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A solution of $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$, when provided in equal amounts at the same concentration, form an excellent buffer. What happens is that the two sodium-phosphate compounds each dissociate independently: \begin{align} \ce{Na2HPO4 &<=> HPO4^2- + 2Na+}\tag1\\ \ce{NaH2PO4 &<=> H2PO4^- + Na+}\tag2 \end{align}

What would happen if you made up a solution that had only half of the reactant in (1)?

Am I right in saying that Le Chatelier's principle is not involved, and that shifts in equilibrium don't occur?

Will the solution just have a lower buffering capacity for bases relative to acids?

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  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. $\endgroup$ – Martin - マーチン Sep 2 '15 at 5:30
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You are not right, Le Châtelier's principle plays a very important role.

Let's have a look at the reactions involving disodium monohydrogen phosphate, as it is an amphoteric substance and therefore a buffer on its own: \begin{align}\ce{ Na2HPO4 + H2O &<=> 2Na+ + H3+O + PO4^3- \\ Na2HPO4 + H2O &<=> 2Na+ + {}^{-}OH + H2PO4^- \\ }\end{align} To a small extent, there will also be the formation of phosphoric acid, but we'll ignore that for the moment.

Let's ignore the counter ion and formulate the equilibrium constants: \begin{align} \ce{HPO4^2- + H2O &<=> H3+O + PO4^3-} & K_1 &=\frac{c(\ce{PO4^3-})\,c(\ce{H3+O})}{c(\ce{HPO4^2-})\, c(\ce{H2O})} & = K_\mathrm{a}\cdot c(\ce{H2O})\\ \ce{HPO4^2- + H2O &<=> {}^{-}OH + H2PO4^-} & K_2 &=\frac{c(\ce{H2PO4^-})\,c(\ce{{}^{-}OH})}{c(\ce{HPO4^2-})\, c(\ce{H2O})}& = K_\mathrm{b}\cdot c(\ce{H2O})\\ \end{align}

The whole reaction can be described with a coupled equilibrium constant: $$K = K_\mathrm{a}\cdot K_\mathrm{b}\cdot c^2(\ce{H2O})$$

At all times in a pure solution there will be phosphate, monohydrogen phosphate, dihydrogen phosphate and maybe even phosphoric acid in solution. Changing any of the concentrations of these species will change the equilibrium.
The same equations can be applied to the sodium dihydrogen phosphate system. The initial concentrations of course will be different.
Depending on the ratio of the two solutions, the pH will also change.

The buffer capacity is of course not only dependent on the overall concentration, but also if it is near the ideal buffer point.

Overall, polyprotic acids are very complicated systems and it solely depends on your understanding of the word "excellent" how the buffer behaves. There are many fine tuned buffer systems for various purposes, this website has quite a few neat tables.

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  • $\begingroup$ To make the above simple (1) pH will change to be more acidic (2) Buffer capacity at that pH will be less than if the ideal solution was made. $\endgroup$ – MaxW Nov 1 '15 at 15:43
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    $\begingroup$ @Max, this is exactly what I tried to avoid with my answer. It does not make things simple, but lazy. $\endgroup$ – Martin - マーチン Nov 2 '15 at 4:03

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