Question

Assuming you weighed out a 2.3684g sample of the unknown, and it took 35.63 mL of a .1025 M HCl titrant to reach the endpoint, what is the weight percent of Na2CO3 and NaHCO3 in your unknown sample?

Answer 1

answer:

molarity = moles / volume of solution in L

So,

Moles of HCl reacted = 0.1025 * 0.03563 = 3.65*10-3 moles

Assume 'x' g of Na2CO3 is present and 'y' g of NaHCO3 in the 25mL aliquot.

So,

moles of Na2CO3 = x/106

moles of NaHCO3 = y/84

1 mole Na2CO3 requires 2 moles HCl, and 1 mole NaHCO3 requires 1 moles HCl

So,

2*(x/106) + 1*(y/84) = 3.65*10^-3

Also, since original volume was 250 mL out of which this aliquot was taken, so actual masses are 10*x and 10*y.

Thus,

10*x + 10*y = 2.3684

Solve the above two equations to get the values of x and y