The average concentration of SO_(2) 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 SO_(2) in water at 20=98 K is 1.3653 moles "litre"^(-1) andthe pK_(a)of H_(2)SO_(3) is 1.92, estimate the pH of rain on that day.

The average concentration of SO_(2) in the atmosphere over a city on a

1.	The average concentration of SO_(2) 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 SO_(2) in water at 20=98 K is 1.3653 moles "litre"^(-1) andthe pK_(a)of H_(2)SO_(3) is 1.92, estimate the pH of rain on that day.
Answer» Solution :`SO_(2)+H_(2)O rarr H_(2)SO_(3)` As rain water is falling from a great height, each drop of rain water will get saturated with `SO_(2)` before it reaches the earth. Hence, `[H_(2)SO_(3)]=[SO_(2)]=1.3653` moles `"litre"^(-1)` `H_(2)SO_(3) hArr 2H^(+) + SO_(3)^(2-)` SUPPOSE at EQUILIBRIUM, `[H^(+)] = X "mol" L^(-1)` Then `[H_(2)SO_(3)] = (1.3653-(x)/(2)), [SO_(3)^(2-)]=(x)/(2) "mol" L^(-1)` `K_(a) = ([H^(+)]^(2) [ SO_(3)^(2-)])/([H_(2)SO_(3)]) = (x^(2) xx x//2)/((1.3653-x//2))=10^(-1.92)` (`:' pK_(a) = 1.92` means `- LOG K_(a) = 10^(-1.92) or log K_(a) = 1.92 or K_(a) = 10^(-1.92)`) Neglecting x/2 in comparison to 1.3653, we get `(x^(3))/(2xx1.3653)=10^(-1.92)` or `x^(3) = 2.7306xx10^(-1.92)` or ` 3 log x = log 2.7306 - 1.92 = 0 . 4348 - 1.92= - 1.4852 or - log x = 0.485, i.e., pH = 0.485`