Question

The average concentration of $S{O}_2$ in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of $S{O}_2$ in water at 298 K is 1.3653 mol L${}^{-1}$ and $p{K}_a$ of $H_{2}S{O}_3$ is1.92. Estimate the pH of rain on that day. (write the value to the nearest integer)

Answer 1

The molar mass of $S{O}_2$ is 64 g/mol.
10 ppm of $S{O}_2$ corresponds to 0.01 g in 1 L.
Hence, the molar concentration of dissolved $S{O}_2$ is $\frac{0.01}{64}=1.5625\times {10}^{-4}$
This also corresponds to the concentration of $H_{2}S{O}_3$ which dissociates as $H_2{S}_2{O}_2\rightleftharpoons H^{+}+HS{O}_4^{-}$
$p{K}_a=1.92,K_a=0.012$
The expression for the equilibrium constant K is $K=\frac{\left[H^{+}\right]\left[HS{O}_3^{-}\right]}{\left[H_{2}S{O}_3\right]}$
Substitute values in the above expression.
$0.012=\frac{x\times x}{1-x}$
$83.33x^2+x-1.5625\times {10}^{-24}=0$
Hence, $x=0.0001188$
$pH=-log\left[H^{+}\right]=-log0.0001188=3.925$
Hence, the pH of rain on that day was 3.925.