How to answer syllabus dotpoint (1 Viewer)

[micuzzo]

Hey every1... i need some help answering this dot point: 'Describe the difference between a strong and a weak acid in terms of an equilibrium between the intact molecule and its ions'

I know the first part about the strong and weak but do not know how to tie it in with equilibrium...

can someone please help???

Thank You

 

[Pwnage101]

um, since u kinow the difference between a strong and weak acid, you would know that a weak acid only partially dissociates forming H3O+, ie when weak acids react with water, they are in equilibrium, when strong acids react, they fully ionise (at HSC level chemistry, that is)

eg: HCl is a strong acid and thus fully ionises (notice the full arrown, not the equlibrium arrow)

HCl_(aq) + H₂O_(l) ---> Cl^-_(aq) + H₃O⁺_(aq)


CH₃COOH, on the other hand, is a weak acid, and thus only partially (~5%) ionises, and thus is in equilibrium (with the revers reaction heavily favoured)

CH₃COOH_(aq) + H₂O_(l) <---> CH₃COO^-_(aq) + H₃O⁺_(aq)

 

[micuzzo]

um, since u kinow the difference between a strong and weak acid, you would know that a weak acid only partially dissociates forming H3O+, ie when weak acids react with water, they are in equilibrium, when strong acids react, they fully ionise (at HSC level chemistry, that is)

eg: HCl is a strong acid and thus fully ionises (notice the full arrown, not the equlibrium arrow)

HCl(aq) + H₂O_(l) ---> Cl^-_(aq) + H₃O⁺_(aq)


CH₃COOH, on the other hand, is a weak acid, and thus only partially (~5%) ionises, and thus is in equilibrium (with the revers reaction heavily favoured)

CH₃COOH_(aq) + H₂O_(l) <---> CH₃COO^-_(aq) + H₃O⁺_(aq)

um ok... yeh i know the question seems dumb but i use chemcontext2 and it doesn't follow the syllabus dotpoints sequentially... so sometimes i am confused as to which dotpoint relates to what i am reading...

ie when weak acids react with water, they are in equilibrium

does this always hold true

 

[dp624]

yeah it sounds about right.

 

[Pwnage101]

yes, that is the definition of a 'weak' acid

 

[micuzzo]

yeah it sounds about right.

ok if u dont mind me clarifying... its in equilibrium bcoz not all the H ionise to form H+... so basically the reaction must be continuing (and hasnt stopped) and we call it equilibrium...however wen it reaches equilibrium, to the naked eye, it looks like nothing happening but in fact the reaction still continuing

if this dot point has much to do with pH this is maybe the reason i dont really get it...chemcontext is pissing me off bcoz it speaks of strong/weak acids equilibrium before pH

Which book du u guys recomend???

 

[dp624]

ok if u dont mind me clarifying... its in equilibrium bcoz not all the H ionise to form H+... so basically the reaction must be continuing (and hasnt stopped) and we call it equilibrium...however wen it reaches equilibrium, to the naked eye, it looks like nothing happening but in fact the reaction still continuing

if this dot point has much to do with pH this is maybe the reason i dont really get it...chemcontext is pissing me off bcoz it speaks of strong/weak acids equilibrium before pH

Which book du u guys recomend???

Yup, sounds pretty good.
But you dont need to know much about pH in relation to this dot point.

Maybe later on you'll have to relate the two when you learn both.

 

[Pwnage101]

chemistry contexts 2 (school provided) ,conquering chemistry, jacaranda (by geofferey Thickett - newish one, nice colours), dot point chemistry are what i used, all good textbooks - i agree contexts 2 is not the best to structure your learning on, the others are better, contexts 2 is better for a more in-depth look after you've learnt the basics

on the note of equilibrium, you should have learnt in year 11 that equlibrium does not mean the reaction is not progressing, it simply means that the rate of the forward reaction is equal to the rate of the reverse reaction, so there is no net change

