what is the pH of a solution prepared to be 0.43 mol/L carbonic acid?

Answer 1

You need to be careful when you say "a solution prepared to be 0.43 mol/L carbonic acid". That means [H2CO3], including dissociated ions [HCO3-] and very very small amount of [CO3(2-)], has a total concentration of 0.43 mol/L. However, there is another equilibrium:
CO2 + H2O ⇌ H2CO3
The equilibrium constant at 25°C is Kh= 1.70×10−3, that means that the majority of the carbon dioxide is not converted into carbonic acid and stays as CO2 molecules. The CO2 molecules in the solution is NOT included in your given concentration of 0.43 mol/L. To keep [H2CO3] = 0.43M, you really have to keep a very high CO2 pressure above the solution.
Besides what I have said, let us come to your problem considering [H2CO3] ≈ 0.43M.
H2CO3 ⇌ H+ + HCO3-, pKa = 6.36
Since [HCO3-] ≈ [H+], we have:
Ka = 10^-6.36 = [H+]*[HCO3-]/[H2CO3] = [H+]^2/0.43
Thus [H+] = sqrt(0.43*10^-6.36)
pH = -log([H+]) = 0.5*pKa - 0.5*log(0.43)
= 0.5*6.36 - 0.5*log(0.43) = 3.36