# Calculating pH of solution

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?

• 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 – Poutnik May 7 at 13:02
• Ask yourself what Ka2, resp pKa2 means. – Poutnik May 7 at 13:26
• I have updated my first post with a better attempt. – Kdbmvp May 7 at 14:03
• You may want to look up the Henderson-Hasselbach equation. – Buck Thorn May 7 at 14:03
• @Kdbmvp Ask yourself, how is Ka2 for CO2 defined. – Poutnik May 7 at 14:09