Question

In a chemistry lab, acid-base titration was completed with Na2CO3 being titrated with HCl and HCl was titrated with NaOH. In one portion, a standard solution was made of an unkown and it was mixed with 40.00 mL of HCl, the indicator used was Phenophtalein and NaOH was added till an endpoint was used.

Data:

(0.1255 g of unknown + 40.00 mL HCl) was titrated with 15.60 mL NaOH when the end point was reached. Determine the weight percent of CO₃² in this back titration. Concentration of HCl/NaOH are both 0.05M.

Answer 1

Find % CO3-2 in backtitration

Na2CO3 + HCl = NaCl + CO2 + H2O

HCl + NaOH = NaCl + H2O

mmol of HCl used = MV = 0.05*40 = 2mmol of H+

mmol of NaOH used in backtitration = MV = 15.60*0.05 = 0.78

then

mmol of H+ used in Na2CO3 = 2 - 0.78 = 1.22 mmol of H+

then

1 mmol of H+ = 0.5 mmol of Na2CO3

1.22 mmol --> 1/2*1.22 = 0.61 mmol of Na2CO3

mmol of CO3-2 = 0.61

mass of CO3-2 = mol*MW = 0.61*10^-3)(60) = 0.0366 g of CO3-2

then

% CO3-2 = mass of CO3-2 / Total mass = 0.0366 /0.1255 * 100 = 29.16%