Solubility:
Solubility is the measure of how much of a solute will dissolve in a solvent. In general chemistry, we usually talk about
water as the solvent, so we are talking about what compounds will dissolve in water. We also talk a lot about ionic
compounds, so the goal is to determine which ionic compounds are soluble in water and which aren’t. There are many
factors to consider in the solubility of ionic compounds, none of which you need to worry about. Instead, you will be given
solubility rules which you must apply. If it turns out that something is soluble (it does dissolve in water), then you need to
put a subscript (aq) next to it (i.e. NaCl(aq)). The (aq) stands for aqueous and implies that you have a solution, a mixture of
some compound in water. If the compound is insoluble (does not dissolve in water), then you need to put a subscript (s) next
to it. You will be given a list of the rules for the exam and the final, but I recommend memorizing the first rule, because it
has no exceptions and it takes precedence over all of the other rules. If rule #1 does not apply, look up the anion in the
remaining rules. Solubility rules apply ONLY to ionic compounds (and only in water), NOT to acids or molecular
compounds!!!!!!!
Example: calcium chromate – Calcium is not mentioned in the first rule, so you need to find chromate in the
remaining rules. Rule #7 says that chromates are insoluble so it is CaCrO4(s)
Example: rubidium oxide – rubidium is a group 1 metal, so you need look no further; all group 1 metal containing
compounds are soluble, so it is Rb2O(aq)
Example: lead (II) bromide – according to rule #3, most bromide salts are soluble, but there are exceptions (silver,
mercury (I) and lead (II). This means that lead (II) bromide is going to be insoluble: PbBr2(s)
State the solubility of each of the following compounds in water:
1)
(NH4)2S
2)
BaO
3)
Na2CO3
4)
Hg2SO4
5)
Pb(ClO4)2
6)
SrF2
7)
MgC2O4
8)
Li3PO4
9)
AgI
10)
Mn(OH)4
Electrolytes:
Electrolytes are broken down into 3 different types; non-electrolytes, weak electrolytes, and strong electrolytes. You will
have to be able to differentiate between them. It is NOT that hard, it just takes a second to memorize. Seriously, there are no
tricks here, simple memorization and you have it.
Non-electrolytes: Any molecular compound (compounds without metals).
Weak electrolytes: Any weak acid/bases and all insoluble salts.
Strong electrolytes: Any strong acid or any soluble salt
That’s it. You can already tell between insoluble salts and soluble salts (solubility rules). You learned was a molecular
compound was in the second week of class and since I am going to tell you what the strong acids are, you can tell the
difference between strong and weak acids (the 7 strong acids are HCl(aq), HBr(aq), HI(aq), HNO3(aq), HClO3(aq), HClO4(aq), and
H2SO4(aq)).
You try:
1)
(NH4)2S
2)
HC7H6O2
3)
N2O5
4)
Hg2SO4
5)
HClO4
6)
SrF2
7)
Rb2C2O4
8)
H3PO4
9)
AgI
10)
Sr(OH)2
Net ionic equations:
Remember, in order to write a net ionic equation, you must write out a BALANCED molecular equation.
Example: Sodium carbonate is mixed with bismuth (III) nitrate
Translate:
Na2CO3 + Bi(NO3)3
Write symbols for products:
Na2CO3 + Bi(NO3)3  Na+ NO3– + Bi+3 CO3–2
Balance CHARGES of products:
Na2CO3 + Bi(NO3)3  NaNO3 + Bi2(CO3)3
NO MORE TOUCHING THE SUBSCRIPTS!!!!!!!!
Balance the equation:
Initial:
(ONLY CHANGE COEFICIENTS TO BALANCE EQUATION!!!)
Na2CO3 + Bi(NO3)3  NaNO3 + Bi2(CO3)3
Multiply Na2CO3 by 3 so that the carbonates are equal on both sides.
Trial 1:
3 Na2CO3 + Bi(NO3)3  NaNO3 + Bi2(CO3)3
Multiply NaNO3 by 6 so that the nitrates are equal on both sides.
