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How is it that buffers of different compositions can have the same pH? For example, it is possible to prepare 0.01 M phosphate buffer of pH 7.0 and 0.1 M phosphate buffer of pH 7.0? How?

I used Henderson-Hasselbalch equation, but still failed. No way is working. Is it something related to volume? Can it be shown mathematically? Can anyone show it mathematically that it is possible?

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For a phosphate buffer with $\mathrm{pH} = 7,$ the two dominant species are $\ce{H2PO4-}$ and $\ce{HPO4^2-}$. The relevant $\mathrm{p}K_\mathrm{a}$ is $7.2$ (this is $\mathrm{p}K_\mathrm{a2}$). From the Henderson-Hasselbalch equation, using $\mathrm{pH} = 7,$ the ratio of $\ce{HPO4^2-}$ to $\ce{H2PO4-}$ is about $0.631.$ You can see this in the alpha diagram below:

Phosphate alpha diagram

In this figure, the green trace is the $\ce{H2PO4-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14.$ The blue trace is the $\ce{HPO4^2-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14$. So, at $\mathrm{pH} = 7,$ the blue curve is below the green curve, and the ratio of alpha fractions is $10^{-0.2}$, i.e., about $0.631.$ This is directly from the Henderson-Hasselbalch equation. Since all the phosphate species are in the same buffer volume, this is also the ratio of the respective concentrations.

The difference between the $\pu{0.01 M}$ phosphate buffer and the $\pu{0.1 M}$ phosphate buffer is that the $\pu{0.1 M}$ buffer is ten times higher in both $\ce{HPO4^2-}$ concentration and $\ce{H2PO4-}$ concentration. So the ratio is the same as in the $\pu{0.01 M}$ buffer and the $\mathrm{pH}$ is the same for the two buffers. Of course, the more dilute buffer has lower buffer capacity, but that is another issue.

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