The average concentration of SO2 in the atmosphere over a city on a certain day is 10 ppm. when the average temperature is 298 K. Given that the solubility of SO2 in water at 298 K is 1.3653 moles litre–1 and the pKa of H2SO3 is 1.92, estimate the pH of rain on that day.
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We have, as concentration of SO2 in air is 10 ppm or 10×10–6 mole in 1 mole air or 10–5 mole of SO2 per mole of air. The concentration of SO2 in air is substantial and since rain water is falling from enormously great height , and hence each drop of rain water will get saturated with SO2 before it reaches the earth.
Therefore, [SO2] = [H2SO3] = 1.3653 mol/litre
As, SO2 + H2O → H2SO3
Also, H2SO3 ↔ 2H+ + SO32–
C 0 0
C – a/2 a a/2
Therefore, Ka = a2.a/[2×(C – a/2)]
10–1.92 = a3/2C [As, C = 1.3653]
Therefore, a = [1.3653×2×10–1.92]1/3
= [H+]
Therefore, pH = – log a
= 0.49 [Ans.]
Therefore, [SO2] = [H2SO3] = 1.3653 mol/litre
As, SO2 + H2O → H2SO3
Also, H2SO3 ↔ 2H+ + SO32–
C 0 0
C – a/2 a a/2
Therefore, Ka = a2.a/[2×(C – a/2)]
10–1.92 = a3/2C [As, C = 1.3653]
Therefore, a = [1.3653×2×10–1.92]1/3
= [H+]
Therefore, pH = – log a
= 0.49 [Ans.]
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