The dissociation of weak electrolyte (a weak base or weak acid) is expressed in terms of Ostwald's dilution law. An acid is substance which furnishes a proton or accepts an electron pair, where a base is proton acceptor or electron pair donor. Stronger is acid, weaker is its conjugate base. The dissociation constants of an acid `(K_(a))` and its conjugate base `(K_(b))` are related by `K_(w)=K_(a)xxK_(b)`, where `K_(w)` is ionic product of water equal to `10^(-14) at 25^(@)C`. The numerical value of `K_(w)` however increase with temperature. In a solution of an acid or base `[H^(+)][OH^(-)]=10^(-14)`. Thus the `[H^(+)]` in a solution is expressed as: `[H^(+)]=10^(-pH)` and `pH+pOH=14`. Buffer solution are the solutions which do not show appreciable change in the pH on addition of small amount of acid or base.
`SO_(2)` contents in the atmosphere is 10 ppm and the solubility of `SO_(2)` in water is `1.36` mol `litre^(-1)`. If `pK_(a)` of `H_(2)SO_(3)` is `1.92`, the pH of rainwater is:

Options

A. `0.49`
B. `0.39`
C. `0.29`
D. `0.19`

Answer

A

Explanation

Concentration of `SO_(2)` in air is 10 ppm or `10xx10^(-6)` mole air or `10^(-5)` mole `SO_(2)` per mole of air. The concentration of `SO_(2)` in air being substancial and since rainwater is falling ffrom enormously great height so, each drop of rainwater will get saturarated with `SO_(2)` before it reaches earth. Thus,
`[SO_(2)]=[H_(2)SO_(3)]=1.3653 mol//litre`
`SO_(2)+H_(2)Orarr H_(2)SO_(3)`
Also, `{:(H_(2)SO_(3)hArr,2H^(+),+SO_(3)^(2-)),(C,0,0),((C-(a)/(2)),a,a//2):}`
`:. K_(a)=10^(-1.92)= (a^(2).a)/(2xx(C-(a)/(2)))=(a^(3))/(2C)`
`( :' C=1.3653M)`
`:. a=[1.3653xx2xx10^(-1.92)]^(1//3)= [H^(+)]`
`:. pH= -log a=0.49`