Calculating Equilibrium Concentrations of H+ and HSO3 in SO2-Rainwater Reaction

[gunnar]

Hi, can someone please give me a hint to this problem?

The concentration of SO2 in the troposphere over a certain region is 0,16 ppm by volume. The gas dissolves in rainwater as follows:
SO2(g) + H2O(l) >> H+(aq) + HSO-3(aq)
Given that the equilibrium constant for the preceding reaction is 1.3e-2. We assume the reaction does not affect the partial pressure of SO2.

I've tried everything I can think of. I am not sure how to handle HSO3 in the equation, is it the same conc. as H+?

In trouble

 

[vsage]

Assuming the amount of $$H^+$$ and $$HSO_3^-$$ was insignificant compared to after the reaction, it's safe to treat the two reactants as having the same concentration.

 

[gunnar]

OK. I treated them as to be the same conc. But how can I get the pressure of SO2?

 

[GCT]

You need the concentration of S02, try finding it from the data

$$Ka=[x][x]/[(concentration~of~S02)-x]$$, solve for x

$$pH=-log[x]$$

 

[gunnar]

I solved for x and still don't get it right. I got the pressure of SO2 to be 5,6e-5 atm from the information I got. Used P=nRT/V
Since SO2 is 0,16 ppm it must be 0,16 mg in one liter I assume STP conditions. And I got the MW of SO2 to be 64,07 g/mol
Am I wrong?

 

[GCT]

why are you using pressure, it's supposed to be in concentration, I'm pretty sure that you can directly convert to concentration from ppm by volume scale. Don't forget to take the negative log of x, this will be your answer.