Question

in a sodium carbonate/bicarbonate analysis, a 0.4057g sample required 15.65mL of 0.1690 MHCl solution to reach the phenolphtalein end point and a total of 40.30mL to achievve the bromocresol green endpoint. Determine wt % NaHCO3 and Na2CO3 in this sample. Fromula weights of NaHCO3 and NA2CO3 are 84.0 and 106.0g/mol.

Answer 1

The analysis is of NaHCO3 and Na2CO3

We need to remember that.

The reaction of the two bases with acid will be

CO3-2 + H+ --> HCO3-    .......(1)

HCO3-   + H+ --> H2CO3

1) The phenolphthalein end point indicates complete reaction of carbonate with acid to give bicarbonate (1)

2) bromocresol green end point indicates completion of reaction of bicarbonate with acid

So as per the given information

Moles of HCl used till phenolphthlein end point will be equal to moles of carbonates taken

Moles of HCl = Molarity X volume in litres = 0.1690 X 15.65 / 1000 = 0.0026 moles

After this end point the HCl will react with the bicarbonate produced (from carbonates) and bicarbonate already present,

So moles of bicarbonate present = Moles of HCl used = Molarity of HCl X volume in litres

Moles of HCl = 0.1690 X (40.30 - 15.65) / 1000 = 0.0042 moles = Moles of bicarbonate

Moles of bicarbonate originally taken = Total moles of bicarbonate - Moles of bicarbonate formed from carbonate

The moles of bicarbnate formed from carbonate = Moles of carbonate present = 0.0026

Moles of bicarbonate originally taken = 0.0042 - 0.0026 = 0.0016

Now let us calculate the weight taken

a) weight of Na2CO3 = Moles of carbonates X molecular weight of Na2CO3 = 0.0026 X 106 = 0.2756 grams

b) weight of NaHCO3 = Moles of bicarbonate X molecular weight of NAHCO3 = 0.0016 X 84 = 0.1344 grams

Weight percentage of Na2CO3 = Weight of Na2CO3 X 100 / total weight = 67.22 %

Weight percentage of NaHCO3 = 100- 67.22 = 32.78%