0.100 L
=
0.038 M
2.
How many grams are there in 205. mL of a 0.172 M solution of
NaCl?
0.205 L
x
0.172 moles x
1L
58.5 g
1mole
=
2.06 g
3.
125 g
How many litres of 0.500 M MgCl2 solution contain 125 g
MgCl2?
x
1 mole
95.3 g
x
1L
=
0.500 mole
2.62 L
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?
80.0 g
[ CaCl2 ]
=
CaCl2
1.20 M
x
1 mol
111.1 g
= 1.20 M
0.6000 L
Ca2+
1.20 M
+
2 Cl2.40 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-
0.0333 M
0.0667 M
0.100 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.
AlCl3
Al3+
0.133 M
3.00 L
x
0.13333 mol
1L
+
3 Cl0.400 M
x
133.5 g
1 mol
=
53.4 g
Titration Calculation
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.
H2C2O4 +
0.00650 L
0.100 M
[KOH] =
2KOH
0.0100 L
?M

0.00650 L H2C2O4 x 0.100 mole x
1L
0.0100 L
=
K2C2O4
0.130 M
+
2H2O
2 mole KOH
1 mole H2C2O4
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
+
0.2500 L x 0.100 mol
2 NaOH
0.6000 L
+
2 HOH
x 0.0500 mol
1L
1L
I
0.0250 mol
0.0300 mol
C
0.0150 mol
0.0300 mol
E
0.0100 mol
0.0000 mol
[ H2SO4 ]
Na2SO4
=
0.0100 mol
0.8500 L
Total Volume
250.0 mL
=
+ 600.0 mL
0.0118 M
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.
40.0
100.0
60.0
100.0
KCl
K+
0.400 M
0.160 M
CaCl2
Ca2+
0.600 M
0.360 M
_
+
_
Cl
0.160 M
+
_
2Cl
0.720 M
[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
L
0.0100 L x 0.200 mole KOH
L
= 0.00250 mol
= 0.00200 mol
= 0.00450 mol
0.0200 L x 0.100 mole H2SO4
= 0.00200 mol
L
2XOH
+
H2SO4
→
X2SO4
I
0.00450 mol
0.00200 mol
C
0.00400 mol
0.00200 mol
E
0.00050 mol
0.00000
Total Volume
Molarity
=
Bases =
Total Base
Total Acid
+
2HOH
25.0 mL + 10.0 mL + 20.0 mL = 55.0 mL
0.00050 mol
0.0550 L
=
0.0091 M