8.3: Neutralization
- Page ID
- 52179
Skills to Develop
- Explain what is meant by a neutralization reaction.
- Write the balanced equation for the reaction that occurs when an acid reacts with a base.
Neutralization is a reaction between an acid and a base that produces water and a salt. The general reaction for the neutralization reaction is shown below.
\[\text{acid} + \text{base} \rightarrow \text{salt} + \text{water}\]
In this section, we will be writing the products of neutralization reactions.
Neutralization Reactions
Acids are a combination of hydrogen ions (\(\ce{H^+}\)) and an anion. Examples include \(\ce{HCl}\), \(\ce{HNO_3}\), and \(\ce{HC_2H_3O_2}\). Bases can be a combination of metal cations and hydroxide ions, \(\ce{OH^-}\). Examples include \(\ce{NaOH}\), \(\ce{KOH}\), and \(\ce{Mg(OH)_2}\). According to the Arrhenius definitions of acids and bases, the acid will contribute the \(\ce{H^+}\) ion that will react to neutralize the \(\ce{OH^-}\) ion, contributed by the base, to produce neutral water molecules.
All acid-base reactions produce salts. The anion from the acid will combine with the cation from the base to form the ionic salt. Look at the following equations. What do they have in common?
\[\ce{HClO_4} + \ce{NaOH} \rightarrow \ce{NaClO_4} + \ce{HOH}\]
\[\ce{H_2SO_4} + 2 \ce{KOH} \rightarrow \ce{K_2SO_4} + 2 \ce{HOH}\]
(Note: \(\ce{HOH}\) is the same as \(\ce{H_2O}\))
No matter what the acid or the base may be, the produces of this type of reaction will always be a salt and water. The \(\ce{H^+}\) ion from the acid will neutralize the \(\ce{OH^-}\) ion from the base to form water. The other produce is a salt formed when the cation of the base combines with the anion of the acid. Remember, the total charge on the salt MUST be zero. You must have the correct number of cations and anions to cancel out the charges of each.
Example 8.3.1
Complete the following neutralization reactions.
a) \(\ce{H_2SO_4} + \ce{Ba(OH)_2} \rightarrow\)
b) \(\ce{HCOOH} + \ce{Ca(OH)_2} \rightarrow\)
c) \(\ce{HCl} + \ce{NaOH} \rightarrow\)
Solution:
a) The \(\ce{H^+}\) in \(\ce{H_2SO_4}\) will combine with the \(\ce{OH^-}\) part of \(\ce{Ba(OH)_2}\) to make water (\(\ce{H_2O}\) or \(\ce{HOH}\)). The salt produced is what is formed when \(\ce{Ba^{2+}}\) (the cation from the base) combines with \(\ce{SO_4^{2-}}\) (the anion from the acid). These have charges of \(+2\) and \(-2\), so the formula for this compound is \(\ce{BaSO_4}\).
Before it is balanced, the reaction is:
\[\ce{H_2SO_4} + \ce{Ba(OH)_2} \rightarrow \ce{BaSO_4} + \ce{H_2O}\]
After balancing, we get:
\[\ce{H_2SO_4} + \ce{Ba(OH)_2} \rightarrow \ce{BaSO_4} + 2 \ce{H_2O}\]
b) The \(\ce{H^+}\) in \(\ce{HCOOH}\) will combine with the \(\ce{OH^-}\) part of \(\ce{Ca(OH)_2}\) to make water (\(\ce{H_2O}\) or \(\ce{HOH}\)). The salt produced is what is formed when \(\ce{Ca^{2+}}\) (the cation from the base) combines with \(\ce{COOH^-}\) (the anion from the acid). These have charges of \(+2\) and \(-1\), so the formula for this compound is \(\ce{Ca(COOH)_2}\).
Before it is balanced, the reaction is:
\[\ce{HCOOH} + \ce{Ca(OH)_2} \rightarrow \ce{Ca(COOH)_2} + \ce{H_2O}\]
After balancing, we get:
\[2 \ce{HCOOH} + \ce{Ca(OH)_2} \rightarrow \ce{Ca(COOH)_2} + 2 \ce{H_2O}\]
c) The \(\ce{H^+}\) in \(\ce{HCl}\) will combine with the \(\ce{OH^-}\) part of \(\ce{NaOH}\) to make water (\(\ce{H_2O}\) or \(\ce{HOH}\)). The salt produced is what is formed when \(\ce{Na^+}\) (the cation from the base) combines with \(\ce{Cl^-}\) (the anion from the acid). These have charges of \(+1\) and \(-1\), so the formula for this compound is \(\ce{NaCl}\).
Before it is balanced, the reaction is:
\[\ce{HCl} + \ce{NaOH} \rightarrow \ce{NaCl} + \ce{H_2O}\]
The reaction is already balanced, so we are done.
Lesson Summary
- A neutralization reaction between an acid and a base will produce a salt and water.
Vocabulary
- Neutralization: A reaction between an acid and a base that produces water and a salt.