Titration experiment. Standardization of hydrochloric acid using sodium carbonate.

Data processing :

Gross volume of HCl used :

20.7 + 20.6 + 20.2   =   20.5 cm³ + 0.05 cm³.

        3

Uncertainty : | 20.5-20.7 | + | 20.5-20.6 | + | 20.5-20.2 |   x 100  =  2.9%  

                                  20.5

Gross volume of HCl used with percentage of  uncertainty = 20.50 cm³ + 2.9% cm³.

The molarities of HCl after dilution :

M1V1 = M2V2  

(2mL) (11M)  = (X) (250mL)

                X = 0.088 M

Uncertainty : 

V1  = 2.0 mL + 0.5 mL                                       V2  =  250.0 mL + 0.5 mL

     = 0.5 x 100%                                         =   0.5     x 100%

        2.0                                                     250.0

     =  25 %                                                =  0.2 %

Then, add together to form general uncertainty = 25 % + 0.2 %

                                                    = 25.2 %.

So the molarity of HCl after dilution is 0.088 M + 25.2 % M.

Chemical equation : 2HCl + Na2CO3  →  2NaCl + H2O + CO2

The number of moles for Na2CO3  :  1.329  =  0.0125 moles.

                                           106

2moles of HCl+1mole of Na2CO3  → 2moles of NaCl+1mole of water+1mole of CO2

0.025mole of HCl+0.0125mole of Na2CO3  → 0.025mole of NaCl+0.0125mole of      water+0.0125mole of CO2

So, the number of moles for HCl taken is 0.0250 moles.

The molarities of Na2CO3  :     no. of moles

                                   Volume

    0.0125         = 0.05 M.

      0.25

In this reaction, limiting reagent = Na2CO3   

Theoretical moles of Na2CO3  solution :

                                    Moles = Volume x Molarities

                                           = (25cm³) (0.088M)

                                                    1000

                                           = 0.0022 mole.

Actual moles of Na2CO3  used when titration :

                                      Moles = 0.088

                                                  2

                                            = 0.044 mole.

So the theoretical volume of HCl needed to react with 0.025 mole of Na2CO3  :

                               MV of Na2CO3  = (0.05) (25)

                                             = 1.25 (2M)

                                             =    2.5  .

                                                  1000 

                                             = 28.41 cm³.

Percentage error :

                        28.41cm³-20.5cm³ x 100% = 27.8423%

                                28.41cm³

Conclusion : The actual volume of HCl needed to neutralize the Na2CO3  is  28.41cm³ for the titration.