Trial 2:
3 Na2CO3 + Bi(NO3)3  6 NaNO3 + Bi2(CO3)3
Multiply Bi(NO3)3 by 2 so that the bismuths are equal on both sides.
Trial 3:
3 Na2CO3 + 2 Bi(NO3)3  6 NaNO3 + Bi2(CO3)3
Na
CO3
Bi
NO3
Initial
2
1
1
3
Reactants
Trial 2
Trial 1
6
6
3
3
1
1
3
3
Trial 3
6
3
2
6
Na
CO3
Bi
NO3
Initial
1
3
2
1
Products
Trial 1
Trial 2
1
6
3
3
2
2
1
6
Trial 3
6
3
2
6
BALANCED!!!!
Last step in writing the balanced molecular equation is to add in to the state subscripts (aq, l, g, s). Look at the
solubility rules to determine if a salt is soluble. All acids are written as (aq)
Add in state subscripts:
3 Na2CO3(aq) + 2 Bi(NO3)3(aq)  6 NaNO3(aq) + Bi2(CO3)3(s)
So the balanced molecular equation is:
3 Na2CO3(aq) + 2 Bi(NO3)3(aq)  6 NaNO3(aq) + Bi2(CO3)3(s)
The molecular equation is just your typical balanced chemical equation. It shows you possible reactions, but not necessarily
what is really going on in your beaker. The main drawback to the molecular equation is that it shows all compound as single
particles, regardless of state-of-matter, rather than how they are most likely to exist in water.
The ionic equation (aka total ionic equation) on the other hand, is intended to show each compound in the way in which it is
most likely to be found in water. Because strong electrolytes are ionized/dissociated ~100% in water, they are most likely to
be encounter as ions, with little or none of the original particles from the molecular equation to be found. This means that
any strong electrolyte is written as its constituent ions in the ionic equation. For example, MgCl2(aq), a soluble salt and
therefore a strong electrolyte, would be written as Mg2+(aq) and 2 Cl–(aq) in the ionic equation. HClO4(aq), a strong acid and
therefore a strong electrolyte, would be written as H+(aq) and ClO4–(aq).
In contrast to strong electrolytes, weak electrolytes are ionized/dissociated less than ~1% in water. This means that less than
1 out of every 100 particles of a weak electrolyte are going to split into ions while more than 99 particles out of 100 are going
to remain in the form of the original particle from the molecular equation. This means that are most likely going to encounter
this compounds in particle form rather than ion form, so weak electrolytes are written in their original form in the ionic
equation. For example, PbCl2(s), an insoluble salt and therefore a weak electrolyte, would be written as PbCl2(s) in the ionic
equation. HCN(aq), a weak acid and therefore a weak electrolyte, would be written as HCN(aq). In both of these instances, the
compound forms almost no ions, so the ions are not shown in the ionic equation.
Non electrolytes are similar to weak electrolytes except that non electrolytes for NO ions in water. Because non electrolytes
are ionized/dissociated 0% in water, they are found only in their original form from the molecular equation.
The short version of the above is, when writing the ionic equation, strong electrolytes are written as ions, while weak and non
electrolytes are written as molecules/formula units.
To write the ionic equation, we start from the molecular equation (from above)
3 Na2CO3(aq) + 2 Bi(NO3)3(aq)  6 NaNO3(aq) + Bi2(CO3)3(s)
and determine into which electrolyte categories each of the substances falls.
Na2CO3(aq) is a soluble salt (rule 1) and therefor a STRONG ELECTROLYTE
Bi(NO3)3(aq) is a soluble salt (rule 2) and therefor a STRONG ELECTROLYTE
NaNO3(aq) is a soluble salt (rule 1) and therefor a STRONG ELECTROLYTE
Bi2(CO3)3(s) is an insoluble salt (rule 7) and therefor a WEAK ELECTROLYTE*
Now that we know what type of electrolyte each compound is, we can write the ionic equation. This is done by breaking up
ALL STRONG electrolytes into ions. You break up ONLY the strong electrolytes; weak electrolytes and non-electrolytes
are left alone. Make sure not to break up polyatomic ions; i.e. leave NO3-(aq) alone instead of splitting it into nitrogen and
oxygen. Also, make sure you have the correct number of each ion present.
