I found an extremely confusing problem in one of the problems we did in class for general chemistry. This was the question:
How many moles of $\ce{H3O+}$ or $\ce{OH-}$ must you add per liter of $\ce{HA}$ solution to adjust its $\mathrm{pH}$ from 3.15 to 3.65? Assume a negligible volume change.
The answer given was as follows:
Going from $\mathrm{pH}$ 3.15 to $\mathrm{pH}$ 3.65, you are decreasing in the amount of acid in solution.
$\mathrm{pH}$ 3.15 is for a solution with concentration of $\ce{[H3O+]} = 7.08 \times 10^{-4}$.
$\mathrm{pH}$ 3.65 is for a solution with concentration of $\ce{[H3O+]} = 2.24\times 10^{-4}$.
One must add the difference of these in base $\ce{(OH-)}$ in order to neutralize the hydronium ions.
Add $(7.08\times 10^{-4} - 2.24\times 10^{-4}) = 4.84\times 10^{-4}$ of $\ce{OH-}$
Now this seems simple enough. But I tried to do this another way, that is, by converting these $\mathrm{pH}$ values to $\mathrm{pOH}$: 10.85 to 10.35. And then I followed the same procedure outlined above, to get a completely different answer:
$\mathrm{pOH}$ 10.85 gives a concentration of $\ce{[OH-]} = 1.41\times 10^{-11}$
$\mathrm{pOH}$ 10.35 gives a concentration of $\ce{[OH-]} = 4.47\times 10^{-11}$
The difference of which gives a $\ce{[OH-]} = 3.05\times 10^{-11}$
I'm copy-pasting the answer I was given for my query as to how this can be:
Logarithmic scales are not linear. Amounts of base need to affect change at $\mathrm{pH}$ 3.15 are not the same that are needed to cause change at $\mathrm{pOH}$ 10.85.
$\mathrm{pH}$ 3.15 means you have a lot of acid and not much base. When you are considering changing the $\mathrm{pOH}$ you are thinking (incorrectly) that you just have a solution of hydroxide in water. That is not the case! At $\mathrm{pOH}$ 10.85 there is a lot of acid present!
The reason is that at $\mathrm{pH}$ 3.15 (or 3.65) the concentration of $\ce{H3O+}$ is a million fold greater than $\ce{OH-}$, and the amount of hydroxide (at $\mathrm{pH}$ 3.15 or $\mathrm{pOH}$ 10.85) is 10,000 times less than what is present in pure water. Thus by adding just the $3.07 \times 10^{-11}$ moles of $\ce{NaOH}$ that would appear (on paper) to raise the $\mathrm{pOH}$ from 10.85 to 10.35, you are forgetting that the solution is not just water and hydroxide, but there is a lot of hydronium ion that is present. Actually that small amount of $\ce{OH-}$ would immediately be neutralized by the huge amount of acid and you would still have a low / acidic $\mathrm{pH}$. See this worked out below:
$\ce{[H3O+]} \text{ (at $\mathrm{pH}$ 3.15) } = 7.08 \times 10^{-4}$. By adding $3.07 \times 10^{-11}$ moles of $\ce{OH-}$, an equal amount of $3.07 \times 10^{-11}$ moles of $\ce{H3O+}$ is neutralized. Now the $\ce{[H3O+]} = (7.08 \times 10^{-4} – 3.07 \times 10^{-11}) = 7.08 \times 10^{-4}$. You see, nothing very much has changed when you are dealing with such a small amount of base and a large amount of acid.
This is the reality because at low $\mathrm{pH}$ you don’t have to add very much acid to change $\mathrm{pH}$ significantly. At high $\mathrm{pH}$ you don’t have to add much base to change the $\mathrm{pH}$ significantly. But, on the other hand, at low $\mathrm{pH}$ you have to add a lot of base to neutralize the large amount of acid present, and similarly at high $\mathrm{pH}$ you have to add a lot of acid to neutralize the large amount of base present.
Bottom line, just because the $\mathrm{pH}$ and $\mathrm{pOH}$ signify the same solution (in regard to relative amounts of hydronium or hydroxide present), how much base you have to add to change the $\mathrm{pH}$ versus the $\mathrm{pOH}$ is very different.
I understood that we need a lot of base to neutral the acid at a very low $\mathrm{pH}$. And I did not pursue this further in class because the mathematical calculations for this discrepancy were beyond the scope of what needed to be covered in our class; but I couldn't let this go until I understood how the math works here, since by my above logic, I should have gotten the same answer using $\mathrm{pOH}$ as when I did using $\mathrm{pH}$, if not for the nonlinear logarithmic discrepancy. I don't know how to work this out.