CHEM 115
REACTIONS OF OXIDES WITH WATER
I.
Metal Oxides: [i.e., Basic Oxides]
The metal oxides react with water to form metal hydroxides. For this reason, they are referred to as
basic oxides. The reactions of oxides with water are NOT oxidation-reduction reactions – i.e.,
whatever charge the metal has in the oxide, it will have the same charge when it forms its
hydroxide. For transition metals that can have more than one charge state, you can determine
which one it has from the formula for the oxide. [Remember, oxygen forms O2– when as an anion.]
Since these oxides form hydroxides in water, the water solution will be basic – i.e., it will have
OH– ions in it. If the hydroxide is soluble, then the solution will be strongly basic, but if the
hydroxide is insoluble in water, then only a very few free hydroxides will exist in water.
Nonethe-less, the solution will still be basic, but only weakly basic.
The following are some typical examples:
Na2O (s) + H2O (l)  2 NaOH (aq) – Na has its expected charge of +1
Al2O3 (s) + H2O (l)  2 Al(OH)3 (aq) – Al has its expected charge of +3
FeO (s) + H2O (l)  Fe(OH)2 (aq) – Fe is in its +2 charge state
Fe2O3 (s) + 3 H2O (l)  2 Fe(OH)3 (aq) – Fe is in its +3 charge state
II.
Non-Metal Oxides: [i.e., Acidic Oxides]
When non-metal oxides react with water, they form oxy-acids, i.e., acids involving oxy-anions. For
this reason, the solution will become acidic. This is why the non-metal oxides are referred to as
acidic oxides. [If the acid that forms is a strong acid, then the solution will become strongly acidic.
However, if the acid is a weak acid, then the solution will still be acidic, but only weakly acidic.]
Oxy-anions involving non-metals can have differing numbers of oxygen atoms, [e.g., SO32– vs.
SO42– or PO33– vs. PO43–, etc.]. Two examples for phosphorus are shown below:
P4O10 (s) + 6 H2O (l)  4 H3PO4 (aq)
P2O3 (s) + 3 H2O (l)  2 H3PO3 (aq)
Note that the oxidation numbers on phosphorus remain unchanged during the reaction. In the 1st
reaction, phosphorus has an oxidation number of +5 on both sides of the equation and it has an
oxidation number of +3 on both sides of the 2nd reaction.
Thus, the difficulty that arises with non-metal oxides is that we need to determine how many
oxygen atoms will be in the anion in the acid. There are two different methods we can employ to
do this: 1) we can determine the oxidation number of the non-metal in the oxide and then make
sure that that atom has the same oxidation number in the acid that forms; or 2) we can determine
the number of oxygen atoms in the anion through the process of balancing the overall chemical
reaction. The 2nd method “kills 2 birds with 1 stone” – we simultaneously determine the form for
the oxy-anion and the balancing coefficients. With the 1st method, we first determine the form for
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the oxy-anion and then balance the chemical reaction. We will illustrate both methods in a couple
of examples that are shown below:
Example 1: N2O5 + H2O
Method 1: Nitrogen forms anions with –1 charges – i.e., they have the form of NOx–, where
the number of oxygen atoms, x, can vary. In N2O5, each N atom has an oxidation number of
+5, since each O atom has to have oxidation number of –2. To determine the number of O
atoms in the anion, we can set up the usual equation for the NOx– anion, substituting in +5 for
the ox. no. on N and –2 for the one on O:
–1 = (ox. no. N) + x (ox. no. O) = (+5) + x (–2)
–6 = x (–2)  x = 3
Thus the anion that forms will be the NO3– anion. We need 1 H+ to offset the –1 charge when
we write the form for the acid. Thus we can now write down the unbalanced equation:
N2O5 (s) + H2O (l)  HNO3 (aq)
Finally, we balance the reaction equation. We get
N2O5 (s) + H2O (l)  2 HNO3 (aq)
Method 2: Since nitrogen forms anions with –1 charges, we know that the acid will have the
form of HNOx, where the number of O atoms, x, can vary. Let’s write down the overall
reaction equation with this form for our product:
N2O5 (s) + H2O (l)  HNOx (aq)
Now, let’s balance the reaction:
We need 2 N atoms on the right to balance the 2 N atoms on the left, giving
1 N2O5 (s) + H2O (l)  2 HNOx (aq)
We have 2 H atoms on the right, so we need 1 H2O on the left to ensure the H atoms are
balanced:
1 N2O5 (s) + 1 H2O (l)  2 HNOx (aq)
We still haven’t balanced the O atoms. We can now determine the value of x that will
ensure that they are balanced. We have a total of 6 O atoms on the left. Since we have
2x O atoms on the right, x = 3. We have now determined the value of x and ensured the
equation is balanced, all at the same time:
N2O5 (s) + H2O (l)  2 HNO3 (aq)
Personally, I think that Method 2 is the easiest one to use, since it is not necessary at any point to
determine the oxidation number on the non-metal. Its value is guaranteed to remain unchanged, if
we properly balance the chemical reaction. Note, however, that you MUST know what change is
formed by the oxy-anions, for either method. I.e., you must know that the S oxy-anions and the
C oxy-anions have –2 charges; the Cl, Br and I oxy-anions have –1 charges; the P oxy-anions have
–3 charges, etc.
