What is the $\mathrm{pH}$ when you dissolve equal amounts of $\ce{NaHCO3}$ and $\ce{Na2CO3}$ in water?
When both of the compounds are dissolved, we will have equal amounts of $\ce{HCO3-}$ and $\ce{CO3^2-}.$ But how do we go from this to actually calculating the $\mathrm{pH}$?
In the solution manual, it is given that $\mathrm{pH} = \mathrm pK_\mathrm{a2} = 10.33$ (It seems like the $\mathrm pK_\mathrm{a2}$ value of $\ce{CO2}$ has been used here — why?), but I don't understand how they can just conclude with that.
My attempt
So, we will start with equal amounts of $\ce{HCO3-}$ and $\ce{CO3^2-}$. We have the following equilibrium: $$\ce{HCO3- <=> H+ + CO3^2-}$$ We can set up an equilibrium expression (and cancel) until we are left with $K_\mathrm{a2} = [\ce{H+}]$ and $-\log K_\mathrm{a2} = \mathrm pK_\mathrm{a2} = \mathrm{pH} = -\log[\ce{H+}]$.
Now, what I don't understand is why are we using the $\mathrm pK_\mathrm{a2}$ of $\ce{CO2}$? Why that value?
