Calculating concentration of sulfur dioxide in ppm

How do I calculate the concentration of sulfur dioxide in ppm in air if I am given the pH of the acid rain it produces?  We can assume there is no oxidation of the gas.

Let's start with the obvious, the lower the pH, the greater the concentration of dissolved SO2 since the SO2 reacts with water according to the equation:

SO2 (ag)  +  H2O    H2SO3   H+  + HSO3-

If you know the pH, you could calculate the [H+]   ([H+] = antilog (-pH)

Then if you know the [H+] (and assume the [HSO3-] is the same), and the Ka value (1.5 x 10-2 for sulfurous acid) you could calculate the [H2SO3] which is the same as [SO2].

At that point, you could perhaps use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid, which may be written as:   

p = k x c

where k is a temperature-dependent constant (for example, 769.2 L•atm/mol for dioxygen (O2) in water at 298 K), p is the partial pressure (atm), and c is the concentration of the dissolved gas in the liquid (mol/L).

Of course at that point you would need to know the value of k for SO2, and I'm not familiar with using that, or even where to find it.

Hope that gets you started.  I'd be interested in seeing how the problem is completely solved when you are finished.

Acid rain is produced from SO2 (sulfur dioxide) which forms H2SO4 (sulfuric acid) when it reacts with the moisture, oxygen, and water vapor in ambient air.  Sulfuric acid is a diprotic acid, meaning that there are two protons that will be dissociated when H2SO4 reacts with water.  Through a series of reactions SO2 is converted to H2SO4 using the following stiochemistretic relationship:
          1 mole SO2(g) ==> 1 mole H2SO4 

pH as you know is represented by the following equation:
          pH = -log([H+])     

Given pH of the acid rain you can calculate the concentration of H+ in the solution
        [H+] = 10^(-pH)

Now you have the [H+] in the solution.  Remember that H2SO4 is a diprotic acid and it follows the following dissociation equation in water             
          (1)    SO2 ==> H2SO4 2 H^+ + SO4^-2

Using stoichemistry of (1) we can calculate the concentration of H2SO4 at any given pH.  Assume we have a pH of 4.2 in the acid rain, then the [H+] in solution would be:
          [H+] = 10^-4.2 = 6.3*10^-5 mol/L

          [H2SO4] = [H+] / 2      in our example [H2SO4] = 3.15*10-5 mol/L

        [H2SO4] = [SO2]          in our example [SO2] = 3.15*10-5 mol/L

It important to remember that you're calculating the concentrations in the liquid acid rain not the concentrations of the vapor phase.  Also remember that you're calculating the concentration in units of moles/L.  The question asks for concentration in air in ppm, which is also mg/L.  So, to get to the final answer we need to do two things 1. Calculate the mass of SO2 using the molar mass and 2. Convert the concentration in solution to concentration in air.  Let's  deal with the conversion of concentration in solution to concentration in air. 

Using the Ideal Gas Law we can determine that 1  mole of anything occupies 22.4 L when in the gas phase at standard temperature and pressure. 
          (2)    PV=nRT ==> V = (nRT)/(P)  = [(1 mol)*(0.08206 L*atm*K^-1*mol^-1)*(273.15 K)]/[(1 atm))] = 22.4 L

We have [SO2] in the acid rain.  Let's assume we have 1 mole of acid rain.  From (2) we know that 1 mole of liquid becomes 22.4 L of gas.  So, to convert the concentrations calculated from pH values we use the relationship the 1 mole of SO2 will occupy 22.4 times more volume as a gas then it would as a liquid. 
          [SO2] in air = [(3.15*10^-5 mol SO2/1 mole acid rain)*(1 mole acid rain)]*(1 mole SO2/22.4L) =  1.406*10^-6 moles/L in air

Now convert the concentrations to the correct units         
          [SO2] in air (mg/L) = [SO2]in air (mol/L) * Molar mass * conversion of g to mg
          [SO2] in air (mg/L) = [1.406*10^-6 (mol/L) * 64.048 g SO2/mol * 1000 mg/1g
          [SO2] in air (mg/L) = 9.020*10-2 mg/L = 9.020*10-2 ppm

Hope this helps in your chemistry endeavors.

         

I'm afraid I need to take exception to several statements made by Sparky above.

First, although some of the S02 may be converted into sulfuric acid, this happens after the SO2 (g) reacts with O2 in the air to produce SO3.  The SO3 gas then reacts with water to form sulfuric acid (H2SO4)

2SO2   +  O2   -->  2SO3    (slow)
SO3   +  H2O   --> H2SO4  (sulfuric acid)  (rapid)

However, the production of SO3 (and therefore H2SO4) is not immediate.

However, the SO2 can also react directly with water in the air to produce the weaker sulfurous acid (H2SO3).

       H2O   +  SO2  -->  H2SO3  (sulfurous acid)  (rapid)

I believe, given the wording of the question, that we are to assume that we are suppose to solve the problem based on the assumption that the SO2 goes directly to sulfurous acid rather than the sulfuric acid.