–2–
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Example 2 illustrates the application of Method 2, using the reaction of P4O10 with H2O:
Example 2: P4O10 (s) + H2O (l)  ?
We know that phosphorus oxy-anions all have –3 charges, so the acid that forms must have
the following form: H3POx. We will now write the overall equation using this form:
P4O10 (s) + H2O (l)  H3POx (aq)
Next, we balance the non-metal atom, P. There are 4 on the left so we need 4 of the acid:
1 P4O10 (s) + H2O (l)  4 H3POx (aq)
All of the H atoms are determined on the right – there are 12 H atoms there – so we need 12
H atoms on the left, i.e., 6 H2O molecules:
1 P4O10 (s) + 6 H2O (l)  4 H3POx (aq)
The only atom that has not yet been balanced is the O atom. There are 10 in the oxide and 6
total from the water molecules, giving a total of 16 O atoms on the left. We need to ensure
that there are 16 on the right. Thus 4x = 16, and thus x = 4. We now have the completed and
balanced equation:
P4O10 (s) + 6 H2O (l)  4 H3PO4 (aq)
III. Practice Problems and Solutions:
Try the following practice problems on your own, making sure that you predict the correct product
in each case and the correct balancing coefficients. Note that some of these are acidic oxides and
some are basic oxides, it is up to you to figure out which is which. The answers are given on the
next page. Don’t look at them until after you have tried each problem on your own.
As an added exercise, name the product that is formed in each case.
1.
SO3 (g) + H2O (l) 
2.
SO2 (g) + H2O (l) 
3.
Cr2O (s) + H2O (l) 
4.
P2O3 (s) + H2O (l) 
5.
MnO2 (s) + H2O (l) 
6.
SiO2 (s) + H2O (l) 
7.
Cl2O7 (g) + H2O (l) 
8.
Br2O3 (g) + H2O (l) 
9.
Cu2O (s) + H2O (l) 
10. Ni2O3 (s) + H2O (l) 
11. N2O3 (s) + H2O (l) 
12. CO (g) + H2O (l) 
–3–
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IV. Answers to Practice Problems
1.
SO3 (g) + H2O (l)  H2SO4 (aq) – sulfuric acid
2.
SO2 (g) + H2O (l)  H2SO3 (aq) – sulfurous acid
3.
Cr2O3 (s) + H2O (l)  2 Cr(OH)3 (aq) – chromium (III) hydroxide
4.
P2O3 (s) + 3 H2O (l)  2 H3PO3 (aq) – phosphorous acid
5.
MnO2 (s) + 2 H2O (l)  Mn(OH)4 (aq) – manganese (IV) hydroxide
6.
SiO2 (s) + H2O (l)  H2SiO3 (aq) – siliconic acid
7.
Cl2O7 (g) + H2O (l)  2 HClO4 (aq) – perchloric acid
8.
Br2O3 (g) + H2O (l) 2 HBrO2 (aq) – bromous acid
9.
Cu2O (s) + H2O (l)  2 Cu(OH)2 (aq) – copper (I) hydroxide
10. Ni2O3 (s) + 3 H2O (l)  2 Ni(OH)3 (aq) – nickle (III) hydroxide
11. N2O3 (s) + H2O (l)  2 HNO2 (aq) – nitrous acid
12. CO (g) + H2O (l)  H2CO2 (aq) – carbonous acid
–4–
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