HSO4- acidic or neutral

[ibr]

Hi Chad,

I was just reading my acid base notes and it seems that during the lecture you said HSO4- is an acidic salt but in the answers to one of the problem sets for UNIT 5 GC, you said that even though HSO4- is a NEUTRAL salt, its conj base,,, SO4 2- is basic.... i get the second part since its obviously coming from a weak base but what about the HSO4- anion?


Is it acidic ( since it can donate a H) or is it neutral...since it comes from a string acid ( H2so4)

Kind of confused. Thanks

[ibr]

Hey again,

I am really having a lot of trouble with the concept of equivalence points... for example, you said that a weak acid strong base produce a basic salt at equivalence point which is why the ph > 7 and you gave the example of NaOH titrating HF to form NaF... but what else is in the solution at that point apart from Na+ and F-?

What served to aggravate my confusion was a destroyer question, " a solution of NaOh and HF are titrated to the end point ...What does the final solution contain?"

The answer was H+, OH-, H20 HF, and Na+ and F-....and they said this was because HF doesnt usually dissociate in water,....

This seems very neutral of a solution to me though...Is there a difference between an Endpoint and equiv point.

Also, would HCl be present at endpoint too or if HF just one exception we should know?

Sorry about all the questions, DAT in 2 weeks and suddenly all the things I thought I knew have become super confusing

[chad]

Hi Chad,

I was just reading my acid base notes and it seems that during the lecture you said HSO4- is an acidic salt but in the answers to one of the problem sets for UNIT 5 GC, you said that even though HSO4- is a NEUTRAL salt, its conj base,,, SO4 2- is basic.... i get the second part since its obviously coming from a weak base but what about the HSO4- anion?


Is it acidic ( since it can donate a H) or is it neutral...since it comes from a string acid ( H2so4)

Kind of confused. Thanks

I looked back at the Day 5 problem set and the only question I could see even mentioned HSO4- was #30 and the answer correctly calls it an acidic anion. Which question were you specifically refering to? If I've got an error I'd like to correct it because HSO4- is definitely an acidic anion.

[chad]

Hey again,

I am really having a lot of trouble with the concept of equivalence points... for example, you said that a weak acid strong base produce a basic salt at equivalence point which is why the ph > 7 and you gave the example of NaOH titrating HF to form NaF... but what else is in the solution at that point apart from Na+ and F-?

What served to aggravate my confusion was a destroyer question, \" a solution of NaOh and HF are titrated to the end point ...What does the final solution contain?\"

The answer was H+, OH-, H20 HF, and Na+ and F-....and they said this was because HF doesnt usually dissociate in water,....

This seems very neutral of a solution to me though...Is there a difference between an Endpoint and equiv point.

Also, would HCl be present at endpoint too or if HF just one exception we should know?

Sorry about all the questions, DAT in 2 weeks and suddenly all the things I thought I knew have become super confusing

At the equivalence point in the titration of NaOH and HF, the NaOH and HF will have neutralized each other leaving H2O and Na+ and F-. But F- is a base and will react to produce a little OH-and HF in the solution (F- + H2O --> OH- + HF).
The end point is a little different than the equivalence pt in a titration. The end point is the best approximation of the end point in a titration as you won't typically hit the equivalence point exactly. But treat it pretty much the same. So note you have H2O, and if you have H2O you have to have H+ and OH- too as it autoionizes (the whole basis of the pH scale). And now we see above that we'll have Na+ and F- and that the F- will partially dissociate (weak base) to form a little OH- and HF thus explaining why all these species can be present at equilibrium.
Hope this helps!

[ibr]

Also, I read in destroyer than the surface area does not affect vapor pressure. Isnt vapor pressure sort of like evaporation...and the more the surface area the more vapors?

The other thing destroyer claims is that the more the number of reactants we have the LESS likely is a furitful collision in an elementary sep o the elementary step with the least reactants is most likely. I find this very counterintuitive and was hoping you might be able to shed some light?

Thanks a lot

[ibr]

I looked back at the Day 5 problem set and the only question I could see even mentioned HSO4- was #30 and the answer correctly calls it an acidic anion. Which question were you specifically refering to? If I've got an error I'd like to correct it because HSO4- is definitely an acidic anion.

I was referring to that same question. It seems I just miswrote it in my notes while reviewing the answer... Really sorry about the confusion.. there is no error in the key you posted.

[chad]

Also, I read in destroyer than the surface area does not affect vapor pressure. Isnt vapor pressure sort of like evaporation...and the more the surface area the more vapors?

The other thing destroyer claims is that the more the number of reactants we have the LESS likely is a furitful collision in an elementary sep o the elementary step with the least reactants is most likely. I find this very counterintuitive and was hoping you might be able to shed some light?

Thanks a lot

The vapor pressure is not affected by the surface area. At a given temperature, there will be established an equilibrium concentration (partial pressure) of a liquid's vapor above the solution. No matter how you change the surface area, this vapor pressure (concentration) is fixed by the equilibrium constant at that temperature. I think that you're thinking that like above a lake you'd have a lot more water vapor or something like that. The total number of moles of water vapor above a lake might be higher than say over a covered glass of water but the vapor pressure (concentration) would be the same.

[ibr]

Thanks Chad. That clears up a lot. I am still kind of wondering about the other thing : destroyer claims is that the more the number of reactants we have the LESS likely is a furitful collision in an elementary sep o the elementary step with the least reactants is most likely. I find this very counterintuitive and was hoping you might be able to shed some light?

[chad]

If there were 2 reactants in an elementary step, then these 2 reactants would have to collide in that step and 2 molecules colliding is a likely event.
If there were 3 reactants in an elementary step, then these 3 reactants would have to collide in that step and 3 molecules colliding simultaneously is not such a likely event. Think of 3 cars colliding at the same exact moment all head-on into each other. 2 cars colliding happens all the time. But 3 cars colliding at exactly the same time isn't such a likely occurence.
If we start adding even more reactants...say 8 reactants. Imagine 8 cars all simultaneously converging on the same point in space and colliding. This is just never going to happen right?
It works the same way in chemical reactions. Unimolecular and bimolecular elementary reactions are common but termolecular (3 reactants molecules) are pretty rare and anything more is non-existent as far as I know.