Question

Struck on part v :(

The concentration of ozone in the atmosphere may be determined by bubbling air through a solution of acidified potassium iodide. Iodine is formed in solution, the concentration of which may be determined by titration with a solution of sodium thiosulfate of known concentration. The equations for the reactions are:

O 3 + 2I – + 2H + O 2 + H 2 O + I 2 Equation 1

I 2 + 2S 2 O 3 2– 2I – + S 4 O 6 2– Equation 2

In an experiment to determine the concentration of ozone in a sample of air, 100 m 3 of air was bubbled through 100 cm 3 of a solution containing an excess of acidified potassium iodide. The resulting solution was titrated against a solution of sodium thiosulfate of concentration 0.0155 mol dm –3 . The volume of sodium thiosulfate solution required for complete reaction was 25.50 cm 3 . (i) Calculate the number of moles of sodium thiosulfate that react. Answer: Amount of thiosulfate = 0.0155 x 25.50 x 10^−3= 3.9525 x 10^-4 mol ii) Calculate the number of moles of iodine that reacted with the sodium thiosulfate. Ans) 3.9525 x10^−4/2 = 1.97625 x 10^−4(mol) iii) Use equation 1 to deduce the number of moles of ozone that reacted with the acidified potassium iodide. Ans) Amount of iodine = Amount of ozone iv) Calculate the volume of ozone, measured in m^3, present in the original sample of air. Assume that all gas volumes were measured at room temperature and pressure and that the molar volume of any gas under these conditions is 0.024 m^3mol–1. Ans) =1.97625 x 10^−4x 0.024= 4.743 x 10^−6(m^3)in 100 m^3) v) Calculate the concentration of ozone in the sample of air in units of parts per million (ppm) by volume.

Answer 1

(v) From part (iii) above, we have deduced that we 1.97625*10^-4 mol ozone gas.

Molar mass of ozone is 48 g/mol.

Therefore, mass of ozone present is (1.97265*10^-4 mol)*(48 g/1 mol) = 9.486*10^-3 g = (9.486*10^-3 g)*(1 mg/10^-3 g) = 9.486 mg.

Again, from part (iv) above, we have deduced that 1.97625*10^-4 mol ozone is present in 100 m³ of our air sample.

In other words, 100 m³ of the air sample contains 9.486 mg ozone.

We know that 1 m³ = 1000 L.

Therefore, 100 m³ = (100 m³)*(1000 L/1 m³) = 100000 L = 1.0*10⁵ L.

We define concentration in ppm as mg of sample in 1 L of solvent (air in this case).

Therefore, we can say

1.0*10⁵ L air sample contains 9.486 mg ozone.

Therefore, 1 L air sample will contain (9.486 mg)*(1 L/1.0*10⁵ L) = 9.486*10^-5 mg ozone.

Therefore, concentration of ozone is 9.486*10^-5 ppm (ans).