Common ion effect and temperature

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The discussion centers on the influence of common ion concentration on the dissociation of weak acids and the resulting equilibrium shifts. It questions whether adding more H+ to a less dissociated weak acid solution has a more significant effect than to a more dissociated one, particularly at varying temperatures. The participants emphasize the complexity of such systems, cautioning against relying solely on thumb rules for predictions. They suggest that calculations based on the known equilibrium constant (Ka) can clarify the extent of dissociation changes. The experiment aims to determine the Ka of hypochlorous acid (HOCl) at different temperatures, noting substantial pH changes with temperature increases.
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I have been researching extensively, but am unable to find if the extent of the common ion effect is influenced by the concentration of the common ion already in the solution? For example, if there are two solutions of weak acid which have dissociated to a different extent, would the addition of more H+ affect the one that is less dissociated more substantially than the one with many H+ already in solution?

I was wondering if it was like the entropy change at various temperature - if the temperature is high, the change in entropy will have less of an effect.
 
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In general common ion effect is only a rule of thumb. System always proceeds to equilibrium, so that the reaction quotient equals equilibrium constant.

qwerty1232 said:
if there are two solutions of weak acid which have dissociated to a different extent

Why are these solutions dissociated to different extent? They must be different for that, so you can directly compare them.
 
The extent of dissociation of a weak acid might be different at different temperatures. So if the same concentration of strong acid was added to solutions with different concentration of H+ already, would there by any difference in the extent to which the reverse reaction proceeds (to restore equilibrium due to common ion effect)?
 
Try to calculate it, it is not that hard. You will have to define what you mean by the reaction extent.

Please note that - as I signaled earlier - it doesn't make much sense to analyze complex system with a thumb rules. That's the easiest way of getting wrong answers, as in most cases thumb rules work only in some conditions are meet.
 
Seems to me it depends a lot on what you mean by "affect more", that is vague. For example, do you mean will it have a larger percentage effect? If so, relative to what? In any case, as Borek said, it is easy enough to calculate, particularly if you keep the amount of added acid small.
 
Do you mean calculate it by working out the ka? This is what I have been trying to do, but unfortunately cannot due to having so many chemicals in the solution (we were performing a titration).

Basically, we were titrating NaOCl to find the concentration of HOCl/OCl- at various temperatures, and sulfuric acid was added in order to catalyze the oxidation of these (by I-). What I would like to know is if the solution which was less dissociated (cooler temperature) would have been more likely to reform HOCl than the solution which was more dissociated?
Does that make sense?
 
You should assume Ka is known.

I still don't know what the experiment is, but I have again a felling that you are investigating things that are unrelated to the real problem. Hypochlorous acid is so weak it will be dissociated in about 10-3% at pH 2 - and if memory serves me well your pH is even lower than that.
 
The experiment is basically to find the ka of HOCl at various temperatures. However, we've been unable to find each of the concentrations separately, so have just used a pH meter to investigate the extent of dissociation. However, the change in pH was fairly substantial as temperature increased (from about pH0.7-0.3 as temperature increased from 8C to 40C). I am just trying to work out why the change was large, and how to relate this to ka.

I really appreciate all your help in answering all my questions. Please trust me when I say I have not tried to find these answers everywhere myself.

If you have any more ideas, they would be greatly appreciated. Many thanks again.
 
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