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.

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:

The solubility of BaSO_4 in water at 298K is 1.1 xx10^(-5) mol L^(-1) . Calculate the solubility product of Baso_4 at 298K.

The solubility of AgCl at 298 K is 1.3 xx 10 ^(-5) " mol " L^(-1) Calculate the value of K_(sp) at this temperature.

The solubility of AgCl in water at 298 K is 1.06xx10^(-5) mole per litre. Calculate its solubility product at this temperature.

The solubility of AgCl in water at 298 K is 1.06xx10^(-5) mle per litre. Calculate its solubility product at this temperature.

The solubility of AgCl in water at 298 K is 1.06xx10^(-5) mle per litre. Calculate its solubility product at this temperature.

The solubility of CaF_2 in water at 298 K is 1.7 xx 10 ^(-3) grams per 100 cm ^(3) .Calculate the solubility product of CaF_2 at 298 K.