How to prepare a standard solution of an acid

   A standard solution is a solution which has a known concentration, which means the amount in mole or mass concentration of the substance which is dissolved in a known volume of a solvent is known.

   If you want to prepare a standard solution of an acid, the knowledge of the relationship between amount, molar concentration and volume is very necessary

                                     C = n/V

Where: n = amount (mole)

              V = volume (dm³)

             C = molar concentration (mole per dm³)

Experiment has shown that

       C₁V₁ = C₂V₂

Since the amount of solute in a solution is not affected by increase in the volume of the solvent. This formula becomes the basic principle of dilution. It is also known as dilution formula.

  Information required for preparation of standard solution of acid is listed below:

·         Specific gravity of the acid

·         Percentage purity of the acid

·         Molar mass of the acid

·         Concentration  of the required dilute acid

·         Volume of the dilute solution required

Specific gravity of a substance is the ratio of density of the substance to the density of water at a specified temperature. Specific gravity shows what mass of a substance is contained in 1cm³ of a solution. For example, if the specific gravity of a substance is 1.67, it means that 1.67g of that substance is contained in 1cm³ of water.

Molar mass of a substance is mass of one mole of the substance expressed in grams

The Percentage Purity and Specific gravity of common Acids

Acid	Molar mass (g/mol)	Percentage purity	Specific gravity
HCl	36.5	32% -36%	1.18
H2SO4	98	96% -98%	1.84
HNO3	63	65% -72%	1.047- 1.052

 

Example question 1

What volume of concentrated hydrochloric acid would be required to prepare 2 dm³ of 0.6mol/dm³ solution? (Specific gravity =1.18, percentage purity =36%, molar mass =36.5 g/mol)

Answer

Molar mass of HCl = 36.5g/mol

Specific gravity = 1.18

1cm³ of the stock of solution contains = 1.18g of HCl

1000cm³ solution contains = 1.18 x 1000 =1180g

Mass of pure hydrochloric acid in 1dm³ of solution = 36/100 x1180 = 424.8g

Amount of pure hydrochloric acid = 424.8/36.5 = 11.64mol

Using dilution formula

          C₁ V₁ =C₂ V₂

C₁ = 11.64 mol

V₁ = unknown

C₂ = 0.6mol/dm³

V₂ = 2000 cm³ (2dm³)

11.64 x V₁ = 0.6 x 2000

V₁   = 1200/11.64 = 103.1cm³

103.1cm³ of hydrochloric acid would be diluted to 2dm³ in order to produce 0.6mol/dm³

Example question 2

Calculate the volume of concentrated trioxonitrate (V) acid required to prepare 2dm³ of 0.5mol/dm³ of solution. (Specific gravity =1.052, Percentage purity = 72%, molar mass = 63g/mol)

Answer

Molar mass of HNO₃ = 63g/mol

Specific gravity of HNO₃ = 1.052

1cm³ of the stock solution contains = 1.052g

1000cm³ solution contains =1.052 x 1000 = 1052g

Mass of pure HNO₃  in  1dm³ solution = 72/100 x 1052 = 757.4g

Amount of HNO₃ in 1dm³ of solution = 757.4/63 = 12.02 mole

Using dilution formula

             C₁ V₁ = C₂ V₂

C₁ = 12.02 mol

V₁= unknown

C₂ = 0.5 mol/dm³

V₂ = 2000cm³ (2dm³)

12.02 x V₁ = 0.5 x 2000

12.02V₁ = 1000

V₁ = 1000/12.02 = 83.2cm³ 

83.2 cm³ of the concentrated acid would be diluted to 2dm³ in order to produce 0.5 mol/dm³ of solution.