Question

A 2.2222 gram sample containing soda ash is dissolved in a 100 mL volumetric flask. A 24.98 mL aliquot is analyzed and required 45.67 mL of 0.1234 M HCl, before it was boiled, cooled, and back-titrated with 8.88 mL of 0.0987 M NaOH. What is the percent Na2CO3?

Answer 1

Number of moles of NaOH required to back titrate the excess HCl is , n = Molarity * volume in L

n = 0.0987 M * 8.88 mL*10^-3L/mL

   = 8.76*10^-4 moles

NaOH + HCl ---> NaCl + H2O

1 mole of NaOH reacts with 1 mole of HCl

8.76*10^-4 moles of NaOH reacts with 8.76*10^-4 moles of HCl

initial number of moles of HCl , n' = Molarity * Volume in L

n' = 0.1234 moles * 45.67 mL*10^-3L/mL

= 5.63*10^-3 moles

Number of moles of HCl reacted with Na2CO3 ,N = n'-n = 5.63*10^-3 moles- 8.76*10^-4 moles = 4.754*10^-3 moles

Na2CO3 + 2HCl ---> 2NaCl + 2H2O

2 moles of HCl reacts with 1 mole of Na2CO3

4.754*10^-3 moles of HCl reacts with ( 4.754*10^-3 moles)/2 = 2..377*10^-3 moles of Na2CO3

Mass of Na2CO3 = Number of moles * Molar mass

                          = 2..377*10^-3 mole * 106 g/moles

                          = 0.252 g

24.98 mL of sample contains 0.252 g

100 mL of sample contains (100*0.252)/24.98 g = 1.009 g

So percent of Na2CO3 = ( mass of Na2CO3/mass of sample) *100

                                  = ( 1.009g / 2.2222 g) *100

                                  = 45.4 %