It is important to note that this does not mean that it is 50% ractants 50% products

i'll give an example

say i have the reaction I_2(g) <----> I_2(s)

i start off with 100 molecules of the solid iodine. So initially the reaction will go to the left. Once the gas molecules are produced, some will turn back solid. At some point, it will reach equilibrium - not necessarily when there are 50 gas molecules and 50 solid molecules, but when the rate of gas molecules turning to solid is equal to the rate of solid molecules going to gas

for example, this may be when there are 75 gas molecules and 25 solid molecules. The reaction still continues, however, as gas molecules are still turning to solid, and vice versa, however at this point for every gas molecule that changes into solid, a solid molecule changes into gas, and so the rastes of both the forward and reverse reactions are equal

 

[micuzzo]

chemistry contexts 2 (school provided) ,conquering chemistry, jacaranda (by geofferey Thickett - newish one, nice colours), dot point chemistry are what i used, all good textbooks - i agree contexts 2 is not the best to structure your learning on, the others are better, contexts 2 is better for a more in-depth look after you've learnt the basics

on the note of equilibrium, you should have learnt in year 11 that equlibrium does not mean the reaction is not progressing, it simply means that the rate of the forward reaction is equal to the rate of the reverse reaction, so there is no net change

It is important to note that this does not mean that it is 50% ractants 50% products

i'll give an example

say i have the reaction I_2(g) <----> I_2(s)

i start off with 100 molecules of the solid iodine. So initially the reaction will go to the left. Once the gas molecules are produced, some will turn back solid. At some point, it will reach equilibrium - not necessarily when there are 50 gas molecules and 50 solid molecules, but when the rate of gas molecules turning to solid is equal to the rate of solid molecules going to gas

for example, this may be when there are 75 gas molecules and 25 solid molecules. The reaction still continues, however, as gas molecules are still turning to solid, and vice versa, however at this point for every gas molecule that changes into solid, a solid molecule changes into gas, and so the rastes of both the forward and reverse reactions are equal

Yeah i understand the equilibrium process... it just that my text book didnt clarify that when weak acids ionise in water the reaction is an equilibrium one... but im pretty sure i understand it now... as i said earlier, to my knowledge, it is basically because not all the Hydrogen is ionised so therefore the reaction must still be occuring (probably because H3O+ is forming and also undergoing equilibrium to produce water at which point more Hydrogen from the acid is ionised) and continues until equilibirium, at which point a reaction is still occuring but now there are equal amounts of reactants and products at any given time so to the naked eye it looks like nothing is occuring... is that right??

Sorry for essay...

 

[Pwnage101]

Yeah i understand the equilibrium process... it just that my text book didnt clarify that when weak acids ionise in water the reaction is an equilibrium one... but im pretty sure i understand it now... as i said earlier, to my knowledge, it is basically because not all the Hydrogen is ionised so therefore the reaction must still be occuring (probably because H3O+ is forming and also undergoing equilibrium to produce water at which point more Hydrogen from the acid is ionised) and continues until equilibirium, at which point a reaction is still occuring but now there are equal amounts of reactants and products at any given time so to the naked eye it looks like nothing is occuring... is that right??

Sorry for essay...

No!!!! There does not necessarily need to be equal amounts of reactants and products, but the rate of the forward and reverse reactions are equal, so to the naked eye it looks like nothing is changing/occurring

 

[micuzzo]

No!!!! There does not necessarily need to be equal amounts of reactants and products, but the rate of the forward and reverse reactions are equal, so to the naked eye it looks like nothing is changing/occurring

oops sorry i meant to say the rate not amount (really i did) but is everything else correct??? thats my main issue

 

[Pwnage101]

oops sorry i meant to say the rate not amount (really i did) but is everything else correct??? thats my main issue

yes, everything else is correct

 

[micuzzo]

yes, everything else is correct

Pwnage101 you have been an exceptional help!

I very much appreciate it...

Thank You very much dude!!!!

 

[Pwnage101]

Pwnage101 you have been an exceptional help!

I very much appreciate it...

Thank You very much dude!!!!

No worriez

post up any questions you have whenever, and i'll b glad to help