Exp. 8 Analysis of a Mixture of Carbonate and Bicarbonate by larryp

VIEWS: 630 PAGES: 1

									                       Exp. 8 Analysis of a Mixture of Carbonate and Bicarbonate

A. A solid sample containing an unknown amount of sodium carbonate (FW=105.988) and
   sodium bicarbonate (FW=84.0066) (and a soluble inert binder) was dried in a desiccator.
   2.50 grams of this sample was added to a 250 mL volumetric flask.
   Distilled water (boiled and cooled to 20 ºC) was used to dissolve the sample and then to increase the
   water level to the volumetric flask calibration mark.

B. 25.0 mL of the solution prepared in part A required 25.0 mL of 0.100 M HCl to reach the
                                                    −                      −
   bromocresol green end point. Let x = mmol CO3 2 and let y = mmol HCO3 in the 25.0 mL aliquot.
   Define the total alkalinity = TA = mmole H+(aq) needed to reach the bromocresol green end point.
                            −                           −
   x(2 mmol H+/mmol CO3 2 ) + y(mmol H+/mmol HCO3 ) = 25.0 mL x 0.100 M HCl = 2.50 mmol H+
   2x + y = 2.50 mmol (= TA)

C. 50.0 mL of 0.100 M NaOH was added to another 25.0 mL of the solution prepared in part A.
   50.0 mL x 0.100 M = 5.00 mmol OH– was added.
   10 mL of 10 wt % BaCl2 was added to precipitate the carbonate.
   40.0 mL of 0.100 M HCl was immediately used to titrate the excess OH–.
   40.0 mL x 0.100 M HCl = 4.00 mmol H+ x (1 OH–/H+) = 4.00 mmol OH– was left over.
                              −            −
   mmole OH– used to form CO3 2 from HCO3 = 5.00 – 4.00 = 1.00 mmol OH–
                       −              −
   1.00 mmol OH– (1 HCO3 /1 OH–) = HCO3 = y

D. 2x + (1.00 mmol) = 2.50 mmol
                 x = 0.750 mmol

                                                              −
E. The chemical amounts of carbonate and bicarbonate        CO3 2 = x = 0.750 mmol/25.0 mL
                                                              −
   (in 25.0 mL) are:                                       HCO3 = y = 1.00 mmol/25.0 mL

F. The chemical amounts of carbonate and bicarbonate in 250 mL are:
          −
      CO3 2 = 0.750 mmol/25.0 mL x 250 mL = 7.50 mmol
          −
       HCO3 = y = 1.00 mmol/25.0 mL x 250 mL = 10.0 mmol

G. The amounts of sodium carbonate and sodium bicarbonate in the original sample are:
      Na2CO3 = 7.50 mmol x (105.998 mg/mmol) = 794.985 mg in the 2.50 g sample.
      NaHCO3 = 10.0 mmol x (84.0066 mg/mmol) = 840.066 mg in the 2.50 g sample.

       wt % Na2CO3 = (0.794985 g/2.50 g) x 100 wt % = 31.8 wt % (3 S.F.)
       wt % NaHCO3 = (0.840066 g/2.50 g) x 100 wt % = 33.6 wt % (3 S.F.)

H. Additional Information:
   Ka1 = 4.45 x 10–7
   Ka2 = 4.69 x 10–11

								
To top