This is the exact question I faced on an exam. Calculate the 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
$$\ce{Na2CO3 ->2Na+ + CO3^2-}$$
I suppose nothing that can contribute to the pH of a solution happens to $\ce{Na+}$ ions and we proceed with $\ce{CO3^2-}$ which has a concentration of $0.05\ \pu{M}$
$$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$
To calculate 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?
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