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.

Answer

B, C

Explanation

Average concentration of `SO_(2)` in atmosphere is `10 ppm`, i.e.,
`(10)/(10^(6)) = 10^(-5) mol L^(-1)`
Molar concentration of `SO_(2)` in atmosphere and solubility of `SO_(2)` in water, i.e.,
`= 1 xx 10^(-5) xx 1.3653 = 1.3653 xx 10^(-5) mol L^(-1)`
`SO_(2) +H_(2)O rarr H_(2)SO_(3)`
`[SO_(2)] = [H_(2)SO_(3)] = 1.653 xx 10^(-5)M`
`{:(H_(2)SO_(3)hArr,H^(o+)+,HSO_(3)^(Theta)),(1.3653 xx10^(-5),0,0),((1.3653xx10^(-5)-a),a,a):}`
`K_(a) = ([H^(o+)][HSO_(3)^(Theta)])/([H_(2)SO_(3)])`
`pK_(a) = 1.92`
`- log K_(a) = 1.92`
`K_(a) =- antilog 1.92`
`= (a^(2))/((1.3653 xx10^(-5)-a))`
`- "antilog" 1.92 = (a^(2))/((1.3653xx10^(-5)-a))`
Tek negative log on both sides:
`1.92 =- 2log a+ log (1.3653 xx 10^(-5)-a)`
Neglecting `a`,
`1.92 =- 2 log [H^(o+)] +log 1.3653 xx 10^(-5)`
`=2pH + log 1.3653 xx 10^(-5)`
`pH ~~ 3.3925`.