Now one might wonder why this is significant.  The answer to that question lies in the fact that sulfuric acid (H2SO4) is a strong acid, while sulfurous acid (H2SO3) is a weak acid.  That is H2SO4 will essentially ionize completely.
                    H2SO4  -->   2 H+   +  SO4-2

In the case of sulfurous acid, the ionization of the first H+ ion is incomplete, and the ionization of the second H+ occurs to such a small degree that it can probably be ignored.

                        H2SO3      H+   +   HSO3-   (Ka= 1.5 x 10-2)
                        HSO3-       H+   +  SO3-2   (Ka = 10-8)

As a result of this, I believe that the following statement from Sparky's analysis is incorrect:

                       [H2SO4] =  [H+]/2

Nor is it valid to say that
                       [H2SO4] = [SO2]

Secondly,  let's take a look at Sparky's statement below:

Using the Ideal Gas Law we can determine that 1  mole of anything occupies 22.4 L when in the gas phase at standard temperature and pressure. 
          (2)    PV=nRT ==> V = (nRT)/(P)  = [(1 mol)*(0.08206 L*atm*K^-1*mol^-1)*(273.15 K)]/[(1 atm))] = 22.4 L
We have [SO2] in the acid rain.  Let's assume we have 1 L of acid rain.  From (2) we know that 1 L of liquid becomes 22.4 L of gas

While I agree that 1 mole of any gas at STP is equal to 22.4 Liter (as demonstrated by equation (2), that is very different than saying that 1 Liter of liquid becomes 22.4 L of gas.  1 Liter of rain water is approximately 1000 grams, or 55.5 moles and if vaporized would occupy a volume of 1243 Liters at STP.

My last criticism of Sparky's solution is that it doesn't address the final question.  While I agree that one must determine the concentration of SO2 in the rain water, the question is what is the concentration of SO2 in ppm of air that leads to that pH.  I don't believe that converting the liquid rainwater to a gas is going to be very helpful.

Instead, we need to look at Henry's law which describes the relationship between the solubility of a gas and it's concentration in a liquid exposed to the gas.  See my solution above.

If we are able to find the partial pressure of the SO2 in air (which we can assume has a pressure of 1 atm), we can find the mass of both the SO2 and the air, and from there the ppm.

Okay, I have fixed the mistakes that spock pointed out.  Thanks again!  Though my wording might be a little off the equations are correct.  Acid rain is formed by SO2 from the series of reactions that Spock pointed out.  SO2 reacts indirectly within the smoke stack to form H2SO4.  Wikipedia has a good reaction flow.  here's the link,

http://en.wikipedia.org/wiki/Acid_rain

And the actual flow of equations (Wikipedia). 

In the gas phase sulfur dioxide is oxidized by reaction with the hydroxyl radical via a intermolecular reaction:

    SO2 + OH· ? HOSO2·

which is followed by:

    HOSO2· + O2 ? HO2· + SO3

In the presence of water sulfur trioxide (SO3) is converted rapidly to sulfuric acid:

    SO3(g) + H2O(l) ? H2SO4(l)

So, as you can see at no point is H2SO3 formed.  However HSO3. is formed which converts to H2SO4 quite rapidly.  Equilibrium constants though helpful in most situations can not be used here because the volume of air is infinite compared to the volume of gases emitted from the smoke stack.  The reactions always proceed to the right.  Thus, 1 mole of SO2 forms 1 mole of H2SO4.  This H2SO4 falls out as acid rain.  Since pH is calculated in Molarity of H+ it is possible to determine the concentration of sulfuric acid at any pH and thus the concentration of SO2 in the gases leaving the smoke stack that formed acid rain. 

The final answer is correct for the assumed pH of 4.2.  Thanks again Spock!

A bit further down in the Wikipedia article you quoted it says:

Sulfur dioxide dissolves in water and then, like carbon dioxide, hydrolyses in a series of equilibrium reactions:

SO2 (g)+ H2O ? SO2·H2O
SO2·H2O ? H++HSO3-
HSO3- ? H++SO32-

You might also want to take a look at http://www.upei.ca/~physics/p261/projects/acidprec1/Sources%20of%20Acid%20Rain.htm

As for the Ka of H2SO3, this refers to the dissociation of the H2SO3 according to the following equation:
                        H2SO3 (aq)      H+ (aq)    +  HSO3- (aq)

This equilibrium would occur within each raindrop (which represents a closed system) and has nothing to do with the volume of air.

I will be perfectly honest in that I am not sure what the ppm would be.  I don't think it's realistic to suppose that 100% of the SO2 will be converted into H2SO4 for the reasons I've outlined above.

But even if I accept that assumption as being correct, which I agree would indicate a [SO2] in the rainwater of 3.15 x 10-5 M (as you have indicated).  I still don't follow how you can translate that into a ppm of 9.02 x 10-2.  It appears that you are assuming that the [SO2] in the rainwater as being equal to [SO2] in air and I don't believe that to be a correct assumption.

I really hope that Inajam will get the answer from his/her professor and share it with all of us because I think that it is a very interesting problem.