The solubility of ${SO}_{2 (g)}$ in water is 1.65 mole/atm per litre of water and ${Ka}_{1} (H_{2} {SO}_{3})$ is $10^{- 4}$ . If ${SO}_{2}$ content over a city is 20ppm by volume and if it is dissolved in rain water then th pH of rain water is

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9.25

6.55

5.85

4.24

answer is D.

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Detailed Solution

$P_{{SO}_{2} (air)} = volume friction \times 1 atm$

$= \frac{20}{10^{6}} \times 1 = 2 \times 10^{- 5} atm Conc . Of {SO}_{2} (or)) H_{2} {SO}_{3} = 1 .65 \times 2 \times 10^{- 5} = 3 .3 \times 10^{5} M {[H^{+}]}_{WA} = \sqrt{K_{a} \times C} = \sqrt{10^{- 4} \times 3 .3 \times 10^{- 5}} = \sqrt{33 \times 10^{- 10}} {[H^{+}]}_{WA} = 5 .745 \times 10^{- 5} pH = 5 - \log 5 .745 = (5 - 0 .78) = 4 .24$