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I am confused as to why the [H3O+] of a buffer changes very slightly when diluted (at least in an intuitive sense).

Consider:

$\ce{HNO2 + H2O <=> H3O+ + NO2-}$

  • [$\ce{HNO2}$] = 0.45 M and [$\ce{NO2-}$] = 0.30 M yields [$\ce{H3O+}$] of $6.8 \times 10^3 M$. After solving: $4.5 \times 10^{-4} = \frac{x(0.30+x)}{0.45-x}$ where the ionization constant of nitrous acid is $4.5 \times 10^{-4}$

If these concentrations were to be divided by two (the solution is diluted by a factor of 2)

  • [$\ce{HNO2}$] = 0.225 M and [$\ce{NO2-}$] = 0.15 M yields [$\ce{H3O+}$] of $6.7 \times 10^3 M$. After solving: $4.5 \times 10^{-4} = \frac{x(0.15+x)}{0.225-x}$

Mathematically this makes sense, but I can not intuitively reason why a buffer would behave this way in the "real world." Why can buffers resist changes in pH so effectively when diluted?

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  • $\begingroup$ In the real world it doesn't quite work out that way. The equilibrium depends not on the concentration but the activity of the chemical species. In dilute solutions the two are pretty much equal. But about 0.1 molar and above the activity is less than the concentration. In essence the ions are not just floating around the solution bare. Because water molecules are polar, each ion has multiple spheres of water molecules oriented towards the ion. So in essence, at higher concentrations, the water molecules get used up and the oriented spheres start to interact with each other. $\endgroup$ – MaxW Dec 3 '17 at 1:17
  • $\begingroup$ Buffers also don't stay constant when greatly diluted because of the autodissociation of water. But that leaves a great range where they are effective. Also a buffer doesn't typically need to maintain an exact pH, but rather +/- 1 pH unit. $\endgroup$ – MaxW Dec 3 '17 at 1:29
  • $\begingroup$ The solutions of the equations are wrong. They should be $x = 6.725 \cdot 10^{-4}$ for the first one, and $x = 6.700 \cdot 10^{-4}$ for the second one, both corresponding to a pH of about 3.17. The mathematical reason is pretty clear I suppose (essentially, $x = [ H_3 O^+ ]$ is much smaller than the acid and anion concentrations you're considering). But IMO that's also the intuitive reason: if on dilution the ratio between the concentration of the acid and anion species stays ~constant, from the Henderson-Hasselbach equation you know that $[ H_3 O^+ ]$ will also stay ~constant. $\endgroup$ – user6376297 Dec 3 '17 at 9:30
  • $\begingroup$ BTW, if you want to do a more accurate calculation that is valid for lower concentrations too (but not 100% exact, due to the activity as MaxW observed), you need to write the mass balance for each species, the charge balance (including $[OH^- ]$) and the equilibrium constants (including water), and eliminate all unknowns except $[ H_3 O^+ ]$. If you do things right, you will be left with 1 equation in $[ H_3 O^+ ]$ and known constants, which you can solve (numerically) to obtain the pH. This will show you that below a certain concentration, dilution does change the pH a lot. $\endgroup$ – user6376297 Dec 3 '17 at 9:36

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