H_2CO_3, or carbonic acid, is a weak acid formed from carbon dioxide reacting with water.
CO_2(g) + H_2O(l) rightleftharpoons H_2CO_3 (aq)
Being a weak acid, it will only partially dissociate in water, and has a dissociation constant, K_a, of 4.3 times 10^-7 according to [This table.](http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf) Really, carbonic acid is diprotic, meaning it can dissociate twice, so we have a second K_a value for the second dissociation: K_a=4.8 times 10^-11. Which will also contribute to the pH. (albeit to a smaller extent than the first dissociation)
Lets set up the dissociation equation for K_a of the first dissociation:
K_a= ([H_3O^+] times [HCO_3^(-)])/([H_2CO_3])
Now, let's plug in our values for the concentration of the carbonic acid, along with the K_a value.
4.3 times 10^-7=([H_3O^+] times [HCO_3^(-)])/(5.0 times 10^-2)
2.15 times 10^-8=([H_3O^+] times [HCO_3^(-)])
Now, we can assume that [H_3O^+] =[ HCO_3^(-)] as they exist in a 1:1 ratio in the solution. This allows to take the square root out of the expression ([H_3O^+] times [HCO_3^(-)]) to find the respective concentrations:
sqrt(2.15 times 10^-8) approx (1.47 times 10^-4)=([H_3O^+] = [HCO_3^(-)])
Now, in the second dissociation, the [HCO_3^(-)] ion will act as the acid, and therefore the concentration of this species, which we found in the first dissociation, will be the value of the denominator in the new K_a expression:
K_a= ([H_3O^+] times [CO_3^(2-)])/([HCO_3^-])
4.8 times 10^-11= ([H_3O^+] times [CO_3^(2-)])/([1.47 times 10^-4])
approx 7.04 times 10^-15=[H_3O^+] times [CO_3^(2-)]
sqrt(7.04 times 10^-15) approx 8.39 times 10^-8=[H_3O^+] = [CO_3^(2-)]
So the concentration of Oxonium ions, [H_3O^+], which determine the pH, is approximately (1.47 times 10^-4)+(8.39 times 10^-8) approx 1.47 times 10^-4.
In other words, the second dissociation was so small that it could be have been considered negligible- But let's be thorough.
Now, using the equation to find pH we can calculate the pH of this solution.
pH=-log[H_3O^+]
pH=-log[1.47 times 10^-4]
pH approx 3.83
