Lesson 7:
Acids and Bases



While pH is used to record the acidity or alkalinity of natural waters, we use a measurement known as normality to show the concentration of the much stronger acid and base solutions we use in the lab.  Normality is based on molarity, but also takes into account a characteristic of acids and bases which we will call "equivalents" and will describe in the next section. 

Normality, equivalents and equivalent weight are all related terms typically used in titrations when the titration reaction is unknown or just not used. Consequently, definitions for these terms vary depending on the type of chemical reaction that is being used for the titration. The two most common types of reactions for which normality is used are acid-base reactions and redox (reduction-oxidation) reactions.

The basic unit for normality related conventions is the equivalent. Equivalents are comparable to moles and used to relate one substance to another. Normality is a measure of concentration equal to the gram equivalent weight per liter of solution. Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.

We have already determined in a previous lesson that molarity of a solution refers to its concentration (the solute dissolved in the solution). The normality of a solutionr refers to the number of equivalents of solute per Liter of solution. The definition of chemical equivalent depends on the substance or type of chemical reaction under consideration. Because the concept of equivalents is based on the reacting power of an element or compound, it follows that a specific number of equivalents of one substance will react with the same number of equivalents of another substance. When the concept of equivalents is taken into consideration, it is less likely that chemicals will be wasted as excess amounts. Keeping in mind that normality is a measure of the reacting power of a solution, we use the following equation to determine normality:



Example 1:

If 2.0 equivalents of a chemical are dissolved in 1.5 L of solution, what is the normality of the solution?



Example 2:

A 800-mL solution contains 1.6 equivalents of a chemical. What is the normality of the solution?

First convert 800 mL to Liters:

800 mL / 1000 mL = 0.8 L


Now calculate the normality of the solution:



The reactive capacity of a chemical species, the ions or electrons, depends on what is being transferred in a chemical reaction. In acid-base reactions, an equivalent is the amount of a substance that will react with one mole of hydrogen ions. In oxidation-reduction (redox) reactions, where electrons are either gained or lost in a chemical reaction, it is one mole of electrons. Finding equivalents depends on the chemical species under consideration.

The oxidation state of an element describes the number of electrons transferred in reactions. For example, the oxidation, or valence states, of the following elements are equal to the number of equivalents:

Calcium: (Ca+2) ion: valence of 2; number of equivalents = 2

Aluminum: (Al+3) ion: valence of 3; number of equivalents = 3

For acids, an equivalent is the number of hydrogen ions a molecule transfers. In acids it is straightforward to find equivalent units. Look at the number directly after the hydrogen (H) in the chemical formulas below. The number provides the number of equivalents per mole of that acid:

Hydrochloric acid (HCl): equivalents = 1

Sulfuric acid (H2SO4): equivalents = 2

Phosphoric acid (H3PO4): equivalents = 3

Nitric acid (HNO3): equivalents = 1


For bases, it is the number of hydroxide ions (OH-) provided for a reaction, such as:

Sodium hydroxide (NaOH): equivalents = 1

Barium hydroxide (Ba(OH)2): equivalent = 2


One equivalent of an acid reacts with one equivalent of a base. For the acid HCl and base NaOH, both with one equivalent, they have the same reactivity. For H2SO4, with two equivalents, and NaOH, it will take twice the amount of the NaOH to react with the sulfuric acid. Mixing equal equivalents of acidic and basic solutions will result in a neutral solution.

Altough molarity can be used to measure the concentration of acids, it is a relatively unuseful measurement for understanding neutralization reactions.  Why?  Because not every acid or base can add (or remove) the same number of hydrogen ions from solution. 



Equivalent Weight

The equivalent weight can be thought of as the weight (or mass, to be precise) of a substance that will contain a single reactive proton (or hydrogen ion) or a single reactive hydroxide ion. The former case applies to acids, which are proton donors, while the second applies to bases, which are proton acceptors. The reason the concept of equivalent weight is needed is that some compounds can donate or accept more than one proton, meaning that for every mole present, the substance is in effect doubly reactive. The equivalent weight can be determined by:



Equivalent weight is defined as the ratio of molar mass of a substance to the valence of the substance. Valence is also denoted as equivalence factor. The valence is the number of hydrogen atoms in an acid, or hydroxide atoms in a base, and for salt, charge present in ionic forms. For reduction/oxidation (redox) reactions, it is the number of electrons than an oxidizing or reducing agent can accept or dontate that are counted as valence or equivalence factor. When the equivalent weight is expressed in grams using molar mass in grams, it is called gram equivalent weight.

Let's get some practice determining the equivalent weight for the following formulas:


Molar mass of H2SO4 = 98 g/mol

Looking at the formula, there are 2 hydrogen atoms, so "n" will be 2 when determine the equivalent weight:



Molar mass of NaCl = 58.5 g/mol

Looking at the formula, because there are no hydrogen or hydroxide atoms, the number of equivalents is 1, because there will always be at least one equivalent before a rection can occur.



Molar mass of NaOH = 40 g/mol

In looking at the formula, there is 1 hydroxide (OH) atom, so the equivalent is 1.


Now let's look at a salt (a salt determines its equivalents differently because there are no hydrogen or hydroxide atoms involved, so we look at the charge):


Molar mass of Na2CO3 = 106 g/mol

The salt Na2CO3 ionizes to form 2Na+ and CO3-2, so the charge present on both is 2.



