How to calculate pH of the Na2CO3 solution given ambiguous Ka values This is the exact question I faced on an exam:

Calculate the $\mathrm{pH}$ of $\pu{0.05 M}~\ce{Na2CO3}. \\(\ce{H2CO3}: K_\mathrm{a_1}= 4\times 10^{-7},~ K_\mathrm{a_2}= 4.7\times 10^{-11})$

Solution
$$\ce{Na2CO3 -&gt;2Na+ + CO3^2-}$$
I suppose nothing that can contribute to the $\mathrm{pH}$ of a solution happens to $\ce{Na+}$ ions and we proceed with $\ce{CO3^2-}$ which has a concentration of $\pu{0.05M}$
$$\ce{CO3^2- + H2O &lt;=&gt; HCO3- + OH-}$$
To calculate $\mathrm{pH}$, I need to first figure out concentration of $\ce{OH-}$ ions, and to do so I have to know $K_\mathrm{b}$ dissociation constant for $\ce{CO3^2-}$
On equilibrium concentration of species are as follows: $$[\ce{CO3^2-}] = \pu{0.05 M}-x,[\ce{HCO3-}] = [\ce{OH-}] = x$$
So, we have:
$$\dfrac{x^2}{0.05-x}=K_\mathrm b$$
All I need to know is $K_\mathrm{b}$ value for $\ce{CO3^2-}$ ion. I tried to derive $K_\mathrm{b}$ from $K_\mathrm{a}$ values using $K_\mathrm{a} \times K_\mathrm{b} = 1\times 10^{-14}$ but apparently obtained the incorrect answer. What method should I use to do it the right way?
The  carbonate ion is the Conjugate base of the weak acid 
$\ce{HCO_3^-}\ (K={4.7\times10^{-11}})$, 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:
$\ce{CO_3^{-2} + H_2O &lt;=&gt; HCO_3^-  + OH^-}$
$$\ce{K_b}=\frac{[OH^-][HCO_3^-]}{[CO_3^{-2}]} =\frac{K_w}{Ka}=\frac{10^{-14}}{4.7\times 10^{-11}}=\ce{10^{-3.7}}$$
Neglecting the $\ce{OH^-}$ produced by the autoprotolysis of water, it is valid to make the usual assumption that $\ce{[OH^-]}={[HCO_3^-]}$,and thus
  $$\dfrac{[OH^-]^2}{0.05-{[OH^-]}}=K_b= \ce{10^{-3.7}}$$
The equilibrium expression must be solved as a quadratic and yields the root $\ce{[OH^-]}$=0.00306 Which corresponds to pOH = 2.5 or pH = 11.5