So, the ionic equation looks like this:
6 Na+(aq) + 3 CO32–(aq) + 2 Bi3+(aq) + 6 NO3–(aq)  6 Na+(aq) + 6 NO3–(aq) + Bi2(CO3)3(s)
That is it for the ionic equation. Each of the compounds is represented in the form in which it is most likely to be found in
water. Now, remember what it means to be reaction. A reaction is the result of a chemical change which means that in order
to have reacted, the substance must change chemically. In other words, in order to be part of the reaction, the substance must
be different on the reactant and product sides of the equation. If the particle is identical on both sides of the reaction, then it
was not part of the reaction and is then referred to as a spectator.
The final step is to write the net ionic equation. The net ionic equation only shows those particles that participated (not to be
confused with precipitated) in the reaction. To write the net ionic equation, the ionic equation is rewritten excluding all
spectators. The easiest way to do this is to neatly cross out anything that is IDENTICAL on both sides of the ionic equation,
like so:
6 Na+(aq) + 3 CO32–(aq) + 2 Bi3+(aq) + 6 NO3–(aq)  6 Na+(aq) + 6 NO3–(aq) + Bi2(CO3)3(s)
The ions that are crossed out are all spectators. Rewrite anything that is not crossed out. This is the NET IONIC
EQUATION!!
3 CO32–(aq) + 2 Bi3+(aq)  Bi2(CO3)3(s)
Examples:
Write the net ionic equation of the reaction between silver sulfate and calcium chloride yielding silver chloride and calcium
sulfate.
Balance the equation:
Ag2SO4 + CaCl2  2AgCl + CaSO4
Add in state subscripts:
Ag2SO4(aq) + CaCl2(aq)  2AgCl(s) + CaSO4(aq)
Molecular equation:
Ag2SO4(aq) + CaCl2(aq)  2 AgCl(s) + CaSO4(aq)
Ionic equation:
2 Ag+(aq) + SO4–2(aq) + Ca+2(aq) + 2 Cl–(aq)  2 AgCl(s) + Ca+2(aq) + SO4–2(aq)
Cross out spectators:
2 Ag+(aq) + SO4–2(aq) + Ca+2(aq) + 2 Cl–(aq)  2 AgCl(s) + Ca+2(aq) + SO4–2(aq)
Ag+(aq) + Cl–(aq)  AgCl(s)
Net ionic equation:
(note: the remaining coefficients reduced to the lowest common denominator)
*
WE is a small lie that will be corrected in 2nd semester gen. chem. This lie does NOT change the process however.
Okay, so what about the states on molecular compounds? For most of these, I will have to tell you what the state is. If it is a
molecular compound, whether it is (aq), (g), (l), or (s), it does not change or break up EVER (molecular compounds are nonelectrolytes)!!. You should know that CO2 is a gas and water is a liquid unless it is otherwise stated.
Remember, strong acids are strong electrolytes and will therefore ionize (i.e. HCl(aq)  H+(aq) + Cl-(aq)). If the acid that is
present is NOT one of these 7 acids, then it is a WEAK acid (weak electrolyte) and DOES NOT EVER BREAK UP IN
THE N.I.E.!!!!! (i.e. HCO2H(aq) stays HCO2H(aq))
Write the N.I.E. for the reaction between calcium hydroxide and hypochlorous acid forming calcium hypochlorite and water.
