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?

  • 1
    $\begingroup$ We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework. Don't worry, they're not banned. But, we require a minimal effort. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. See homework $\endgroup$ – Poutnik May 7 at 13:02
  • $\begingroup$ Ask yourself what Ka2, resp pKa2 means. $\endgroup$ – Poutnik May 7 at 13:26
  • $\begingroup$ I have updated my first post with a better attempt. $\endgroup$ – Kdbmvp May 7 at 14:03
  • $\begingroup$ You may want to look up the Henderson-Hasselbach equation. $\endgroup$ – Buck Thorn May 7 at 14:03
  • $\begingroup$ @Kdbmvp Ask yourself, how is Ka2 for CO2 defined. $\endgroup$ – Poutnik May 7 at 14:09

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