Titration titre volume question

Here is the exam question:
The pH of a barium hydroxide solution is lower at 50 °C than at 10 °C
At 50 °C a 25 cm3 sample of this barium hydroxide solution was neutralised by
22.45 cm3 of hydrochloric acid added from a burette.
Deduce the volume of this hydrochloric acid that should be added from a burette to
neutralise another 25 cm3 sample of this barium hydroxide solution at 10 °C
I thought that the answer would be greater than 22.45 cm3 because at 10 C, the pH of the barium hydroxide solution is higher. Therefore, there would be a greater concentration of hydroxide ions in the solution. Therefore, wouldn't a greater volume of hydrochloric acid be required for neutralisation? The actual answer is just 22.45 cm3.
Thank you in advance.

Here is the exam question:
The pH of a barium hydroxide solution is lower at 50 °C than at 10 °C
At 50 °C a 25 cm3 sample of this barium hydroxide solution was neutralised by
22.45 cm3 of hydrochloric acid added from a burette.
Deduce the volume of this hydrochloric acid that should be added from a burette to
neutralise another 25 cm3 sample of this barium hydroxide solution at 10 °C
I thought that the answer would be greater than 22.45 cm3 because at 10 C, the pH of the barium hydroxide solution is higher. Therefore, there would be a greater concentration of hydroxide ions in the solution. Therefore, wouldn't a greater volume of hydrochloric acid be required for neutralisation? The actual answer is just 22.45 cm3.
Thank you in advance.

If you heat the solution of Ba(OH)2, you aren’t actually changing how much solute there is. As such, the amount of the acid required to react with all of it is no different. It doesn’t require any calculations, but it was quite a sneaky question.
Whilst [OH^-] increases when Ba(OH)2 (aq) is heated, these hydroxide ions come from the ionisation of water (e.g H2O <=> H^+ + OH^-) and so you are also increasing [H^+] by the same amount. Essentially any extra hydroxide ions you get are effectively already neutralised and so it doesn’t take more acid to react with them all.

I'm going to chip in with what I hope clarifies TN's fine answer.
The point of titration is to add just the right amount of (in this case) acid to the alkali until the mixture is neutral. Neutral being defined as when [H+] = [OH-].
Now lets say in a certain volume of water at 50oC there are 100 H+ ions and therefore 100 OH- ions, i.e. the water is neutral. All these ions come from this equilibrium: H2O <-> H+ + OH-
If you add in some Ba(OH)2 which equates to 10,000 OH- ions. If you add 10,000 H+ ions (from the added HCl) then you'd end up with 10,000 H2O, leaving you with the 100 H+ and 100 OH- you started with.
If you started at 10oC with say 10 H+ and 10 OH-... added 10,000 OH- and then 10,000 H+... you'd have the equal amounts of H+ and OH- that you started with.
i.e. the OH- and H+ from the water dissociation don't interfere with the amount of acid needed to do the titration.
Well... it made sense in my head.
And yeah I know Le Chatelier will have a thing or two to say about how many H+ and OH- from the water dissociation when you add a bunch of Ba(OH)2, but it doesn't really matter, since they will both be affected by the same amount, which is kinda the whole point of this rant.

I thought the hydroxide ions originate from barium hydroxide and not the water

I kind of understand but I don’t think I entirely do either. So when you are stating the number of hydrogen and hydroxide ions does this translate to the concentrations of Ba(OH)2 and HCl?

I thought the hydroxide ions originate from barium hydroxide and not the water

I need to log on more often, two fine people have beaten me to the reply
Changing T will affect the position of H2O <-> OH- + H+. Imagine having some pure hot water. The pH will be less than 7 since the equilibrium shifts to the right and [H+] has increased, but it is still neutral.
Adding a bunch of Ba(OH)2 increases [OH-] and makes it alkaline as now [OH-] >> [H+]. Adding a bunch of HCl to that neutralises the OH- until once again [OH-] = [H+] and we call it neutral. We don't care what the exact value of [H+] and [OH-] are, I just care that they're equal.
The amount of H+ needed to neutralise it must be equal to the amount of extra OH- added, not the total amount of OH- (i.e. from both the Ba(OH)2 AND the water dissociation... otherwise that would leave just the H+ from the water dissociation and the final mixture would be acidic).

