pH of 0.5M K2CO3

$\ce{H2CO3}$ ionises as 
\begin{align}
\ce{H2CO3 + H2O &<=> H3O+ + HCO3-} & K\_\mathrm a &= \pu{4.0\*10^{-7}}\\
\ce{HCO3- + H2O &<=> H3O+ + CO3^2-} & K\_\mathrm a &= \pu{5\*10^{-11}}\\
\end{align}


The question is to find out the $\mathrm{pH}$ of $\pu{0.5 M}$ $\ce{K2CO3}$ solution. 


Since $\ce{K2CO3}$ is a salt of weak acid and strong base hence its pH is given by 
$$\mathrm{pH} = \frac{\mathrm pK\_\mathrm w + \mathrm pK\_\mathrm a + \log c}{2}.$$


I am unsure of which $\mathrm pK\_\mathrm a$ I should consider. Any help shall be appreciated. 




Write down the actual (equilibrium) reaction that develops between carbonate ion and water when one proton is exchanged (only full-fledged strong acids or bases might exchange a second proton to any significant extent). Which conjugate acid/base appears? $\ce {H2CO3/HCO3-}$ or $\ce {HCO3-/CO3^{2-}}$? When you see which one it is, use that dissociation constant.


 1

The carbonate ion is the Conjugate base of the weak acid 
$\ce{HCO\_3^-}\ (K\_{a(HCO\_3^-)}={5\times10^{-11}})$, so this solution will alkaline. Given the concentration of this solution ,the pH should be sufficiently high to preclude the formation of any significant amount of $\ce{H\_2CO\_3}$ , so the solution of this problem as a solution of a monoprotic weak base:
$\ce{CO\_3^{-2} + H\_2O <=> HCO\_3^- + OH^-}$
$$\ce{K\_{b(CO\_3^{-2})}}=\frac{[OH^-][HCO\_3^-]}{[CO\_3^{-2}]} =\frac{K\_w}{K\_{a(HCO\_3^-)}}=\frac{10^{-14}}{5\times 10^{-11}}=\ce{2\times10^{-4}}$$
Neglecting the $\ce{OH^-}$ produced by the autoprotolysis of water, it is valid to make the usual assumption that $\ce{[OH^-]}={[HCO\_3^-]}$,and little of the carbonate goes to bicarbonate,so$\ce{[CO\_3^{-2}]}={[CO\_3^{-2}]\_0}=0.5$ ,so thus
 $$\dfrac{[OH^-]^2}{0.5}=K\_{b(CO\_3^{-2})}= \ce{2\times10^{-4}}$$
$\ce{[OH^-]}={10^{-2}}$ Which corresponds to $\ce{pOH = 2}$ or $\ce{pH = 12}$


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