The average concentration of `SO_(2)` in the atmosphere over a city on a cetrain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of `SO_(2)` in water at 298 K is `1.3653` mol `litre^(-1)` and the `pK_(a)` of `H_(2)SO_(3)` is `1.92`, estimate the pH of rain on that day.

Explanation

`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`