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Standard solutions: concept, preparation, examples

When talking about the normality of a solution, it refers to the number of equivalents of a solute that it has per liter of solution. But to find this number of equivalents it is necessary to know its equivalent weight , which varies between the elements, the type of chemical compound , or even the reaction that takes place.

This is why normal solutions are generally more complicated to prepare in terms of their theoretical calculations. They are recognized because they have the ‘N’ for normality on their labels. Many acids and bases have been prepared according to this concentration; for example, NaOH 0.01 N.

Where most normal solutions are present are in the reagents used for redox reactions. Generally, they are solutions of salts such as KMnO 4 , CuSO 4 , CrCl 3 , among others.

Preparation of normal solutions

How are normal solutions prepared? Although the steps to follow are not different from those of other solutions, they will be explained below:

Step 1

Find the chemical characteristics of the reagent you want to prepare, using the information that appears on the reagent container label. The information required is the chemical formula of the reagent, its molecular weight, whether the reagent is anhydrous or not, etc.

Step 2

Carry out the necessary calculations for the preparation of the normal solutions. Normality is expressed in equivalents per liter (Eq / L) and is abbreviated with the letter ‘N’.

The calculation is started by dividing the concentration of the solution expressed in grams / liter (g / L) by the equivalent weight expressed in grams per equivalent (g / Eq). But first, the equivalent weight of the reagent must be obtained, taking into account the type of the chemical reagent.

How many grams of sodium carbonate are needed to prepare a liter of a 2 N solution, knowing that it has a molecular weight of 106 g / mol?

By definition, a normal solution (N) is expressed in equivalents / liter (Eq / L). But the number of equivalents must be calculated based on the equivalent weight of the chemical reagent. Then, the initial step of the calculation is to obtain the equivalent weight of Na 2 CO 3 .

The reagent is a salt, so its pEq is:

PM / (Sm x Vm)

The metal in Na 2 CO 3 is Na. The subscript of Na (Sm) is 2 and its valence (Vm) is 1. Therefore, Sm x Vm is equal to 2.

pEq = PM / 2

= 106 g / mol ÷ 2 Eq / mol

= 53 g / Eq

The Na 2 CO 3 solution you want to prepare is 2 N, so by definition it has a concentration of 2 Eq / L. Then, the concentration expressed in g / L can be found by using the mathematical expression:

g / L = Eq / L (N) x pEq (g / Eq)

= 2 Eq / L x 53 g / Eq

= 106

Thus, to prepare 1 liter of a 2N sodium carbonate solution, 106 g of the reagent are required.

Step 3

Weigh the calculated grams of reagent on an analytical or precision balance, carefully so as not to make weighing errors.

Step 4

Dissolve the weighed reagent in a beaker and add an adequate volume of deionized or distilled water, so that the volume in which the reagent dissolves does not exceed the stipulated volume.

Step 5

Pour the content of the beaker into a volumetric flask and add water until it reaches its capacity. Finally, the reagent volume is transferred to a suitable container for storage and use.

Examples of normal solutions

Example 1

How many grams of sodium hydroxide (NaOH) is required to prepare 1.5 liters of a 2N solution, and what volume of 1N HCl is required to completely neutralize the NaOH? Molecular weight NaOH = 40 g / mol.

Part A

The equivalent weight of NaOH is calculated as:

PEq NaOH = PM / No. OH

NaOH is a base that has only one OH.

pEq NaOH = 40 g / mol ÷ 1 Eq / mol

= 40 g / Eq

The number of grams of NaOH needed to prepare a NaOH solution can be obtained by applying the relationship:

g / L of NaOH = Normality (Eq / L) x pEq (g / Eq)

= 2 Eq / L x 40 g / Eq

= 80 g / L

Now, the grams NaOH needed to prepare 1.5 L of a 2 N NaOH solution can be obtained:

g of NaOH = 80 g / L x 1.5 L

= 120 g NaOH

Part b

A characteristic of equivalents is that a number of them react with the same number of other equivalents.

The proposed reaction is a neutralization reaction, in which an acid (HCl) reacts with a base (NaOH) to produce a salt and water. Therefore, an equivalent number of the acid (EqA) reacts with the same equivalent number of a base (EqB) to produce its neutralization.

Knowing that the equivalents are related to normality and volume through the following expression:

Eq = V x N

The volume of HCl required to neutralize NaOH can be determined:

EqA = V A x N A

EqB = V B x N B

EqA = EqB

Then,

A x N A = V B x N B

We solve for V A :

A = V B x N B / N A

In this case, hydrochloric acid (1 N) and sodium hydroxide (2 N) intervene:

A = (1.5 L x 2 Eq / L) / 1 Eq / L

= 3 L

3 liters of a 1 N HCl solution are required to neutralize 1.5 liters of a 2 N NaOH solution.

Example 2

What will be the normality of a solution of calcium chloride (CaCl 2 ) that is prepared by dissolving 120 grams of the reagent in 1.5 liters? Molecular weight of CaCl 2 = 111 g / mol

Let’s first determine the equivalent weight (pEq) of CaCl 2 . CaCl 2 is a salt, therefore:

pEq = PM / (Sm x Vm)

The metal is calcium (Ca), its subscript is 1 (Sm) and its valence is 2 (Vm). Thus, we substitute:

pEq = 111 g / mol / (1 x 2)

= 55.5 g / Eq

Finally, we proceed to determine the normality (Eq / L) of the solution. We can obtain this calculation by applying the appropriate conversion factors:

N = (120 g / 1.5 L) x (Eq / 55.5 g)

= 1.44

Therefore, by definition the normality of the CaCl 2 solution is 1.44 N