Neutralization and Salts

20min Part 1 / Ch2 / Lesson 5

Prerequisites: 1-2-4

Objectives

Explain how neutralization reactions work
Calculate quantities in neutralization reactions
Classify salts and predict solution properties

What Is Neutralization?

Neutralization: $\mathrm{H^+}$ from an acid reacts with $\mathrm{OH^-}$ from a base to form water.

$\mathrm{H^+ + OH^- \rightarrow H_2O}$

A salt is also produced.

Examples: $\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}$ $\mathrm{H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O}$

Quantitative Relationship

Neutralization is complete when: mol of $\mathrm{H^+}$ from acid = mol of $\mathrm{OH^-}$ from base

$a \cdot c_a \cdot V_a = b \cdot c_b \cdot V_b$

$a$ : valence of acid, $b$ : valence of base, $c$ : molarity, $V$ : volume

Quantitative relationship in neutralization

Types of Salts

Salt type	Formed from	Solution pH	Examples
Normal salt (neutral)	Strong acid + Strong base	Neutral	NaCl, $\mathrm{K_2SO_4}$
Normal salt (acidic)	Strong acid + Weak base	Acidic	$\mathrm{NH_4Cl}$
Normal salt (basic)	Weak acid + Strong base	Basic	$\mathrm{CH_3COONa}$
Acid salt	H⁺ remains	Usually acidic	$\mathrm{NaHSO_4}$ , $\mathrm{NaHCO_3}$ *

Salt classification and solution properties

* $\mathrm{NaHCO_3}$ (sodium bicarbonate) is an “acid salt” by name but its solution is weakly basic — an important exception.

Worked Example

Example: Neutralization calculation

How many mL of $0.20\ \mathrm{mol/L}$ NaOH are needed to neutralize $20\ \mathrm{mL}$ of $0.10\ \mathrm{mol/L}$ HCl?

HCl is monoprotic ( $a = 1$ ), NaOH is monobasic ( $b = 1$ )

$1 \times 0.10 \times 20 = 1 \times 0.20 \times V_b$

$V_b = \frac{0.10 \times 20}{0.20} = 10\ \mathrm{mL}$

Check Your Understanding

Q1 What is always produced in a neutralization reaction?

Q2 How many mL of 0.20 mol/L NaOH neutralize 50 mL of 0.10 mol/L H₂SO₄?

Q3 A solution of CH₃COONa is:

Exercises

Q1. $10.0\ \mathrm{mL}$ of acetic acid with unknown concentration is titrated with $0.10\ \mathrm{mol/L}$ NaOH. The equivalence point is reached at $15.0\ \mathrm{mL}$ . Find the molarity of the acetic acid.

Solution

Acetic acid is monoprotic, NaOH is monobasic:

$1 \times c_a \times 10.0 = 1 \times 0.10 \times 15.0$

$c_a = \frac{0.10 \times 15.0}{10.0} = 0.15\ \mathrm{mol/L}$