Making Normal Solutions (N)

Normality is the most common measurement used for showing the concentration of acids and bases.  Normality takes into account both the molarity of the solution and the equivalent content of the acid or base. It is defined as the number of gram equivalent present in per liter solution and can be determined through the following formula:


To calculate normality of the solution, follow these steps:


Making normal solutions can be a bit confusing. Aqueous solutions of acids and bases are often described in terms of their normality rather than their molarity. In order to properly make a Normal solution, the student must understand the difference between a pure reagent and a diluted reagent.

A "1 Normal" solution (1 N) contains 1 "gram equivalent weight" of solute, topped-off to one liter of solution. The gram equivalent weight is equal to the solute's molecular weight (molar mass), expressed as grams, divided by the valence (n) of the solute:


After the equivalent weight (or milliequivalent weight) has been calculated, then the following equation is used:



The equivalent weight of a substance depends upon the type of reaction in which the substance is taking part. Some different types of chemical reactions, along with how to determine a solute's equivalent weight for each reaction, are given below.




Making a Normal Solution With Salts


Calculate the normality of a sodium chloride (NaCl) solution prepared by dissolved 2.9216 grams of NaCl in water and then topping it off with more water to a total volume of 500.0 mL.

First, check the periodic table to determine the molar mass, or molecular weight, of NaCl, which is 58.44.

In looking at the formula (NaCl), the equivalent is 1 because there is room in the molecule for only one replaceable (H+) ion. In other words, one hydrogen atom can replace the sodium atom in NaCl.

Now determine the equivalent weight of NaCl:


You will need to know the milliequivalent weight of NaCl (since the solute volume is in mL):


Next calculate the normality of the sodium chloride solution:


Making Normal Solutions With Pure (Non-Aqueous) Acids

The equivalent weight of an acid is its molecular weight, divided by the number of replaceable hydrogen atoms in the reaction. To clarify this concept, consider the following acids:

Hydrochloric acid (HCl) has one replaceable hydrogen ion (H+). Sulfuric acid (H2SO4) has two replaceable hydrogen ions (2H+). The valences of these acids are determined by their respective replaceable hydrogen ions, displayed below.

Acid Valence
(Replaceable hydrogen ions)
HCl n = 1
HNO3 n = 1
H2SO4 n = 2
HF n = 1


So, for pure HCl, its molecular weight is 36.46, its equivalent weight is 36.46 and therefore a 1N solution would be 36.46 grams of the pure chemical per liter. Note that, in the case of HCl, a 1N solution has the same concentration as a 1M solution.

To make a 1N H2SO4 solution from pure H2SO4, its molecular weight is 98.08, and its equivalent weight is (98.08/2 = 49.04 grams/Liter or 49.04 grams per 1000 mL). So a 1N solution would be 49.04 grams of the pure chemical per liter.




Making Normal Solutions From Pure Alkalis (Bases)

The equivalent weight of a base is defined as "its molecular weight divided by the number of hydrogen ions that are required to neutralize the base". To understand the valences of alkalis, consider the following examples:

The (OH-) ion in Sodium hydroxide (NaOH) can be neutralized by one hydrogen ion. the (OH)2-- ions in Calcium hydroxide (Ca(OH)2) can be neutralized by two hydrogen ions. As was the case with acids, the valences (n) of these bases are determined by their respective replaceable hydrogen ions, displayed below.

Base Valence
(Replaceable hydrogen ions)
NaOH n = 1
Ca(OH)2 n = 2


So, for NaOH, its molecular weight is 40, its equivalent weight is 40, and therefore a 1N solution would be 40 grams of the pure chemical per Liter of water. You will also note, in the case of NaOH, a 1N solution is the same concentration as a 1M solution.

For Ca(OH)2, its molecular weight is 74, its equivalent weight is (74/2 = 37, because n = 2). Therefore, a 1N aqueous solution of Ca(OH)2 is 37 grams of the pure chemical per Liter of water. Of course, many acidic reagents and basic reagents come from the factory in a diluted aqueous form, which is a form of preparation we will not cover in this lesson.



Calculate the normality of a NaOH solution formed by dissolving 0.2 g NaOH to make a 250 mL solution.

First we need to determine the molar mass, which is 40 g/mol.

Then determine the milliequivalent weight (since the solution is in mL):


Now, calculate the normality of the NaOH solution:


Relationship Between Normality and Molarity

Here is Normality in terms of molarity:

Normality = n x Molarity

Where n = number of Hydrogen in acids, or Hydroxides in bases and for salt, charge present in ionic forms.


Calculating Dilutions (Acid-Base Titration)

Once you have calculated the normality of an acid or base solution, you can easily calculate the concentration of any dilutions of that solution.  The formula used is essentially the same as that used for any other dilution calculation:

N1V1 = N2V2


N1 = normality of the first solution
V1 = volume of the first solution
N2 = normality of the second solution
V2 = volume of the second solution


Example 1:

After producing the 0.5 L of a 0.37N solution of calcium hydroxide in the last section, how would you dilute it to form a 0.25N solution? 

(0.37N) (0.50 L) = (0.25N) V2

(0.37N)(0.50 L) / 0.25N = V2

0.74 L = V2

Based on the calculations above, we know that we have to add enough water to the 0.37N solution so that the total volume reaches 0.74 L.  Then we would have a 0.25N solution. 



Example 2:

How many milliliters of 2N NaOH are needed to prepare 300 mL of 1.2N NaOH?

N1V1 = N2V2

(2N)(V1) = (1.2N)(300 mL)

V1 = (1.2N)(300 mL) / 2N

V1 = 180 mL

So, to prepare the 1.2N NaOH solution, you pour 180 mL of 2N NaOH into your container and add water to get 300 mL total volume.



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Part 4: Alkalinity