A 3.900 g sample of a mixture of sodium hydrogen carbonate and potassium chlorid

Question

A 3.900 g sample of a mixture of sodium hydrogen carbonate and potassium chloride is dissolved in 25.80 mL of 0.439 M H2SO4. Some acid remains after treatment of the sample. Write both the net ionic and the molecular equations for the complete reaction of sodium hydrogen carbonate with sulfuric acid. If 34.8 mL of 0.109 M NaOH were required to titrate the excess sulfuric acid, how many moles of sodium hydrogen carbonate were present in the original sample? What is the percent composition of the original sample?

Explanation / Answer

molecular equation:

2NaHCO3 + H2SO4 ----------------> Na2SO4 + 2H2O + 2CO2

ionic equation :

2Na+ + CO3 2- + 2H+ + SO4 2- ----------------> 2Na+ + SO4 2- + 2H2O + 2CO2

net ionic equation :

2HCO3 2- + 2H+ --------------> 2H2O + 2CO2

now problamatic part :

moles of H2SO4 = 25.80 x 0.439 /1000 = 0.01133

moles of NaOH =34.8 x 0.109 / 1000 = 0.00379

H2SO4 + 2 NaOH --------------------> Na2SO4 + 2H2O

1                 2

x                0.00379

x = 0.00379 / 2 = 0.001897

so with NaOH 0.001897 mol H2SO4 reacted .

the moles of H2SO4 that reacted with NaHCO3 = total moles - NaOH

                                                                            = 0.01133 - 0.001897

                                                                            = 0.009433

2NaHCO3 + H2SO4 ----------------> Na2SO4 + 2H2O + 2CO2

2 mol            1 mol

   y                 0.009433

NaHCO3 moles = 0.00791 x 2

NaHCO3 moles       = 0.0189

NaHCO3 molar mass = 84 g/mol

NaHCO3 mass = 0.0189 x 84 = 1.58g

% composition of NaHCO3 in original sample = (1.58 / 3.90) x 100

                                                                              = 40.51%

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