Chemistry 11
Chapter 16 - “Solutions” Worksheet Answer Key
1. Indicate whether the following are ionic compounds or covalent compounds:
a) NaCl ionic
b) CaBr2 ionic
c) SO2 covalent
d) NH3 covalent
e) H2O covalent
f) Sr(OH)2 ionic
g) MgCO3 ionic
h) AgOH ionic
i) ZnCl2 ionic
j) SiO2
covalent/ionic
2. Show the dissociation of the following substances:

K+ (aq) +
Cl- (aq)
b) MgF2 (s) 
Mg2+ (aq)
+
2F- (aq)

Fe2+ (aq)
+
S2- (aq)
a) KCl (s)
c) FeS (s)
d) Hg2SO4 (s)

2Hg+ (aq)
+
SO42- (aq)
e) Ca3(PO4)2 (s)

3Ca2+ (aq)
+
2PO43- (aq)
f) CCl4 (l)

No dissociation (non-polar)
g) Ag2S (s)

2Ag+ (aq)
+
S2- (aq)
h) NH4Cl (s) 
NH4+ (aq)
+
Cl- (aq)
i) (NH4)2Cr2O7 (s)

2NH4+ (aq)
+
Cr2O72- (aq)
j) K2CrO4 (s)

2K+ (aq)
+
CrO42- (aq)
3. Write the net ionic equation for the dissociation of:
a) ammonium sulphate in water
(NH4)2SO4 (s) 
2NH4+ (aq)
+
SO42- (aq)
b) iron (III) carbonate in water
Fe2(CO3)3 (s)

2Fe3+ (aq)
+
3CO32- (aq)
4. Calculate the number of moles of all aqueous ions in the following solutions, assuming each
dissolved substance dissociates completely in solution.
a) 0.60 L of 0.20 mol/L K2SO4
K+ = 0.24 mol
SO42- = 0.12 mol
b) 0.450 L of 0.300 M Na3PO4
Na+ = 0.405 mol
PO43- = 0.135 mol
c) 75.0 mL of 0.160 mol/L MnCl2
Mn2+ = 0.012 mol
Cl- = 0.024 mol
d) 0.950 L of 0.235 M Al2(SO4)3
Al3+ = 0.446 mol
SO42- = 0.669 mol
e) 25 mL of 0.065 M SnCl4
Sn4+ = 1.6 x 10-3 mol
Cl- = 6.5 x 10-3 mol
5. A student has 250 mL of a 0.35 M CuCl2 solution. What volume of 0.15 M solution can be
made from this ? How much water must be added ? (0.580 L or 580 mL)
6. Water is added to 100 mL of 6.0 M KOH solution until the final volume is 1.5 L. What is the
molarity of the new diluted solution ? (0.40 M)
7. Using a solubility table, determine whether the compound is insoluble or soluble in water:
a) Ca(NO3)2 soluble
b) FeCl2 soluble c) Ni(OH)2 insoluble d) AgNO3 soluble
e) Cr2(SO4)3 insoluble f) Al(C2H3O2)3 insoluble g) (NH4)2CO3 soluble
8. Complete and balance; then write the complete ionic and net ionic equations:
a) Na2SO3 (aq) + Ba(NO3)2 (aq) 
2NaNO3 (aq)
+
BaSO3 (s)
complete ionic: 2Na+(aq) + SO32-(aq) + Ba2+(aq) + 2NO3-(aq)
 2Na+(aq) + 2NO3-(aq) + BaSO3 (s)
net ionic:
Ba2+ (aq)
b) NH4Br (aq)
+
+
SO32- (aq)

BaSO3 (s)
Pb(C2H3O2)2(aq)  (NH4)C2H3O2 (aq) + PbBr2 (s)
complete ionic: NH4+(aq) + Br-(aq) + Pb2+(aq) + 2C2H3O2-(aq)
 NH4+(aq) + C2H3O2- (aq) + PbBr2 (s)
net ionic:
Pb2+(aq)
c) Cr2(SO4)3 (aq)
+
+
2Br-(aq)

PbBr2 (s)
3K2CO3 (aq)  Cr2(CO3)3 (aq) + 3 K2SO4 (aq)
complete ionic: 2Cr3+(aq) + 3SO42-(aq) + 6K+(aq) + 3CO32-(aq)
 2Crr2+(aq) + 3CO32-(aq) + 6K+(aq) + 3SO42-(aq)
no net ionic - no precipitate formed
d) Sr(OH)2 (aq)
+
MgCl2 (aq)
 SrCl2 (aq)
complete ionic: Sr2+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq)
 Sr2+(aq) + 2Cl-(aq)
net ionic: Mg2+(aq)
e) (NH4)2S (aq)
+
+
Mg(OH)2 (s)
Mg(OH)2 (s)
+
2OH-(aq)

Mg(OH)2 (s)
+
2NaCl (aq)

2NH4Cl (aq) +
complete ionic: 2NH4+(aq) + S2-(aq) + 2Na+(aq) + 2Cl-(aq)
 2NH4+(aq) + 2Cl-(aq) +
Na2S (aq)
2Na+(aq) + S2-(aq)
no net ionic - no precipitate formed
8. f) The reaction between lithium phosphate and calcium bromide
2Li3PO4 (aq) + 3CaBr2 (aq)

6LiBr (aq)
+
Ca3(PO4)2 (s)
complete ionic: 6Li+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Br-(aq)
 6Li+(aq) + 6Br-(aq) + Ca3(PO4)2 (s)
net ionic: 3Ca2+(aq) + 2PO43-(aq)  Ca3(PO4)2 (s)
9. Use the Solubility Table to predict precipitation formation:
FeCl3
FeCl3
CoCl2
NO
Co(NO3)2
KOH
NaNO3
NO
Fe(OH)3 (s)
NO
NO
Co(OH)2 (s)
NO
Co(OH)2 (s)
NO
CoCl2
NO
Co(NO3)2
NO
NO
KOH
Fe(OH)3 (s)
Co(OH)2 (s)
Co(OH)2 (s)
NaNO3
NO
NO
NO
NO
NO
10. BONUS - see if you can do this dilution question:
A solution is made by mixing 100 mL of 0.200 M BaCl2 and 150 mL of 0.400 M NaCl.
What is the concentration of each ionic species in the final solution ?
[Ba2+] = 0.080 M,
[Na+] = 0.240 M,
[Cl-] = 0.400 M