Relationship between Normality and Molarity of Solutions

Molarity is the number of moles of a solute dissolved in a liter of solution. The normality of a solution is the gram equivalent weight of a solute per liter of solution.

Normality and Molarity are related as follows,

Normality = Molarity x (Molar Mass / Equivalent Mass)

For Acids, the relationship between normality and molarity is given as follows,

Normality = Molarity x Basicity

where, basicity is the number of H⁺ ions a molecule of an acid can give.

For bases, the relationship between normality and molarity is given as follows,

Normality = Molarity x Acidity

where, acidity is the number of OH^- ions a molecule of a base can give.

Given below are some example problems on normality and molarity.

Example 1:

Commercially available concentrated hydrochloric acid contains 38% HCl by mass.

(a) What is the molarity of this solution? The density is 1.19 g/mL.

(b) What volume of concentrated HCl is required to make 1.00 L of 0.10 M HCl?

Solution:

(a) 38% solution, means 38 g of HCl in 100 g of solution.

Then, Mass of the solution= 100 g

Volume of the Solution = 100 g / density = 100 g / (1.19 g / mL) = 84.0 mL

= 84.0 / 1000 L

Molar mass of HCl = 36.5 g mol^-1

No. of Mole of HCL dissolved = 38 / 36.5 mol = 1.04 mol

Molarity of HCL Solution = 1.04 mol / (84/1000)L = 12.38 mol L ^-1

(b) The molarity of conc. HCl sample = 12.38 mol/L

Molarity of HCl solution to be prepared = 0.10 mol/L

Volume of HCl solution to be prepared =1.00 L= 1000mL

Then, using molarity equation, M₁V₁ = M₂V₂

V _HCL = (0.10x1000)/12.38 mL = 8.08 mL

Thus, to obtain 1.0 L of 0.10 M HCl, one should dissolve 8.08 mL of concentrated HCl to make up the volume to 1.0L.

Example 2:

Concentrated nitric acid used as a laboratory reagent is usually 69% by mass of nitric acid. Calculate the volume of the solution which contained 23 g HNO₃. Density of the concentrated HNO₃ solution is 1.41 g cm^-3.

Solution:

Let, mass of conc. HNO₃ sample = 100g

Mass of HNO₃ in 100 g of sample = 69 g

Mass of water in 100 g of sample = 31 g

Density of conc. HNO₃ sample = 1.41 g cm^-3

Volume of 100g of the HNO₃ sample = mass / density = 100g / 1.41 gcm ^-3

= 70.92 cm³

Thus, 69 g of HNO₃ is contained in 70.92 cm³ of conc. HNO₃

1 g of HNO3 is contained in 70.92 / 69 cm³ of conc. HNO₃ 23 g of HNO₃ is contained in (70.92/69) x 23 cm³ of conc. HNO₃ = 23.6 cm³

Thus 23.6 cm³ concentrated HNO₃ sample contained 23 g of HNO₃.