The Henrys law constant for CO2 in water at 25 C is 31 102

The Henry\'s law constant for CO2 in water at 25 C is 3.1 * 10^-2 M . atm -1.

a. What is the solubility of CO2 in water at this temperature if the solution is in contact with air at normal atmospheric pressure? The mole fraction of CO2 in air is 0.000375.

b. Assume that all of this CO2 is in the form of H2CO3 produced by the reaction between CO2 and H2O: CO2 (aq) + H2O (l) => H2CO3 (aq).
What is the pH of this solution?

Solution

A)You are given the mole fraction (X) for CO2 in air.

Use Raults Law equation P(gas)=X(solvent)P(Solvent)

You know that standard pressure is 1atm and in this question we are given fraction of CO2 in air.

P(CO2) =(0.000375)(1atm)= 0.000375atm

We can plug this answer as our partial pressure for CO2 into Solubility equation

S=kP

S(CO2)=(3.1 * 10^-2 M atm^-1)(0.000375atm)

              =1.2*10^-5M

B)For this part the equation is just stated to show that CO2 completely dissociates to H2CO3. From here we need to do a acid/base equation for H2CO3 while looking up our acid dissociation constant (ka) in Appendix D of chemistry textbook. (ka=4.3*10^-7)

          H2CO3(aq) +H2O(l) -> H+(or H3O+ if you prefer) +HCO3-(aq)

Set up you ICE table and you know that your starting concentration of H2CO3 is equal to our solubility from Part A

[H2CO3] =1.2*10^-5M

Our equilibrium expression will be set up as:

   Ka=[H+][HCO3-]/[H2CO3]

   4.3*10^-7=(x)(x)/(1.2*10^-5M-x)        (ignore x in denominator because k has exponent >-5)

[H+]=x=2.27156*10^-6M (round answer at the end to avoid error)

Plug [H+] into pH=-log[H+]

pH=5.64