Question

a 4.4766g sample of petroleum product was burned in a tube furnace, and the SO2 produced was collected in 3% H202. reaction:

SO2 + H2O2 -> H2SO4

A 25ml portion of 0.00923 M NaOH was introduced into the solution of H2SO4 , following which the excess base was back-tirated with 13 33ml of 0.01007 M HCl. calculate the ppm of sulfur in the sample

Answer 1

Ans. Part A: Because 1 mol of HCl completely neutralizes 1 mol NaOH,

The number of moles of base “in excess” = number of moles of HCL required to neutralize it

Or, number of moles of HCl = Molarity of HCl x volume of HCl consumed in Litres

                                    = 0.01007 M x 0.01333 L                                    [1 L =1 1000 mL]

                                    = 0.0001342331 moles

Thus, number of moles of excess NaOH = 0.0001342331 moles

Total number of moles of NaOH added to the H₂SO₄ solution =

Molarity x volume (in L) of total NaOH solution

= 0.00923 M x 0.025 L = 0.00023075 moles

Now,

Number of moles of NaOH neutralized By H₂SO₄ =

                        Total moles of NaOH added to H₂SO₄ solution – moles of excess NaOH left over

                        = 0.00023075 moles - 0.0001342331 moles = 0.0000965169 moles

Part B: We have, number of moles of NaOH consumed by H₂SO₄ solution = 0.0000965169 moles

            H₂SO_4(aq) + 2 NaOH_(aq) ----------> Na₂SO_4(aq) + 2 H₂O_(l)

Since, 1 mol H₂SO₄ (diprotic acid) neutralizes 2 moles of NaOH, the number of moles of acid is equal to half the number of moles of NaOH consumed to neutralize it.

So, number of moles of H₂SO4 in the solution = (1/2) x 0.0000965169 moles = 0.00004825845 moles

Part C: We have, number of moles of H₂SO₄ produced from SO₂ = 0.00004825845 moles

From the stoichiometry of the reaction of formation of H₂SO₄, 1 mol SO₂ produces 1 mol H₂SO₄.

Therefore, the moles of SO₂ produced after combusting 4.4766 gram hydrocarbon = 0.00004825845

Since, 1 mol SO₂ has 1 mol S-atom, thus, number of moles of S-atom is equal to the number of moles of SO₂ = 0.00004825845

Mass of S-atoms in hydrocarbon sample = moles x atomic mass of S

                                    = 0.00004825845 moles x 32.060 g mol^-1

                                    = 0.001547165907 gram

So, 4.4766 g hydrocarbon has 0.001547165907-gram S

Amount of S in 1 kg hydrocarbon sample = (1000 g/ 4.4766 g) x 0.001547165907-gram

                                                            = 0.34561182750301568154402895054282 gram

                                                            = 346 mg

Thus, [S] in ppm = 346 ppm                                           [1 ppm = 1 mg S / kg hydrocarbon]