Chapter 4
Exercises
4.6
a) MgI2 dissociates
H 2O
MgI 2 
→ Mg 2 + ( aq) + 2 I − (aq)
b) Al(NO3)3, dissociates
−
H 2O
Al ( NO3 ) 3 
→ Al 3+ ( aq) + 3 NO3 (aq)
c) HClO4, ionizes
−
HClO4 + H 2O 
→ H 3O + ( aq) + ClO4 (aq)
d) (NH4)2SO4, dissociates
2−
+
H 2O
( NH 4 ) 2 SO4 
→ 2 ( NH 4 )( aq ) + SO4 (aq )
4.12
- using the rules in your notes or the table on page 118.
a) Ni(OH)2
insoluble
b) PbSO4
c) Ba(NO3)2 soluble
d) AlPO4
e) AgC2H3O2 soluble
4.24
insoluble
insoluble
Use table 4.2 or your notes for this question
a) CsOH - strong base
b) H3PO4 - weak acid
c) HC7H5O2 - weak acid
d) H2SO4 - strong acid
4.32
a) calcium carbonate – CaCO3
Nitric Acid – HNO3
CaCO3(s) + 2 HNO3(aq) Æ Ca(NO3)2(aq) + CO2(g) + H2O(l)
Complete Ionic Equation:
CaCO3(s) + 2 H+(aq) + 2 NO3-(aq) Æ Ca2+(aq) + 2 NO3-(aq) + CO2(g) + H2O(l)
Spectator Ions:
NO3-(aq)
Net Ionic Equation:
CaCO3(s) + 2 H+(aq) + Æ Ca2+(aq) + CO2(g) + H2O(l)
b) Iron(II) Sulide – FeS
Hydrobromic Acid – HBr
FeS(s) + 2 HBr(aq) Æ FeBr2(aq) + H2S(g)
Complete Ionic Equation
FeS(s) + 2 H+(aq) + 2 Br-(aq) Æ Fe2+(aq) + 2 Br-(aq) + H2S(aq)
Spectator Ions
Br-(aq)
Net Ionic Equation
FeS(s) + 2 H+(aq) Æ Fe2+(aq) + H2S(aq)
4.42
a) Cu(OH)2(s) + 2 HNO3(aq) Æ Cu(NO3)2(aq) + 2 H2O(l)
Acid – Base Reaction
b) Fe2O3(s) + 3 CO(g) Æ 2 Fe(s) + 3 CO2(g)
Oxidation – Reduction Reaction
Fe+3 in Fe2O3 is reduced
C+2 in CO is oxidized
c) Sr(NO3)2(aq) + H2SO4(aq) Æ SrSO4(s) + 2 HNO3(aq)
Precipitation Reaction
d) 4 Zn(s) + 10H+(aq) + 2 NO3-(aq) Æ 4 Zn2+(aq) + N2O(g) + 5 H2O(l)
Oxidation – Reduction Reaction
Zn in Zn(s) is oxidized
N5+ in NO3- is reduced
4.46
a) Iron metal + Copper(II) Nitrate
Fe(s) + Cu(NO3)2(aq) Æ Fe(NO3)2(aq) + Cu(s)
b) Zinc metal + Magnesium Sulfate
Zn + MgS Æ no reaction
c) Hydrobromic acid + Tin metal
2 HBr(aq) + Sn(s) Æ SnBr2(aq) + H2(g)
d) Hydrogen gas + NiCl2(aq)
H2(g) + NiCl2(aq) Æ no reaction
e) Aluminum metal + Cobalt(II) sulfate
2 Al(s) + 3 Co(SO4)(aq) Æ Al2(SO4)3(aq)+ Co(s)
4.52
a)
moles of solute
Liter of solution
0.145 mol
=
0.750L
= 0.193M
M=
b)
moles of solute
Liter of solution
mol
0.0850 M =
0.125L
mol = 0.0850 M (0.125L )
= 0.0106 mol KMnO4
M=
c)
moles of solute
Liters of solution
0.255 mol
11.6 M =
L
0.255 mol
L=
11.6 M
L = 0.0220 L or 22.0mL
M=
4.64
mass of glycerol
mass
D=
volume
mass
50.000mL
mass = 63.280 g
moles of glycerol
63.280 g
moles =
92.095g / mol
= 0.68712mol
molarity of the solution
mol
M=
L
0.68712mol
=
0.25000 L
=2.7485M
1.2656 g / mL =
4.70
HC2H3O2(aq) + NaOH(aq) Æ H2O(l) + NaC2H3O2(aq)
2.50mL
35.5mL
0.102M
moles NaOH
mol NaOH = 0.0355L(0.102 M ) = 0.00362mol
moles HC 2 H 3O2
1 HC 2 H 3O2
mol
=
1 NaOH
0.00362mol
mol HC2 H 3O2 = 0.00362mol
grams HC2 H 3O2
g HC2 H 3O2 = 0.00362mol (60.05 g / mol )
= 0.217 g HC 2 H 3O2
grams in one quart of vinegar
 1L  1000mL 
1qt 

 = 946.1mL
 1.057 qt  1L 
x
0.217 g
=
2.50mL
946.1mL
x = 82.3g HC 2 H 3O2
4.72
Sr(NO3)2(aq) + Na2CrO4(aq) Æ SrCrO4(s) + 2 NaNO3(aq)
0.0425M
7.52g
0.750L
used 0.100L
Solution Concentration
7.52 g
moles Sr ( NO3 ) 2 =
211.6 g / mol
= 0.03554mol
mol
L
0.03554mol
=
0.750 L
= 0.04739 M
M=
moles of Sr ( NO3 ) 2 used
mol
L
mol
0.04739 M =
0.100 L
mol = 0.004739mol Sr ( NO3 ) 2
M=
moles Na 2CrO4
1 Na 2 CrO4
x
=
1 Sr ( NO3 ) 2 0.004739 mol
x = 0.004739 mol Na 2 CrO4
volume Na 2 CrO4
mol
L
0.004739mol
0.0425M =
L
L = 0.111L or 111mL
M=