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

Amount of `SO_2` in atmosphere =10 ppm
`=10/10^6`
Molar conc. Of `SO_2` in presence of water = Amt. of `SO_2` x Solubility of `SO_2` in water
`=10xx10^(-6)xx1.3653 "mol L"^(-1)`
`=1.3653xx10^(-5) "mol L"^(-1)`
`H_2SO_3` dissociates as :
`H_2SO_3 hArr H^(+)+ HSO_3^(-)`
`{:("Initial conc.",1.3653xx10^(-5), 0,0),("Molar conc. at equiv.",1.3653xx10^(-5)-x , x,x):}`
`K_a=x^2/(1.3653xx10^(-5)-x)`
Now `pK_a =1.92` or - log `K_a` =1.92
`therefore K_a=1.2xx10^(-2)`
Substituting
`1.2xx10^(-2) =x^2/(1.3653xx10^(-5)-x)`
`x^2=1.2xx10^(-2) (1.3653xx10^(-5)-x)`
Solving we get `x=1.364xx10^(-5)`
`therefore pH=-log (1.364xx10^(-5))`
=4.865