Neutralization and salts: write neutralization reactions of an acid with a base to form a salt and water, and identify the salt produced.

What this topic is asking

The Core Curriculum asks you to describe neutralization, the reaction of an acid with a base to form a salt and water, and to identify the salt produced. The Regents tests this as a Part A definition and a Part B-2 "complete the equation and name the salt" task. It builds on the Arrhenius definitions and feeds into titration.

What neutralization is

The general pattern is acid $+$ base $\rightarrow$ salt $+$ water. The reaction is so named because the acidic and basic properties cancel: the $\text{H}^+$ and $\text{OH}^-$ ions, responsible for acidity and basicity, combine to form neutral water. The salt formed may be acidic, basic or neutral in solution, but the core process produces water plus an ionic salt.

How the ions rearrange

So the products are determined by which acid and base react: the acid supplies the anion, the base supplies the cation, and together they make the salt, while $\text{H}^+$ and $\text{OH}^-$ make water. To write the salt's formula, balance the charges of its cation and anion as on the formula-writing page.

Predicting the salt

To find the salt from a neutralization:

1. Identify the cation from the base (the metal, or ammonium).
2. Identify the anion from the acid (what is left after the $\text{H}^+$).
3. Combine them in a neutral ratio (balance the charges).

For example, sulfuric acid ($\text{H}_2\text{SO}_4$) with potassium hydroxide ($\text{KOH}$): the cation is $\text{K}^+$ and the anion is sulfate $\text{SO}_4^{2-}$, so the salt is $\text{K}_2\text{SO}_4$, and the balanced equation is $\text{H}_2\text{SO}_4 + 2\,\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\,\text{H}_2\text{O}$.

Try this

Q1. State the two products of any neutralization reaction. [1 point]

• Cue. A salt and water.

Q2. Identify the salt formed when hydrochloric acid reacts with potassium hydroxide. [1 point]

• Cue. Potassium chloride, $\text{KCl}$ (from $\text{K}^+$ and $\text{Cl}^-$).