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Card1ologist:
Hi all,

I've been a little stuck on a textbook problem concept that I just can't seem to wrap my head around. The question is as follows:

"How many moles of H3O+ or OH− must you add to a 5.6L of strong acid solution to adjust its pH from 4.52 to 5.25? Assume a negligible volume change."

I decided to approach this problem by first converting the provided pH values into pOH, then found [OH−] for both. I then converted [OH−] to nOH− by multiplying by 5.6L. I subtracted the two values (final-initial) to yield ≈8.10*10-9 mol.

This solution was incorrect. The textbook says that the solution is 1.4*10-4 mol, which does make sense looking at the way they solved it, keeping pHs, finding Δ[H3O+] and using the autoionization equation of water to justify Δ[H3O+]=[OH−] needed. But I don't know why my solution doesn't make sense. Wouldn't [H3O+] and [OH−] change proportionally to one another, given that Kw=[H3O+][OH−]? So wouldn't a change in the mols of OH- in the solution as per the change in pOH reflect the moles of OH- added? I would greatly appreciate knowing where I went wrong...

Borek:

--- Quote from: Card1ologist on November 12, 2022, 10:31:33 PM ---Wouldn't [H3O+] and [OH−] change proportionally to one another, given that Kw=[H3O+][OH−]?
--- End quote ---

That would work if the dependence was linear (one thing directly proportional to another).  It is not.

Card1ologist:

--- Quote ---That would work if the dependence was linear (one thing directly proportional to another).  It is not.
--- End quote ---
I see; I figured that might be the case. I am wondering why the dependence isn't linear. If H3O+ and OH− react in a 1:1 ratio, and Kw=[H3O+][OH−], why wouldn't a decrease in [H3O+] result in an increase of the same magnitude in [OH−], given that their product has to equate to Kw?

In any case, thank you so much for the reply!

mjc123:
The product is constant, not the sum. That's why it isn't linear. (It is linear on a log scale, which is why pH + pOH is constant.)

If H+ is in excess over OH-, then when you add OH- it is nearly all consumed by reacting with H+. That's why it doesn't work to simply subtract the OH- concentrations in the two solutions. You have to add more OH- than that to make the change - in fact, equivalent to the difference in [H+] between the two solutions.

Card1ologist:
Ah, that makes sense now! Thank you for the reply!

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