# Chemistry - Calculating pH of rain water

## Solution 1:

The value $$\pu{1.3653 mol L^{−1} atm^{−1}}$$ is the solubility constant (or Henry's law solubility constant), not the solubility. The solubility is defined as the maximum possible concentration (the saturation concentration) of a solute under given solution conditions (e.g. temperature and pressure), whereas the solubility constant $$H^{cp}$$ defines how solute partitions between the gas (concentration in gas here given as the partial pressure in atmospheres — although this is not stated in the OP it can be inferred from the expected answer) and solution phases (concentration in $$\pu{mol/L}):$$

$$H^{cp} = \frac{c}{p}= \pu{1.3653 mol L^{−1} atm^{−1}}$$

Assuming atmospheric pressure of 1 atm then $$\pu{10 ppm}$$ translates into a partial pressure of $$\pu{1e-5 atm}$$ (if the atmospheric pressure is $$p_T$$, then the partial pressure of $$\ce{SO2}$$ is $$p = p_T \cdot 10^{-5}$$) and a solution concentration $$\pu{1.365e-5 M}$$ $$\;(c=H^{cp} p)$$. Proceeding from the definition of $$\mathrm{p}K_\mathrm{a}$$ the degree of dissociation $$(\alpha=0.99886)$$ can be computed and from this the $$\mathrm{pH} = 4.865,$$ which agrees with the expected result.

You should check solubility limits when evaluating this sort of problem, but since the solubility is not given you can ignore this and assume that the limit has not been reached.

## Solution 2:

I agree with Buck Thorn's explanation on $$\ce{SO2}$$ concentration in aqueous phase. Thus, when dissolve in water (or when is added to water), the initial reaction of $$\ce{SO2}$$ with water is shown in the following reaction (Ref.1):

$$\ce{SO2 (g) + H2O (l) -> H2SO3 (aq)}$$

Then, formed $$\ce{H2SO3}$$ would stabilize following equilibrium:

$$\ce{H2SO3 + H2O <=> H3O+ + HSO3-}$$

Suppose initial concentration of $$\ce{H2SO3}$$ is $$c$$ and at equilibrium, $$[\ce{H3O+}]$$ is $$\alpha$$. Thus, concentration of $$\ce{HSO3-}$$ is $$\alpha$$ as well.

$$\therefore \; K_\mathrm{a1} = \frac{[\ce{H3O+}][\ce{HSO3-}]}{[\ce{H2SO3}]}=\frac{\alpha \cdot \alpha}{c-\alpha}= 10^{-1.92}$$

When simplify this equation, you'd get:

$$10^{1.92}\alpha^2 + \alpha - c = 0 \tag{1}$$

Since $$c$$ is $$1.3653 \times 10^{-5}$$ (see Buck Thorn's answer elsewhere), you can solve $$(1)$$ for $$\alpha$$, which is equal to $$2.7308 \times 10^{-5}$$. Thus $$\mathrm{pH}$$ of solution is $$4.564$$.

References:

1. A. Sathasivan, B.S. Herath, S. T. M. L. D. Seneviratne, G. Kastl, “Chapter 14: Dechlorination in Wastewater Treatment Processes,” In Current Developments in Biotechnology and Bioengineering: Biological Treatment of Industrial Effluents; Duu-Jong Lee, Veeriah Jegatheesan, Huu Hao Ngo, Patrick C. Hallenbeck, Ashok Pandey, Editors; Elesvier: Boston, MA, 2017, pp. 359-380.