General Chemistry/Reactions of Acids and Bases

Overview

To summarize the properties and behaviors of acids and bases, this chapter lists and explains the various chemical reactions that they undergo. You may wish to review chemical equations and types of reactions before attempting this chapter.

The following reactions are net ionic equations. In other words, spectator ions are not written. If an ion does not partake in the reaction, it is simply excluded. The spectator ions can be found because they occur on both the reactant and the product side of the equation. Cross them out and rewrite the equation without them. Of course, the coefficients must be equal.

Canceling out the spectator ions explains the net of net ionic equations. The ionic part means that dissolved compounds are written as ions instead of compounds. Acids, bases, and salts are all ionic, so they are written as separate ions if they have dissociated.

Net Ionic Equations

• Soluble salts are written as ions.

e.g.: Na⁺ + Cl^-

• Solids, liquids, and gases are written as compounds.

e.g.: NaCl_(s), H₂O_(l), HCl_(g)

• Strong acids and strong bases are written as ions (because they dissociate almost completely).

e.g.: H⁺ + NO₃^-

• Weak acids and weak bases are written as compounds (because they barely dissociate).

e.g.: HNO₂

As an example, sodium bicarbonate (NaHCO₃) would be written as Na⁺ and HCO₃^- because the salt will dissociate, but the bicarbonate will not dissociate (it's a weak acid).

Neutralization

When an acid and a base react, they form a neutral substance, often water and a salt.

First, let's examine the neutralization of a strong acid with a strong base.

${\hbox{KOH}}+{\hbox{H}}^{+}+{\hbox{Cl}}^{-}$	Solid potassium hydroxide is added to an aqueous solution of hydrochloric acid. Notice how the solid is written as a compound, but the acid is written as ions because it dissociates.
${\hbox{K}}^{+}+{\hbox{OH}}^{-}+{\hbox{H}}^{+}+{\hbox{Cl}}^{-}\to {\hbox{K}}^{+}+{\hbox{Cl}}^{-}+{\hbox{H}}_{2}{\hbox{O}}$	The hydrogen ions will react with hydroxide ions to form water.
${\hbox{H}}^{+}+{\hbox{OH}}^{-}\to {\hbox{H}}_{2}{\hbox{O}}$	Ignoring spectator ions, this is the net ionic equation.

Now, let's see some examples involving weak acids and weak bases.

${\hbox{PO}}_{4}^{3-}+3{\hbox{H}}^{+}\to {\hbox{H}}_{3}{\hbox{PO}}_{4}$	Excess hydrochloric acid is added to a solution of sodium phosphate. Phosphoric acid is weak, so the phosphate ions will react with hydrogen ions. The result is a solution with some, but much less, hydrogen ions, so it is much closer to neutral than either of the original reactants.
${\hbox{PO}}_{4}^{3-}+{\hbox{H}}^{+}\to {\hbox{H}}{\hbox{PO}}_{4}^{-2}$	Equimolar amounts of sodium phosphate and hydrochloric acid are mixed. Notice the difference between this reaction and the previous one.
${\hbox{HCO}}_{3}^{-}+{\hbox{OH}}^{-}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{CO}}_{3}^{-2}$	A strong base is added to a solution of calcium bicarbonate. (Bicarbonate is a weak acid.)
${\hbox{HCO}}_{3}^{-}+{\hbox{H}}^{+}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{CO}}_{2}$	A strong acid is added to a solution of calcium bicarbonate. Gas bubbles appear.

Many reactions result in the formation of gas bubbles or a solid precipitate that will make the solution cloudy. The last equation brings up an interesting application. Many rocks and minerals contain calcium carbonate or calcium bicarbonate. To identify these rocks, geologists can perform the "acid test". A drop of acid is applied, and the presence of gas bubbles indicates carbonate.

Here are more examples of neutralization reactions.

${\hbox{NH}}_{4}{\hbox{Cl}}+{\hbox{OH}}^{-}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{NH}}_{3}+{\hbox{Cl}}^{-}$	Solid ammonium chloride crystals are dissolved into a solution of sodium hydroxide. The smell of ammonia is detected.
${\hbox{NH}}_{3}+{\hbox{H}}^{+}\to {\hbox{NH}}_{4}^{+}$	Ammonia gas is bubbled through a solution of hydrochloric acid. This reaction is essentially the opposite of the previous. In that reaction, ammonium ions react with base to form ammonia gas. In this reaction, ammonia gas reacts with acid to form ammonium ions.
${\hbox{NH}}_{3}+{\hbox{CH}}_{3}{\hbox{COOH}}\to {\hbox{NH}}_{4}^{+}+{\hbox{CH}}_{3}{\hbox{COO}}^{-}$	Ammonia (a weak base) reacts with acetic acid (also weak). The resulting solution is nearly neutral, but it will be slightly basic because ammonia is stronger than acetic acid.
${\hbox{H}}_{2}{\hbox{S}}+2{\hbox{OH}}^{-}\to 2{\hbox{H}}_{2}{\hbox{O}}+{\hbox{S}}^{2-}$	Hydrogen sulfide gas is bubbled into a strong base.
$2{\hbox{H}}^{+}+{\hbox{S}}^{2-}\to {\hbox{H}}_{2}{\hbox{S}}$	A strong acid is added to the above result, and hydrogen sulfide gas is released.

