# How can I prepare two phosphate buffer of different concentrations but give same pH? [closed]

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?

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: 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.