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The concentration of $\ce{SO2}$ in the atmosphere over a city on a certain day is $\pu{10 ppm}$. Given that solubility of $\ce{SO2}$ in water is $\pu{1.3653 mol L-1}$ and $\mathrm{p}K_\mathrm{a}$ of $\ce{H2SO3}$ is $1.92,$ estimate $\mathrm{pH}$ of rain on that day.

$\pu{10 ppm}$ means there is $\pu{10^-2 g}$ of $\ce{SO2}$ in $\pu{1 L}$ of air. This accounts to $\pu{1.25E-4 mol}$ of $\ce{SO2}$ in $\pu{1 L}$ air, which is less than its solubility, which means that all this amount will be dissolved in rain water to make $\ce{H2SO3}.$

Now, using the expression $\frac{4c^2α^3}{(1-α)}$ for $K_\mathrm{a}$ expression of $\ce{H2SO3},$ and substituting $c$ as $\pu{1.25E-4},$ I get $α$ to be approx. $0.99.$

Then $[\ce{H+}]$ is $2cα,$ which comes out to be $\pu{2.475E-4}.$ So $\mathrm{pH}$ is $3.606.$ But the answer given is $4.865.$ Why I am wrong?

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    $\begingroup$ Use Unicode symbol α (works everywhere) or macro \alpha (in math mode, e.g. inside $…$ or $$…$$). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$ – andselisk Dec 11 '19 at 12:54
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    $\begingroup$ Also, one thing that is not immediately obvious: it's often a good idea also to enclose the following punctuation marks in math mode: this way they won't become orphans if the inline math expression is placed on the end of the line. For example, $\alpha^2.$ is more bullet-proof than $\alpha^2$.. $\endgroup$ – andselisk Dec 11 '19 at 13:02
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    $\begingroup$ Note that H2SO3 does not exist even in solution. The equilibrium equation is SO2(aq) + (2) H2O <=> HSO3- + H+ (H3O+) $\endgroup$ – Poutnik Dec 11 '19 at 13:34
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    $\begingroup$ You are wrong when you say that "10 ppm means there is 0.01 g SO2 in 1 L air". One liter air weighs about 1.3 g. So 0.01 g SO2 would be about 0.8%. 10 ppm means 10 mg in 1 kg air, which is about 0.8 cubic meter. $\endgroup$ – Maurice Dec 11 '19 at 14:22
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    $\begingroup$ @dhanesh Vijay ppm stands for "parts per million,i.e 1 in 1 000 000. Unless further specified, it is ambiguous as it can be mass/mass ( w/w ), volume/volume ( v/v ), Mass/volume ( w/v ), molar amount/molar amount ( n/n) and few less common combination. For this ambiguity, it is no longer recommended to be used. $\endgroup$ – Poutnik Dec 11 '19 at 19:02
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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.

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  • $\begingroup$ Can you elaborate how u got the concentration from ppm. $\endgroup$ – dhanesh vijay Dec 13 '19 at 15:28
  • $\begingroup$ @dhaneshvijay Ok, done... $\endgroup$ – Buck Thorn Dec 13 '19 at 15:39
  • $\begingroup$ Sorry I am not aware of Henry law constant yet. Is there anyother way to get the solution concentration without this concept? $\endgroup$ – dhanesh vijay Dec 14 '19 at 5:18
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    $\begingroup$ @dhaneshvijay you don't have to call it Henry's law constant, you can call it solubility constant or partition constant if you like. But no, you can't solve the problem without using the concept afaik. Basically it tells you in what concentration ratio the solute (SO2) will segregate between the gas and liquid phases. Without this concept you would not know to what extent the gas transfers into solution. $\endgroup$ – Buck Thorn Dec 14 '19 at 13:07
  • $\begingroup$ I agree. But $c$ in the equation represents the concentration of solute in aqueous phase, isn't it? $\endgroup$ – Mathew Mahindaratne Feb 12 at 17:20
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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.
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    $\begingroup$ Wikipedia "Raman spectra of solutions of sulfur dioxide in water show only signals due to the SO2 molecule and the bisulfite ion." $\endgroup$ – Poutnik Feb 13 at 4:37
  • $\begingroup$ How did u find c? $\endgroup$ – dhanesh vijay Mar 9 at 4:18
  • $\begingroup$ I started my answer agreeing with Buck Thorn. So I use his solution for $c$, to avoid work it again. $\endgroup$ – Mathew Mahindaratne Mar 10 at 4:51
  • $\begingroup$ But I wanted an explanation without using Henry's law and and also reason why my answer is wrong. So only I announced a bounty but your answer is not worth it. $\endgroup$ – dhanesh vijay Mar 11 at 2:06

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