Experiment: Standardisation Of 0.25 M Hydrochloric Acid

Experiment 1: Standardisation of approximately 0.25 M Hydrochloric Acid
Abstract
The aim of the study was to standardise a solution of hydrochloric acid with an approximate concentration of 2.5 M against sodium carbonate solution. The main steps of this experiment were preparing the sodium carbonate solution, transferring the solution into a conical flask by using a pipette, adding the bromophenol blue indicator and titrating with hydrochloric acid until the first green colour appeared. The concentration of hydrochloric acid was found to be 0.2508 mol L-1.
Introduction
Standardisation is a process which is carried out to determine the actual molarity of a solution. Titration is a technique that is often used in standardisation. In titration, a solution is …show more content…
Number of moles of Na2CO3 = (2.5 g)/([2(22.99)+12.011+3(15.999)] g mol^(-1) ) = (2.5 g)/(105.988 g mol^(-1) ) = 2.359 X 10-2 mol
Molarity of Na2CO3 = (2.359 X 〖10〗^(-2) mol)/(0.25 L) = 9.436 x 10-2 mol L-1
Na2CO3 + 2HCL 2NaCl + CO2 + H2O
The 20 mL aliquot of Na2CO3 solution with concentration 9.436 x 10-2 mol L-1 was reacted completely with 15.05 mL of HCl solution with unknown concentration. Therefore, the molarity of HCl solution can be calculated by (M_A V_A)/(M_B V_B ) = a/b
((9.436 X 〖10〗^(-2) )(0.02))/(M_B (0.01505))= 1/2
MB = 0.2508 mol L-1
The molarity of HCl is 0.2508 mol L-1.
Discussion
The titration was repeated until the difference between the volumes of hydrochloric acid used less than 0.2 mL. The first result is the rough result that gives the rough determination of the endpoint. The second and third results are the accurate readings that are recorded at the endpoint. The mean titre is 15.05 + 0.04 mL. The molarity of sodium chloride solution is
Conclusions
The molarity of HCl is 0.2508 mol L-1.