Question

2. A 4.476-g sample of a petroleum product was burned in a tube furnace, and the SO2 produced was collected in 3% H2O2. Reaction: SO2 + → H2SO1 A 25.00-mL portion of 0.00923 M NaOH was introduced into the solution of H SO The excess base (NaOH) was back-titrated with 13.33 mL of 0.01007 M HCI. Calculate the concentration of S present in the sample in % mass and ppm

Answer 1

Moles = Molarity x Volume (L)

Moles of HCl = 0.01007 M x 0.01333 L = 1.342 x 10^-4 excess

Moles of NaOH = 0.00923 M x 0.025L) = 2.308 x 10^-4 total

So, (2.308 x 10^-4) – (1.342 x 10^-4) = 9.655 x 10^-5 mol reacted

So, moles of H2SO4 = 9.655 x 10^-5 mol

Hence,

(9.655 x 10^-5) mol H₂SO₄ (1 mol S / 1mol H₂SO₄) (32.07g S / 1 mol S)
= .003096 g S

0.003096 g S/ 4.476 g sample (100) = 0.062% S

Now,

1 ppm = 1 mg/kg

S = 0.003096 g = 3.096 mg

Total sample = 4.476 g = 0.004476 Kg

Since,

1000 mg = 1 g and 1000 g = 1 Kg

So, 3.096 mg / 0.004476 Kg = 691.69 mg/Kg = 691.69 ppm = 692 ppm