[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 n_OH− 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 K_w=[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]

Wouldn't [H3O+] and [OH−] change proportionally to one another, given that K_w=[H3O+][OH−]?

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

[Card1ologist]

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

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 K_w=[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 K_w?

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!