# Calculate the concentration of the hydrochloric acid, in mol L^-1.In order to standardise a solution of hydrochloric acid, a student titrated the solution against 20.00 mL aliquots of a standard...

Calculate the concentration of the hydrochloric acid, in mol L^-1.

In order to standardise a solution of hydrochloric acid, a student titrated the solution against 20.00 mL aliquots of a standard solution of sodium carbonate. Methyl orange indicator was used to identify the end point of the reaction:

2HCl (aq) + Na2CO3 (aq) à 2NaCl (aq) + H2O (l) +CO2 (g)

The sodium carbonate solution had been prepared by dissolving 1.236 g of anhydrous Na2CO3 in water and making the solution up to 250.0 mL in a volumetric flask. The titres recorded were 21.56 mL, 20.98 mL, 20.96 mL and 21.03 mL.

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The sodium carbonate solution was prepared by dissolving 1.236 g of anhydrous Na2CO3 in water and making the solution up to 250 mL later.

Sodium carbonate has a molar mass of 105.9884 g/mol. The mass of the salt used for making the solution is 1.236 g. This is equivalent to 1.236/105.9884 moles = 0.01166 moles of sodium carbonate.

0.01166 moles of the salt is dissolved in a volume of water equal to 250 mL or 0.250 L.

The concentration of the solution of sodium carbonate in mole/liter is 0.01166/(1/4) = 0.01166*4 = 0.04664 M

This was used in the titration of the HCl, 20 mL of the Na2CO3 solution was used to neutralize 21.56 mL, 20.98 mL, 20.96 mL and 21.03 mL of the HCl solution. The average volume of the HCl solution was 21.1325 mL.

One mole of Na2CO3 can neutralize 2 moles of HCl. The concentration of the HCl solution was 2*(21.1325/20)*0.4664 = 0.9856 M

The concentration of the HCl solution is 0.9856 mole/liter.