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I read a problem which was as follows:

Equal volumes of $\pu{1M} \ \ce{H3PO4}$ and $\pu{1M} \ \ce{Na3PO4}$ are mixed. The $\mathrm{pH}$ of the resultant mixture is (along with this $K_\mathrm{a1}, K_\mathrm{a2}, K_\mathrm{a3}$ of the acid were mentioned)

The way I approached this problem was as follows: $K_\mathrm{a1} = \frac{[\ce{H2PO4^-}][\ce{H^+}]}{[\ce{H3PO4}]}$, similarly $K_\mathrm{a2} = \frac{[\ce{HPO4^{2-}}][\ce{H^+}]}{[\ce{H2PO4^-}]}$ and so on...

Which gives me $K_\mathrm{a1}\cdot K_\mathrm{a2}\cdot K_\mathrm{a3} = \frac{[\ce{PO4^{3-}}][\ce{H^+}]^3}{[\ce{H3PO4}]} = [\ce{H^+}]^3$ (since $\ce{[H3PO4] = [Na3PO4]}$). Using this, I calculate the $[\ce{H^+}]$ and so the ph which comes out to be $-\log{\sqrt[3]{K_\mathrm{a1}\cdot K_\mathrm{a2}\cdot K_\mathrm{a3}}}$ which is $\frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2} + \mathrm{p}K_\mathrm{a3}}{3}$.

However, it seems from the answer that the system is forming a buffer and $\mathrm{pH}$ is $\mathrm{p}K_\mathrm{a2}$

What am I doing wrong here and how does the $\mathrm{pH}$ come out to be $\mathrm{p}K_\mathrm{a2}$. Is there any such general result or theory for polyprotic acids?

EDIT: I understand now why my solution is wrong. Any help as to how to proceed further will be appreciated.

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    $\begingroup$ Your assumption that the phosphate ion will not hydrolyze makes your attempt incorrect. Final concentration of phosphate will not be equal to it's initial concentration. $\endgroup$ Commented May 28, 2021 at 8:15
  • $\begingroup$ @MathewMahindaratne Doesn't seem to me like they are the same questions because it's hydrolysis is also to be considered in this case since the concentration of Na3PO4 is also given which is not taken in the linked question. Rather the linked question would answer my question if only the concentration of H3PO4 was provided to me $\endgroup$
    – Ashish
    Commented May 28, 2021 at 8:43
  • $\begingroup$ @NisargBhavsar Oh, that makes sense. I understand why it's wrong now $\endgroup$
    – Ashish
    Commented May 28, 2021 at 8:46
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    $\begingroup$ It would be 1 M NaH2PO4 + 1 M Na2HPO4 ( hydrolysis neglected), so pH=pKa2. ( simplification assuming activity coefficient is 1, what is expected in this task context, otherwise not really justified ). It is high concentration, buffers are usually much more diluted solutions, typically 0.05-0.1 M. $\endgroup$
    – Poutnik
    Commented May 28, 2021 at 10:04
  • $\begingroup$ Correction 0.5 M + 0.5 M - simple error. $\endgroup$
    – Poutnik
    Commented May 28, 2021 at 10:57

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