Make an order of magnitude estimate of the dry deposition rate of nitric acid (H

Question

Make an order of magnitude estimate of the dry deposition rate of nitric acid (HNO3) aerosol on a forest, if the atmospheric concentration of HNO3 is 5 µg m-3.

How important is this source of acid compared to the annual deposition of HNO3 in the rain, assuming that the nitrate concentration in rain is 60 µM and annual precipitation averages 115 cm.

Similarly, make an order-of-magnitude estimate of the dry deposition rate of aerosol sulfate on a forest, if the atmospheric concentration is 8 µg m-3. Compare with the wet input resulting from rain that averages 80 µM SO4.

Comment on which acid appears to have the bigger impact on the forest.

Explanation / Answer

average rain = 115 cm = 1.15 m

Now, let the area of influence = 1 m2

Thus, total rain water collected = 1.15 m3 = 1150 litres

Now, Molar concentration of HNO3 in the rain = 6*10-5 M = 6*10-5 moles of HNO3 per litre of rain

Thus, total moles of HNO3 collected = (6*10-5)*1150 = 0.069

Molar mass of HNO3 = 63 g/mole

Thus, mass of HNO3 = moles*molar mass = 4.347 g per 1.15 m3

Thus, per m3 deposition of HNO3 = 4.347/1.15 = 3.78 g...............(1)

Similarly,

Molar concentration of H2SO4 in the rain = 8*10-5 M = 8*10-5 moles of H2SO4 per litre of rain

Thus, total moles of H2SO4 collected = (8*10-5)*1150 = 0.092

Molar mass of H2SO4 = 98 g/mole

Thus, mass of H2SO4 = moles*molar mass = 9.016 g per 1.15 m3

Thus, per m3 deposition of H2SO4 = 9.016/1.15 = 7.84 g...............(2)

Clearly from (1) & (2) H2SO4 will have greater impact.

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