#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. 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#