Solubility and Ionic Reactions
Name:
Period:
AP Chemistry
RF Mandes, PhD, NBCT
Date:
PART 1
State whether each compound is soluble ( S ) or insoluble ( I ).
Compound
Solubility
Compound
1
AgNO3
26
LiCH3CO2
2
BaSO4
27
NH4NO3
3
NaCl
28
Mg(ClO3)2
4
K3PO4
29
Na2CO3
5
K2CO3
30
SrSO4
6
Fe(NO3)3
31
NH4Cl
7
K2SO4
32
Na2S
8
NaOH
33
Ca(OH)2
9
MgO
34
KNO3
10
KI
35
BaCO3
11
NaClO4
36
Mg(NO3)2
12
FeS
37
Na2O
13
AgCl
38
FeCO3
14
Na2SO4
39
NH4OH
15
Sn(NO3)2
40
CaPO4
16
MgBr2
41
KClO4
17
CuO
42
Li2SO4
18
NH4ClO3
43
NH4C2O4
19
KOH
44
AgBr
20
Mg(CH3CO2)2
45
K2CrO4
21
PbI2
46
Ba(OH)2
22
Li2CO3
47
CuNO3
23
CaSO4
48
PbSO4
24
Pb(NO3)2
49
CaCl2
25
Fe2O3
50
KClO3
Solubility
Solubility and Ionic Reactions
Part 2 Predicting Products and Writing Ionic and Net-Ionic Equations
1st
2nd
3rd
4th
5th
6th
generally
Dissociate (separate) the reactants into individual ions because they are dissolved in water.
Switch partners, but keep them as ions.
Write the new formulas.
Evaluate if products are soluble or insoluble
Write soluble compounds as ions and insoluble as formulas in the ionic equations
Cancel spectator ions and write what remains as the net ionic equation
AbBa
+

CdDc
AdDa
1st
Aa+ + Bb-
CbBc____
3rd
Cc+ + Dd-
+
+
2nd
Aa+ + Dd-
+
Cc+ + Bb-
Predict the products for each of the following double replacement reactions.
ex.
Na2CO3
+

BaCl2
NaCl

Na1+ + CO32-
+

+
Ba2+ + Cl1-

Na1+ + Cl1-
+
1st product (soluble – insoluble), so write as (ions – formula)
2nd product (soluble – insoluble), so write as (ions – formula)
ionic eq:
Na1+ + CO32- + Ba2+ + Cl1-  Na1+ + Cl1- + BaCO3
net ionic eq:
CO32- + Ba2+
1.
+
AlCl3

Na2O
BaCO3


+

+

+

+
ionic eq:
net ionic eq:
2.
Ni (OH)3 +
Cu(NO3)2

+
ionic eq:
net ionic eq:
BaCO3


+
Ba2+ + CO32-
Solubility and Ionic Reactions
3.
Na2SO4
+
Ba(OH)2


+

+

+

+
ionic eq:
net ionic eq:
4.
BaCl2
+
K2CO3


+

+

+
ionic eq:
net ionic eq:
5.
Sr(ClO3)2 +
MgSO4


+

+

+
ionic eq:
net ionic eq:
6.
Pb (CHCO2)2 +
FeI3


+

+

+
ionic eq:
net ionic eq:
7.
BaCO3
+
HNO3

+
ionic eq:
net ionic eq:


+