a). The concentration of ozone in urban air can be determined by reacting an aqu

Question

a). The concentration of ozone in urban air can be determined by reacting an aqueous solution of potassium iodide (KI). The equation for this process is given below. Determine the ozone concentration in a 10 dm3 sample of air at 20C and one atmosphere if 12 g of KI was required to consume all of the ozone.

2KI + O3 + H2O ===> I2 + O2 + 2KOH

Molar volume at 20C = 24 dm3

Molar masses K = 39.1 g mol-1; I = 126.9 g mol-1 .................................. ( 8 marks)

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Explanation / Answer

According to the question 10L of air reacts with 12 g of KI

Now 12 g KI = 12 / 166 moles = 0.0723 moles of KI ( mol wt of KI is 166)

Consider the reaction , 2KI + O3 + H2O ===> I2 + O2 + 2KOH

As is evident, 2 moles of KI react with 1 mole of O3

thus, 0.0723 moles of KI react with (0.0723/2) 0.036 moles of O3

Also, at NTP 1 mole O3 occupies 24 L

So, 0.036 moles of O3 occupy 24 x 0.036 L = 0.864 L

The volume of air sample = 10 L, Volume of O3 in this air = 0.864 L

Thus % of O3 in air = (0.864/10 )X 100 = 8.64 %

Note: There seems to be some misprint in the exact units of mass of KI. In case the units are mg or micro grams, then we just need to multiply the answer by 10-3 and 10-6 respectively. Also in these cases, expressing the concentration in ppm or ppb makes more sense.

For eg. If % of O3 in air = 0.00864 % = 86.4/106 = 86.4 ppm

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