Molecular equation:
Ca(OH)2 (aq) + 2 HClO(aq)  Ca(ClO)2 (aq) + 2 HOH(l)
SE
WE
SE
NE
Ionic equation:
Ca+2(aq) + 2 OH– (aq) + 2 HClO(aq)  Ca+2(aq) + 2 ClO– (aq) + 2 HOH(l)
(HClO is a weak acid and weak electrolyte, so it does NOT break up)
Cross out spectators:
Ca+2(aq) + 2 OH– (aq) + 2 HClO(aq)  Ca+2(aq) + 2 ClO– (aq) + 2 HOH(l)
Net ionic equation:
OH– (aq) + HClO(aq)  ClO– (aq) + HOH(l)
Write the N.I.E. for the reaction between barium hydroxide and nitric acid which gives barium nitrate and water.
Molecular equation:
Ionic equation:
Ba(OH)2 (aq) + 2 HNO3 (aq)  Ba(NO3)2 (aq) + 2 HOH(l)
SE
SE
SE
NE
Ba+2(aq) + 2 OH– (aq) + 2 H+ + 2 NO3– (aq)  Ba+2(aq) + 2 NO3– (aq) + 2 HOH (l)
(HNO3 is a strong acid)
Net ionic equation:
OH– (aq) + H+(aq)  HOH (l)
Write the N.I.E. for the reaction between ammonium sulfate and potassium iodide giving potassium sulfate and ammonium
iodide.
Molecular equation:
(NH4)2SO4(aq) + 2 KI(aq)  K2SO4(aq) + 2 NH4I(aq)
Ionic equation:
2 NH4+(aq) + SO42– (aq) + 2 K+(aq) + 2 I– (aq)  2 K+(aq) + SO42– (aq) + 2 NH4+(aq) + 2 I– (aq)
Net ionic equation:
NO REACTION (when everything is a strong electrolyte, everything is a spectator and there is no reaction)
Write the net ionic equation for each of the following reactions (assume all are in water solution):
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)
15)
16)
17)
18)
19)
20)
21)
aluminum hydroxide and sulfuric acid react
ammonium sulfide and lithium oxide
potassium hydroxide and selenic acid react
hydrobromic acid and silver nitrate
nitric acid and calcium hydroxide react
rubidium chromate and chromium (III) acetate are combined
ammonium phosphate and manganese (IV) sulfate 
cesium arsenate and potassium sulfite 
magnesium oxide and hydroiodic acid 
lithium sulfate and barium nitrate react
sodium hydroxide and benzoic acid (HC7H5O2) react to give sodium benzoate (NaC7H5O2) and water
solid magnesium metal and acetic acid react to form magnesium acetate and hydrogen gas
iron (III) carbonate and chloric acid to form iron (III) chlorate, carbon dioxide, and water
sodium fluoride and calcium nitrate 
titanium (III) iodide and mercury (I) acetate 
bismuth (V) chlorate and aluminum nitrate react
solid calcium bicarbonate and hydrobromic acid  calcium bromide, water, and carbon dioxide
dihydrogen monosulfide gas and cadmium chloride 
lithium oxalate and calcium perchlorate 
strontium hydroxide and hydrochloric acid react
oxalic acid and barium hydroxide react
22)
23)
24)
25)
26)
27)
28)
29)
30)
31)
32)
33)
34)
35)
36)
37)
38)
ammonium carbonate + copper(II) nitrate 
lead(II) hydroxide + hydrochloric acid 
barium carbonate + hydrochloric acid 
aluminum + tin(IV) sulfate  aluminum sulfate + tin
potassium + gold(III)chloride  potassium chloride + gold
ammonium phosphate + sodium nitrate 
silver nitrate + potassium iodide 
magnesium sulfate + lithium bromide 
strontium chloride + sodium carbonate 
nitric acid + calcium carbonate 
zinc + barium bromide  barium + zinc bromide
ammonium chromate + barium chloride 
copper(II) sulfate + sodium sulfide 
iron (II) chloride + ammonium phosphate 
potassium oxalate + calcium nitrate 
silver nitrate + sodium carbonate 
sodium hydroxide + copper(II) nitrate 