I am unable to calculate the $\mathrm{pH}$ of a solution made of $\pu{0.1 M}$ $\ce{H3PO4}$ and $\pu{0.05 M}$ $\ce{Na3PO4}$.

How to set the approximation and which ions are present at the equilibrium?


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  • $\begingroup$ Hint here. $\endgroup$ – Todd Minehardt Sep 21 '16 at 22:24
  • $\begingroup$ Hint 2 - Phosphoric acid can lose three protons. Write each of the reactions down with pKa's. For such polyprotic acids (in general...) only two species will be important. so it's either (1) H3PO4 & H2PO4- (2) H2PO4- & HPO4-2 or (3) HPO4-2 & PO4-3 see ion.chem.usu.edu/~sbialkow/Classes/3600/Overheads/H3A/… $\endgroup$ – MaxW Sep 22 '16 at 0:43
  • $\begingroup$ @Andreade Welcome to ChemistrySE. This appears to be a homework question, and you must demonstrate some effort towards solving the problem. You have already received 2 hints, that should get you started. Feel free to take the tour and get familiar with the site. $\endgroup$ – getafix Sep 22 '16 at 1:29
  • $\begingroup$ Thank you for the hints, I know it is a polyprothic acid but I can't manage to understand which ions leave in solution. I think due to the acid(h3po4) base(na3po4) reaction, only H2PO4- and HPO4-- leave and make a buffer solution, but I'm not sure about that. $\endgroup$ – Andreade Sep 22 '16 at 6:58

You have the the following six independent equations in order to specify the unknown concentrations of six species as present in an aqueous solution consisting of $\pu{0.1 M}$ $\ce{H3PO4}$ and $\pu{0.05M}$ $\ce{Na3PO4}$:

\begin{align} K_1 &= \frac{[\ce{H+}][\ce{H2PO4-}]}{[\ce{H3PO4}]} = 7.5 \times 10^{−3} \tag{1}\\ K_2 &= \frac{[\ce{H+}][\ce{HPO4^2-}]}{[\ce{H2PO4-}]} = 6.2 \times 10^{−8} \tag{2}\\ K_3 &= \frac{[\ce{H+}][\ce{PO4^3-}]}{[\ce{HPO4^2-}]} = 3.6 \times 10^{−13} \tag{3}\\ K_\mathrm{w} &= [\ce{H+}][\ce{OH-}] = 1 \times 10^{−14} \tag{4}\\ 0.15 &=[\ce{PO4^3-}] + [\ce{HPO4^2-}] + [\ce{HPO4-}] + [\ce{H3PO4}]\tag{5}\\ [\ce{H+}] &= 3[\ce{PO4^3-}] + 2[\ce{HPO4^2-}] + 2[\ce{HPO4-}] + [\ce{OH-}] -0.15\tag{6} \end{align}

  • 1
    $\begingroup$ @Charlie Crown I appreciate your edit $\endgroup$ – Adnan AL-Amleh Feb 26 at 17:12

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