Um...how can a decrease in pH to one which is less than 7 and an increase in hydrogen ion concentrations still result in the water being neutral? I thought that, if water is neutral, the concentration of hydrogen ions = concentration of hydroxide ions?

Please tell me if I'm wrong. So, the decrease in temperature of the barium hydroxide solution is not affecting the concentration of hydroxide ions from the barium hydroxide solution in any way? But it is affecting the concentration of hydroxide ions in the water (ie increasing it), so this causes the H^+ concentration to increase by the same amount? And the hydroxide ions from the barium hydroxide solution and the hydroxide ions from the water can be treated as two separate systems? Did you know that water was involved in this question because it stated barium hydroxide solution?
I now have additional questions about this:
1. When the [OH-] and [H^+] both increase, do they combine together or remain as separate ions?
2. If the barium hydroxide is in solution, it dissociates to form barium ions and hydroxide ions? So wouldn't the hydroxide ions from the barium hydroxide "mix" (I did not know a better word to say) with the hydroxide ions from the water?
3. Let's say barium hydroxide was insoluble in water and instead was soluble in ethanol. Would the answer to the exam question still be the same?

A solution is only neutral if [H^+] = [OH^-], not necessarily if pH = 7.
Water ionises to form both H^+ and OH^- in equal amounts, as per the equation
H2O (l) <=> H^+ (aq) + OH^- (aq)
This ionisation process is endothermic, so heating up an aqueous solution will cause the equilibrium shown to go further to the right. This will result in [H^+] and [OH^-] increasing equally.
As such, the pH of the solution will decrease because pH is simply defined as pH = -log[H^+] and the bigger [H^+] is allowed to get, the lower the pH becomes. This is independent of how [OH^-] changes, since [OH^-] doesn’t show up in the definition for pH.
If the solution happened to be neutral beforehand, then [H^+] and [OH^-] will still be equal and so the solution will remain neutral. The pH will decrease, however. You can apply a similar principle if the solution was acidic or basic to begin with, since [H^+] and [OH^-] must necessarily increase equally and so one of these kinds of ion will outnumber the other.

Changing the temperature of the Ba(OH)2 solution does not change the extent to which the Ba(OH)2 dissociates - once it’s dissolved, it’s already completely dissociated and that won’t change. However, a change in temperature does affect how much water dissociates - regardless of how much water dissociates, you make equal amounts of H^+ and OH^- as a result.
You can regard the hydroxide ions from the water and the hydroxide ions from Ba(OH)2 as two separate systems. In this particular question, it’s the most appropriate thing to do.
The fact barium hydroxide is a solution was one giveaway that water was involved. The other giveaway was the change in temperature.

I thought neutral meant pH 7 so thank you for clarifying that. So a solution with pH of 0 can be considered neutral because [H^+] = [OH^-]? So this question was testing knowledge on the ionisation of water and nothing to do with titration reaction between Ba(OH)2 and HCl?

I am beginning to understand. So for point 3 even if a decrease in temperature decreased the pH of the barium hydroxide dissolved in a solvent, the molecules in the solvent would be acting in the same way as water?

I’m not entirely sure I understand the question you are trying to ask, but if the pH changes in any way because of a temperature change, it’s almost certainly because the position of the ionisation equilibrium is changing.

In theory, if you were able to make a solution where [H^+] = [OH^-] = 1 mol dm^-3 (e.g pH = 0), that would be a neutral solution (though this isn’t possible under normal conditions since water’s neutral pH at its boiling point is about 6.13)
I suppose so - the trap they seemed to set was that since pH always decreases if you raise the temperature, some candidates would almost certainly think you need to change the volume of acid required to bring it back to neutral.

I meant that, in the same way that position of equilibrium in the ionisation of water changes when the barium hydroxide solution (dissolved in water) was heated, if the barium hydroxide had been dissolved in another solvent which does not have water, and the mixture was heated, then the molecules of the solvent would be acting similarly to water. Maybe I need to look more into ionisation equilibrium.