Calculate the pH of Na2CO3
It suffices to consider only the reaction

$$\ce{CO3^{2-} + H2O &lt;=&gt; HCO3- + OH-}$$

with $k_{B_1} = 2.1 \cdot 10^{-4}$, since the reaction

$$\ce{HCO3- + H2O &lt;=&gt; H2CO3 + OH-}$$

with $k_{B_2} = 2.3 \cdot 10^{-8}$ will only contribute a small amount of $\ce{OH-}$.

$$k_{B_1} = \frac{[\mathrm{{HCO_{3}}^{-}}] [\mathrm{OH^-}]}{[\mathrm{{CO_{3}}^{2-}}]} = \frac{x \cdot x}{c - x} \approx \frac{x \cdot x}{c}$$

$$x = [\mathrm{OH^-}] = \sqrt{k_{B_1} \cdot c} = 6.455 \cdot10^{-3}$$

which gives a pH of 11.8.