Calculate the concentration of all species in a 0.165 m solution of h2co3.

(1) H2CO3 + H2O --> HCO3- + H3O+ [Ka = 4.5E-7]  

(2) HCO3- + H2O --> CO32- + H3O+ [Ka = 4.7E-11]  

Observe that the two Ka values differ from each other drastically. This will mean that in terms of pH, ONLY THE FIRST IONIZATION WILL AFFECT pH.  

(1) H2CO3 + H2O --> HCO3- + H3O+ [Ka = 4.5E-7]  

I0.165M0  

C-x+x  

E0.165-xx  

4.5E-7 = x^2/(0.165-x) <-- use the appoximation rule because Ka is very small  

2.72E-4 = x  

Thus, from equation (1), we see that the [H3O+] = 2.72E-4M, [H2CO3] = 0.165M, and [HCO3-]= 2.72E-4M. Note that THIS step will determine the pH of the polyprotic acid  

To calculate the last species, just transfer these values the next equation  

(2) HCO3- + H2O --> CO32- + H3O+ [Ka = 4.7E-11]  

I2.72E-40  

C  

E  

4.7E-11 = (x*2.72E-4)/2.72E-4  

4.7E-11 = x = [CO32-]  

To calculate the OH- from H3O+, just find the pH of H3O+, subtract it from 14 to find the pOH, and find the [OH-] from that.  

i think that the answer is c