Question

The average concentration of ${SO}_2$ in the atmosphere over a city on a certain day is $10ppm$, when the average temperature is $298K$. Given that the solubility of ${SO}_2$ in water at $298K$ is $1.3653$ $mol$ ${litre}^{-1}$ and The $p{K}_a$ of $H_2{SO}_4$ is $1.92$, the $pH$ of rain on that day was:

• A. $pH=4.165$
• B. $pH=4.865$
• C. $pH=5.865$
• D. $pH=5.065$

Answer 1

The correct option is B $pH=4.865$

Concentration of ${SO}_2$ in air is $10ppm$ or $10$mole in ${10}^6$ mole air or ${10}^{-5}$ mole ${SO}_2$ per mole of air. The concentration of ${SO}_2$ in air being susbstantial and since, rain water is falling from enormously great height so, each drop of rain water will get saturated with ${SO}_2$ before it reaches earth.
Now the given concentration or solubility of ${SO}_2$ at $298K$ is $1.3653M$.
This value of solubility corresponds when $P_{{SO}_2}=1$atm
Thus, according to Henry's law:
[${SO}_2$] dissolved in water $\propto P_{{SO}_2}$ in gas phase
[${SO}_2$] dissolved in water $\propto P_{{SO}_2}$ in air
$\therefore$ $\left[{SO}_2\right]$ dissolved in water $=1.3653\times {10}^{-5}M$
$\underset{11-\alpha }{{SO}_2}+H_{2}O\rightleftharpoons \underset{0\alpha }{H}+\underset{0\alpha }{H{SO}_3^{-}}$
$K_a={10}^{-1.92}=\frac{ca.ca}{c(1-\alpha )}=\frac{c{\alpha }^2}{1-\alpha }$ ($\alpha$ is small)
or $0.012=\frac{1.3653\times {10}^{-8}\times {\alpha }^2}{1-\alpha }(\because {pk}_b=1.92,\therefore K_b=0/012)$
$\therefore$ $1.3653\times {10}^{-5}{\alpha }^2+0.012\alpha -0.012=0$
$\therefore$ $\alpha =1$ (solving quadratic equation)
$\therefore$ $\left[H^{+}\right]=c\alpha =1.3653\times 1=1.3653$
$\therefore$ $pH=4.865$