helllpp (1 Viewer)

[nit]

Originally posted by m_isk
yeh nothing will happen. THis is because in acid/base reactions as such, it is the acid/base SOLUTION which reacts, they don't react as solids. eg if u used litmus paper with a solid acid (i.e no water) it would not change colour. Why? there's no soln. so the answer is C

You are right about nothing happening, but that's not the reason. There is water present in ther form of the acid rain, and the Na2SO4 will dissolve in it due to its ionic nature. The reason is that the second reaction is given as one that goes completely to the right - ie it is not given as an equilbrium. Thus Le Chatelier's Principle is useless in this case regardless of the fact that we now have more [SO4]2minus ions present. Thus there would be no change in the [HSO3]1minus , H+, SO2 conc.

 

[~GrOoVy~]

you do need to know the standard hydrogen electrode, it's a syllabus dot point! just know
- its purpose (allocated the E0 value of 0.00V so it could be used as a standard to measure other E0 values )
- what it is made of (platinum electrode in a 1M H+ solution with Hydrogenhas bubbled through i think from memory)
- how it works: (this half cell is usually hooked up to another half cell, it is designnated the E0 value of 0.00V, so anything that oxidises is given a positive E0 value, and anything that reduces is given a negative E0 value)
- the conditions in which it operates: temperature, molarity of h+ solution and pressure)

i HAVE seen past trial exam paper questions where you need to know the above

my teacher never taught us about these standard hydrogen electrode so i assumed that it was not in the syllabus...

annywayzz thanks for answering my question eveyone

 

[~*HSC 4 life*~]

my teacher never taught us about these standard hydrogen electrode so i assumed that it was not in the syllabus...

annywayzz thanks for answering my question eveyone

ok maybe i'm wrong, but generally never rely on your teacher rely on the syllabus

my teachers skipped soooo much stuff, but at the end of the day, the baord of studies expects you to know it

 

[~*HSC 4 life*~]

You are right about nothing happening, but that's not the reason. There is water present in ther form of the acid rain, and the Na2SO4 will dissolve in it due to its ionic nature. The reason is that the second reaction is given as one that goes completely to the right - ie it is not given as an equilbrium. Thus Le Chatelier's Principle is useless in this case regardless of the fact that we now have more [SO4]2minus ions present. Thus there would be no change in the [HSO3]1minus , H+, SO2 conc.

oh ok, i understand that...what about if the second reaction was in equilibrium and denoted by a double arrow, would that change anything?...just curious cause i don't know

 

[nit]

Well if the 2nd reaction was denoted as an equilbrium then Le Chatelier's principle would apply and thus the second reaction would tend to shift to the left due to the increase in bisulfite ion concentraion. At this point, the situation becomes slightly tricky - not only are you increasing the bisulfite ion conc, but you're also reducing the proton concentration in the solution - thus, depending on the amount of H+ ion lost and [HSO3]1minus ion gained, the first reaction will either shift to the right or to the left - it's hard to say with Kc's and concentrations. So, you may form sulfur dioxide and you might not, as well. I have a feeling that the 2nd reaction was intentionally denoted as one that goes completely to the right in order to avoid this confusion.

 

[m_isk]

You are right about nothing happening, but that's not the reason. There is water present in ther form of the acid rain, and the Na2SO4 will dissolve in it due to its ionic nature. The reason is that the second reaction is given as one that goes completely to the right - ie it is not given as an equilbrium. Thus Le Chatelier's Principle is useless in this case regardless of the fact that we now have more [SO4]2minus ions present. Thus there would be no change in the [HSO3]1minus , H+, SO2 conc.

yeh thanks for clearing that up nit..just one other thing, is what i initally said correct (but not correct in this instance) or is it a pile of crap?