Anhydrides

An anhydride is a substance that does not contain water. More specifically, it is a substance that reacts with water to form an acid or base. Anhydrides are usually in the form of a gas that dissolves into water and reacts to form an acid or base. They can also be solids that will react with water.

${\hbox{N}}_{2}{\hbox{O}}_{5}+{\hbox{H}}_{2}{\hbox{O}}\to 2{\hbox{H}}^{+}+2{\hbox{NO}}_{3}^{-}$	Gaseous dinitrogen pentoxide is bubbled through water to form nitric acid.
${\hbox{N}}_{2}{\hbox{O}}_{3}+{\hbox{H}}_{2}{\hbox{O}}\to 2{\hbox{HNO}}_{2}$	Dinitrogen trioxide is mixed with water to form nitrous acid.

The main difference between those two equations is the fact that nitrous acid is weak and thus does not dissociate, whereas nitric acid is strong and dissociates into ions.

Here are a few more examples of anhydride reactions.

${\hbox{K}}_{2}{\hbox{O}}+{\hbox{H}}_{2}{\hbox{O}}\to 2{\hbox{K}}^{+}+2{\hbox{OH}}^{-}$	Solid potassium oxide is added to water to form a strong base.
${\hbox{P}}_{2}{\hbox{O}}_{5}+3{\hbox{H}}_{2}{\hbox{O}}\to 2{\hbox{H}}_{3}{\hbox{PO}}_{4}$	Phosphorus(V) oxide powder is mixed into water to form a weak acid.

It is important to remember which acids are strong and which are weak. Review this if necessary.

For example, sulfur dioxide gas (acidic anhydride) is bubbled through a solution of calcium hydroxide (basic).

${\hbox{SO}}_{2}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{H}}_{2}{\hbox{SO}}_{3}$	First, determine the reaction of the anhydride with water.
${\hbox{H}}_{2}{\hbox{SO}}_{3}+{\hbox{Ca}}({\hbox{OH}})_{2}\to {\hbox{CaSO}}_{3}+2{\hbox{H}}_{2}{\hbox{O}}$	Then, determine the reaction of the acid and base. This is a double replacement reaction.
${\hbox{SO}}_{2}+{\hbox{H}}_{2}{\hbox{O}}+{\hbox{H}}_{2}{\hbox{SO}}_{3}+{\hbox{Ca}}({\hbox{OH}})_{2}\to {\hbox{H}}_{2}{\hbox{SO}}_{3}+{\hbox{CaSO}}_{3}+2{\hbox{H}}_{2}{\hbox{O}}$	Add the two reactions together.
${\hbox{SO}}_{2}+{\hbox{Ca}}^{2-}+2{\hbox{OH}}^{-}\to {\hbox{CaSO}}_{3}+{\hbox{H}}_{2}{\hbox{O}}$	Cancel out spectators. Also, calcium hydroxide should be ionized (but calcium sulfite is a solid precipitate). This is the final net ionic equation.

Here are more examples.

${\hbox{CaO}}+2{\hbox{H}}^{+}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{Ca}}^{2+}$	Calcium oxide crystals (basic anhydrides) are added to a strong acid. Notice that it does not matter what the acid is (nitric, sulfuric, etc.) because it is strong and this reaction only requires the hydrogen ions. In other words, the anions of the strong acid are spectators and are not written.
${\hbox{SO}}_{2}+{\hbox{OH}}^{-}\to {\hbox{HSO}}_{3}^{-}$	Excess sulfur dioxide gas is bubbled into a dilute solution of strong base. The base is the limiting reactant.
${\hbox{SO}}_{2}+2{\hbox{OH}}^{-}\to {\hbox{H}}_{2}{\hbox{O}}+{\hbox{SO}}_{3}^{-}$	Sulfur dioxide gas is bubbled into an excess of basic solution.

Remember that water is involved in these reactions, but it is not written if it occurs on both sides of the equation.

