Question

1. 0.1M HCl solution is prepared by 120-fold dilution of concentrated HCl. it is standardized by titrating 0.1876g of dried primary standard sodium carbonate: CO₃^2-+2H⁺=H2O+CO2 The titration required 35.86mL acid.

a)Calculate the molar concentration of the HCl

b) what is the substance that is being standardized?

c) what is the primar standard?

2.0.4671g sample containing sodium bicarbonate was dissolved and titrated with standard 0.1067M HCl solution, requiring 40.72mL. The reaction is HCO3-+H+=H2O+CO2

Calculate the percent of sodium bicarbonate in the sample.

3. 0.2638g sample is analyzed by titrating the sodium carbonatae with the standard 0.1288M HCl, requiring 38.27mL. The reaction is CO32-+2H+=H2O+CO2

Calculate the percent of sodium carbonate in the sample.

Answer 1

1)

a) 0.1876g of sodium carbonate is 0.1876g/106 g/mol = 0.001769 moles

The reaction between sodium carbonate and HCl goes by the equation

CO₃^2-+   2H⁺-----> H₂O+CO₂

So 1 mole of sodium carbonate needs 2 moles of acid

So the amount of acid needed is 0.001769 x 2 = 0.003539 moles

since the amount of acid needed for the titration is 35.86 mL the concentration of the acid will be

(c x volume)/1000 = 0.003539

c x 35.86 mL = 3.539

c = 3.539/35.86

c = 0.0987 M is the actual molar concentration of the HCl.

b) The concentration of the hydrochloric acid is being standardised so this is a secondary stahdard as it needs a primary standrd to get the correct concentration.

c) The primary standard is anhydrous sodium carbonate.

Please post the other questions separately.