# Discrepancies in using pOH vs pH to solve H+/OH- concentration change problem

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.

• The question is poorly formulated, as the task has the solution dependent on the HA. If assuming it is fully dissociated strong acid, than the task is easy ( unless we introduce the hell by involving activities ) Mar 18, 2022 at 15:54
• The explanation is also not so clear. The key is that as you add base, the $\ce{H+ + HO- <=> H2O}$ equilibrium must be maintained. With a large excess of H+ relative to HO-, that means that when you add HO-, essentially all of the HO- reacts with H+, so the HO- concentration changes only a tiny bit while the H+ concentration changes a lot. You can work this out with an ICE table. You'll see that the approximation that the H+ concentration changes by the amount of HO- added is nearly correct, and that the HO- concentration changes only a tiny bit Mar 18, 2022 at 16:13
• When the pOH is greater than 7, the concentration of $\ce{OH-}$ ions has nothing to do with the amount of added $\ce{NaOH}$ solution. These ions have only one origin : the autoprotolysis of water. Mar 18, 2022 at 16:15

The most precise answer to a problem like this requires consideration of the equilbrium of water dissociation: $$\ce{H2O <=> H+ + HO-}$$, where $$K_w = [\ce{H+}][\ce{HO-}]=10^{-14}$$.

Although it is not specified in the problem, we also have to assume that HA is a strong acid that is fully dissociated at the initial pH.

As you correctly calculated, the initial pH of 3.15 indicates that $$[\ce{H+}]=7.08\times10^{-4}$$ M and $$[\ce{HO-}] = 1.41\times 10^{-11}$$ M. The final pH of 3.65 indicates that $$[\ce{H+}]=2.24\times10^{-4}$$ M and $$[\ce{HO-}] = 4.47\times 10^{-11}$$ M.

Now let's consider what happens when you add base to the initial solution. There are two possibilities for every $$\ce{HO-}$$ that goes in. It can either react with $$\ce{H+}$$ to make water (eliminating both ions from solution and decreasing the $$\ce{H+}$$ concentration) or it can remain unreacted and result in an increase in the $$\ce{HO-}$$ concentration. We'll thus define two variables: let $$x$$ be the total amount of $$\ce{HO-}$$ added (that is, the answer to the problem) and let $$a$$ be the amount that reacts with $$\ce{H+}$$.

It is then easy to see that change in $$[\ce{H+}]$$ between the initial and final conditions is simply $$-a$$, and the change in $$[\ce{HO-}]$$ must be $$x-a$$. Since we already have numerical values for the initial and final concentrations, solving for $$a$$ and $$x$$ is trivial algebra and gives us a final result of $$a= 4.84\times10^{-4}\text{ M and. . . }$$ $$x=4.84\times10^{-4}\text{ M}$$

The fact that both values are the same to three digits tells us that the difference $$x-a$$ is so small relative to $$x$$ or $$a$$ that it is negligible and thus $$x\approx a$$. That is why you can calculate the amount of base added ($$x$$) simply using the change in the pH (after converting to $$-a$$, the change in $$[\ce{H+}]$$). You're implicitly assuming that $$x\approx a$$.

When you try the same thing with pOH, it doesn't work because you are implicitly assuming that $$x=x-a$$, which is obviously false.

• This makes a lot of sense now. Thank you so much @Andrew!!
– jj_a
Mar 20, 2022 at 0:18
• @jj_a: If you like the answer you should accept it. Just use the green check mark to do so. Mar 20, 2022 at 20:57

The explanation is simple, and has nothing to do with logarithms. [Here I assume we are working with a strong acid and a strong base.]:

You are thinking the change in the amount of $$\ce{OH^-}$$ should be equal to the amount of $$\ce{OH^-}$$ you added. But that can't be the case, since most of the $$\ce{OH^-}$$ is used up in neutralizing the $$\ce{H^+}$$! I.e., most of the $$\ce{OH^-}$$ you add combines with $$\ce{H^+}$$ to give you $$\ce{H2O}$$. The increased amount of $$\ce{OH^-}$$ $$(\pu{3.05*10^{-11} mol})$$ is merely what's left over following the neutralization, and is determined by the final $$\ce{[H^+]}$$ and the value of $$K_w$$.

So you can see your misunderstanding has nothing to do with logarithms. The logarithms are only used to convert concentrations to pH, i.e., to change the mode of presentation of the concentrations. You would have had the same misunderstanding if you had calculated the change in $$\ce{[OH^-]}$$ without using pH or pOH. You still would have been confused why the change in the amount of $$\ce{OH^-}$$ present was much less than the amount of $$\ce{OH^-}$$ you added.

If you don’t want to consider the autodissociation of water explicitly, you have to figure out what the major species is. In the absence of weak acids and bases, the major species (other than water) is hydronium ions at acidic $$\mathrm{pH}$$ and hydroxide ions at basic $$\mathrm{pH}$$.

It gets a bit more complicated if your question asks you to go from acidic to basic, or basic to acidic, e.g. from $$\mathrm{pH \ 3}$$ to $$\mathrm{pH \ 12}$$. In that case, you would add $$\pu{0.001 mol L-1}$$ strong base to get to neutral, and another $$\pu{0.01 mol L-1}$$ base to get to $$\mathrm{pH \ 12}$$.

In all of these considerations, you are ignoring the minor species, hoping that you are making a negligible mistake in doing so.