# Standardization of Hydrochloric Acid using Sodium Carbonate

Standardization of Hydrochloric Acid using Sodium Carbonate

Determine the normality of hydrochloric acid using standard sodium carbonate solution containing 5.1 gL-1

Principle:

The determination of normality is based on the reaction between sodium carbonate and hydrochloric acid as follows.

Na2CO3 +2HCl => 2NaCl + H2O + CO2

It is clear from the equation that one mole of sodium carbonate (two equivalents) requires two moles of hydrochloric acid for complete reaction. Therefore, the equivalent mass of hydrochloric acid is same as its molecular mass.

Eq. mass of Na2CO3 = Molecular Mass/2 = 106/2 = 53

Eq. mass of HCl = Molecular Mass = 36.5

A known volume of standard sodium carbonate solution is titrated against the given hydrochloric acid using methyl orange as indicator. From the titre value, the normality and the mass per litre of hydrochloric acid are calculated.

 Indicator Methyl orange End point Yellow to pale orange red In burette HCl solution In pipette Na2CO3 solution

Procedure:

A clean burette is rinsed and then filled with the acid solution without air bubbles. A clean pipette is rinsed with the sodium carbonate solution. 20 mL of the standard sodium carbonate solution is then pipette out into a conical flask. One or two drops of methyl orange indicator are added to it to impart a golden yellow colour to the solution. It is then titrated against hydrochloric acid taken in the burette.

Wash the sides of the conical flask with water using a wash bottle near the end point. At the end point, the colour of the solution changes from golden yellow to pale orange red. The titration is repeated to get concordant values. The titre values are tabulated as shown below. From the values, the normality and the mass per litre of hydrochloric acid are calculated.

Na2CO3 x HCl

 Exp No: Volume of Na2CO3 Burette reading Volume of HCl used Initial Final 1 2 3 20 20 20

Calculation:

Volume of sodium carbonate solution, V1 = 20 mL

Normality of Na2CO3 solution, N1 = Mass per litre/Eq. Mass = 5.1/53 = 0.0962

Let volume of hydrochloric acid = V2 mL

Let normality of hydrochloric acid = N2

N1 x V1 = N2 x V2

Therefore, Normality of HCl, N2 = 0.0962 x 20/ V2

Mass per litre of HCl = N2 x 36.5 g

Result:

Normality of HCl = _______

Mass per litre of HCl = _______ g