Question

A 4.924-g sample of a petroleum product was burned in a tube furnace, and the SO2 produced was collected in 4% H2O2.

Reaction:

SO2(g) + H2O2 --> H2SO4

A 24.00-mL portion of 0.00840M NaOH was introduced into the solution of H2SO4, following which the excess base was back-titrated with 15.60 mL of 0.01101M HCl. Calculate the sulfur concentration in the sample in parts per million.

Answer 1

Given data,

Mass = 4.924 g

Molarity of NaOH = 0.00840 M

Molarity of HCl = 0.01101 M

We know,

Moles = Molarity x Volume (L)

Moles of HCl = 0.01101 M x 0.01560 L

= 1.71x 10^-4 excess

Moles of NaOH = 0.00840 M x 0.024L)

= 2.01 x 10^-4 total

So, (2.01 x 10^-4) – (1.71x 10^-4 ) = 3.0x 10^-5 mol reacted

So, moles of H2SO4 = 3.0 x 10^-5 mol

Hence, (3.0 x 10^-5) mol H₂SO₄ (1 mol S / 1mol H₂SO₄) (32.07g S / 1 mol S) = 0.0009621 g S

0.0009621 g S/ 4.924 g sample (100) = 0.019% S

We know, 1 ppm = 1 mg/kg

S = 0.0009621g = 0.9621 mg

Total sample = 4.924g (Since,1000 mg = 1 g and 1000 g = 1 Kg)

= 0.004924 Kg

S0,

0.9621 mg / 0.004924 kg = 195.38 mg / kg

= 196 ppm

Therefore, the concentration of sulfur is 196 ppm