How do I determine pH, pOH, [H+] and [OH-], if all I have is the concentration of H2CO3 as 3.51 x 10^-2 M?

Carbonic acid is a weak acid and when in solution, the following reaction takes place:

`H_2CO_3 (aq) -> H^+ (aq) + HCO_3^(-) (aq)`

The dissociation constant for this reaction, Ka1 = 4.3 x 10^-7.

Since both the products are in a 1:1 molar ratio, we can assume that at equilibrium, the concentration of each of them is x M. The leftover concentration of carbonic acid at equilibrium is 3.51 x 10^-2 -x M.

Using the equation for dissociation constant:

`Ka_1 = ([H^+][HCO_3^-])/[H_2CO_3]`

`4.3 xx 10^-7 = (x xx x)/(3.51 xx 10^-2 - x)`

solving this equation, we get, [H+] = 1.23 x 10^-4 M

pH = -log [H+] = - log (1.23 x 10^-4) = 3.91

since pH + pOH = 14

pOH = 14 - pH = 10.1 

and pOH = -log [OH-]

solving this equation, we get [OH-] = 8.12 x 10^-11 M

Hope this helps.