${\hbox{CaO}}+{\hbox{CO}}_{2}\to {\hbox{CaCO}}_{3}$	Solid calcium oxide (basic anhydride) is exposed to dry ice gas (acidic anhydride). The resulting solid is a salt.
${\hbox{CaO}}+{\hbox{SO}}_{3}\to {\hbox{CaSO}}_{4}$	Solid calcium oxide is exposed to a stream of sulfur trioxide gas. The resulting solid is a neutral salt.

Hydrolysis

A salt of a weak acid and strong base dissociates and reacts in water to form OH^-. A salt of a strong acid and weak base dissociates and reacts in water to form H⁺. This process is called hydrolysis.

In this first example, aluminum nitrate is dissolved in water.

${\hbox{Al}}({\hbox{NO}}_{3})_{3}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Al}}^{3+}+3{\hbox{NO}}_{3}^{-}+{\hbox{H}}_{2}{\hbox{O}}$	First, the salt dissociates in the water. It isn't necessary to write H2O in this reaction.
${\hbox{Al}}^{3+}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Al}}({\hbox{OH}})^{2+}+{\hbox{H}}^{+}$	Now, at least one of the ions will react with water. You know that nitric acid is strong, so the nitrate ion will not take an H+ ion from water. Instead, the aluminum ion will react with water, releasing a hydrogen ion.
${\hbox{Al}}({\hbox{NO}}_{3})_{3}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Al}}({\hbox{OH}})^{2+}+{\hbox{H}}^{+}+3{\hbox{NO}}_{3}^{-}$	This is the net ionic equation. The resulting solution is acidic.

The solution is acidic not because nitric acid is strong, but because aluminum is a weak base.

Here is an easier example.

${\hbox{NaNO}}_{2}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Na}}^{+}+{\hbox{NO}}_{2}^{-}+{\hbox{H}}_{2}{\hbox{O}}$	First, the salt dissociates. Again, the H2O need not be written.
${\hbox{NO}}_{2}^{-}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{HNO}}_{2}+{\hbox{OH}}^{-}$	Sodium ions will not react with water. Even if they did, they would form NaOH, which is a strong base, so it would immediately dissociate. Instead, the NO2 reacts with water. Being the conjugate of a weak acid, the nitrite ions will accept a proton from water to form nitrous acid (weak) and hydroxide ions (basic).
${\hbox{NaNO}}_{2}+{\hbox{H}}_{2}{\hbox{O}}\to {\hbox{Na}}^{+}+{\hbox{HNO}}_{2}+{\hbox{OH}}^{-}$	This is the net ionic equation for the hydrolysis of sodium nitrite. The resulting solution is basic.

Lewis Acids/Bases

Lewis acids accept an electron pair. Lewis bases donate an electron pair. Together they react and bond to form an adduct.

Lewis acids/bases do not require the presence of water. However, H⁺ can be thought of as a Lewis acid because it accepts electron pairs. OH^- can donate an electron pair, making it a Lewis base.

${\hbox{BF}}_{3}+{\ddot {\hbox{N}}}{\hbox{H}}_{3}\to {\hbox{F}}_{3}{\hbox{B}}-{\hbox{NH}}_{3}$	Boron trifluoride (Lewis acid) is exposed to ammonia (a Lewis base, as shown by the electron pair over the N). The electron pair is shared between the nitrogen and boron, creating a bond. (The boron trifluoride is written backwards as F3B only to demonstrate the B-N bond. Its structure has not changed.)
${\hbox{B}}_{2}{\hbox{H}}_{6}+2{\hbox{H}}^{-}\to 2{\hbox{BH}}_{4}^{-}$	Diborane accepts the two electrons from H- and forms a Lewis adduct.

Practice Problems

Write the net ionic equations for the following. Make note of any solid precipitates or gas bubbles that would form.

1. Equimolar solutions of sodium biphosphate and potassium hydroxide are mixed.
2. Equimolar solutions of sodium biphosphate and hydrochloric acid are mixed.
3. Excess sulfur dioxide gas is bubbled into a dilute solution of sodium hydroxide. Acid is then added.
4. Aluminum chloride is dissolved into water.
5. Sodium fluoride is dissolved into water. Strong acid is then added.
6. Solid calcium oxide is exposed to a stream of sulfur trioxide gas. If the resulting compound is dissolved, will the solution be acidic, basic, or neutral?
7. Gaseous hydrogen chloride is bubbled into a solution of silver nitrate.
8. Ammonium chloride crystals are dissolved in water. Sodium hydroxide is then added.
9. Calcium hydroxide crystals are dissolved into a solution of sodium bicarbonate.
10. Phosphine gas is sprayed onto pebbles of aluminum trichloride. (Hint: these are Lewis acids/bases.)

Answers to Practice Problems