A 1.9450 g sample containing only Na2CO3 (F. M. 105.99) and NaHCO3 (F. M. 84.01) is titrated with 35.31 mL of 0.8724 M HCl. Calculate the percentage sodium carbonate and sodium bicarbonate.

1.0979g of Na2CO3

0.8471g of NaHCO3

Explanation:

NaHCO3 and Na2CO3 reacts with HCl as follows:

NaHCO3 + HCl → NaCl + H2O + CO2

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

Moles of HCl required were:

0.03531L * (0.8724mol / L) = 0.03080 moles of HCl

With this information we can write 2 equations:

1.9450g = X + Y (1)

X is the mass of NaHCO3 and Y is mass of Na2CO3

Also we can write:

0.03080 moles = X/84 + 2Y/106 (2)

Where 84 is molar mass of NaHCO3 and 106 is molar mass of Na2CO3, 2 is due the reaction (2 moles of HCl react per mole of Na2CO3)

Replacing (1) in (2):

0.03080 moles = (1.9450-Y)/84 + 2Y/106

0.03080 = 0.023155 - Y/84 + Y/53

0.007645 = 0.00696316Y

Y = 1.0979g of Na2CO3

And mass of NaHCO3 =

1.9450g - 1.0979g =

0.8471g of NaHCO3