Chemistry questions

outnumbered

Hey guys,
So does anyone know why the acid is placed in the burette rather than the base? I can't seem to recall it and It's not in the notes

Hazardous Garden Of Eden

A lot of bases (like NaOH) would attack the glass etc..
and cause the tap to stick.

Hells bells...

outnumbered

Hazardous Garden Of Eden wrote: »

A lot of bases (like NaOH) would attack the glass etc..
and cause the tap to stick.

Hells bells...

Okay. Thanks a lot!

outnumbered

2001 Hl
Sodium carbonate is not a Primary standard but anhydrous sodium carbonate is. why is this the case?

Hazardous Garden Of Eden

outnumbered wrote: »

2001 Hl
Sodium carbonate is not a Primary standard but anhydrous sodium carbonate is. why is this the case?

Sodium carbonate cannot be prepared in a 100% pure, stable and soluble solid form so that it can be weighed out and readily dissolved in water to provide a solution of accurately known concentration, but the anhydrous can.

Or something to that effect.

outnumbered

outnumbered wrote: »

2001 Hl
Sodium carbonate is not a Primary standard but anhydrous sodium carbonate is. why is this the case?

As sodium carbonate cannot be made up into a solution of known concentration as we don't know the degree of hydration which implies we can't know the Molarity of the solution we make.

outnumbered

Thanks guys!
another question: examinations.ie seems to be down so I can't look up the Marking but I'll post the q here.
washing soda
2.51g washing soda dissolved made up to 250cm^3. Molarity found by titrating 25 of this against 0.1 HCL. Mean titre is 20.
okay so i calculated the concentration to be 0.05m/l
the value of X wont work out for me?? Can ye try this?

outnumbered

Hazardous Garden Of Eden wrote: »

Sodium carbonate cannot be prepared in a 100% pure, stable and soluble solid form so that it can be weighed out and readily dissolved in water to provide a solution of accurately known concentration, but the anhydrous can.

Or something to that effect.

The exact wording was this: Na2CO3XH20 is not a ps but NA2CO3 may be used as a PS. why is this the case?
Are they asking why is NA2CO3 A ps?

outnumbered

outnumbered wrote: »

Thanks guys!
another question: examinations.ie seems to be down so I can't look up the Marking but I'll post the q here.
washing soda
2.51g washing soda dissolved made up to 250cm^3. Molarity found by titrating 25 of this against 0.1 HCL. Mean titre is 20.
okay so i calculated the concentration to be 0.05m/l
the value of X wont work out for me?? Can ye try this?

Actually it's okay. I got it! Had the m/l wrong! :P
I thought that the mass was always 286? no?
also why am i quoting myself? :P

outnumbered

outnumbered wrote: »

Thanks guys!
another question: examinations.ie seems to be down so I can't look up the Marking but I'll post the q here.
washing soda
2.51g washing soda dissolved made up to 250cm^3. Molarity found by titrating 25 of this against 0.1 HCL. Mean titre is 20.
okay so i calculated the concentration to be 0.05m/l
the value of X wont work out for me?? Can ye try this?

25x/1 = 20(0.1)/1
x = 2/25 = 0.08M

0.08 in 1L = 0.02 in 250cm^3

2.51/0.02 = 125.5 - 106 (rmm of washing soda) = 19.5/18 = 1x?

That doesn't sound right at all but I'm doing it all from memory sorry

outnumbered

outnumbered wrote: »

Actually it's okay. I got it! Had the m/l wrong! :P
I thought that the mass was always 286? no?
also why am i quoting myself? :P

Could you post your solution please?!

outnumbered

Deleted User wrote: »

Could you post your solution please?!

m1v1/n1 = m2v2/n2
ie. 20(0.1)/2 = 25m1/1 then cross multiply giving you M1 as 0.04m/l

then you say 0.04m/l thus 0.01m/250
so 0.01m liberates 2.51g
1m liberated ?
2.51/0.01 giving you 251 grams.
NA2CO3XH20 = 251
106 + 18X= 251
X= 8.1

ChemHickey

outnumbered wrote: »

m1v1/n1 = m2v2/n2
ie. 20(0.1)/2 = 25m1/1 then cross multiply giving you M1 as 0.04m/l

then you say 0.04m/l thus 0.01m/250
so 0.01m liberates 2.51g
1m liberated ?
2.51/0.01 giving you 251 grams.
NA2CO3XH20 = 251
106 + 18X= 251
X= 8.1

Don't forget that the substance could be effloresced and have lost some water to the atmosphere. It's a trick the Chief Examiner could/probably will throw in sometime {again} just because everyone, yourself included, is so used to working with 10 and whole numbers like that which are used in every example. It was one thing our teacher made us look out for and had us ready to answer a question on.

If it comes up on the paper as an answer like that (an answer not equal to 10 or the normal w.o.c) a follow-up question could easily be "Give a reason for your result in the previous part"

Ans: The substance has lost some water (effloresced) to the air/atmosphere and therefore the water of crystallisation is lower than expected.

outnumbered

outnumbered wrote: »

m1v1/n1 = m2v2/n2
ie. 20(0.1)/2 = 25m1/1 then cross multiply giving you M1 as 0.04m/l

then you say 0.04m/l thus 0.01m/250
so 0.01m liberates 2.51g
1m liberated ?
2.51/0.01 giving you 251 grams.
NA2CO3XH20 = 251
106 + 18X= 251
X= 8.1

I knew I had the stoichiometry wrong!

outnumbered

ChemHickey wrote: »

Don't forget that the substance could be effloresced and have lost some water to the atmosphere. It's a trick the Chief Examiner could/probably will throw in sometime {again} just because everyone, yourself included, is so used to working with 10 and whole numbers like that which are used in every example. It was one thing our teacher made us look out for and had us ready to answer a question on.

If it comes up on the paper as an answer like that (an answer not equal to 10 or the normal w.o.c) a follow-up question could easily be "Give a reason for your result in the previous part"

Ans: The substance has lost some water (effloresced) to the air/atmosphere and therefore the water of crystallisation is lower than expected.

Just seen it in my notes there haha.
thanks for the tip on the follow up question!

outnumbered

Deleted User wrote: »

I knew I had the stoichiometry wrong!

haha my explanation is diabolical

outnumbered

outnumbered wrote: »

haha my explanation is diabolical

I understood completely
Just a quick question, does anyone remember doing redox titrations of group 7 elements? I don't and if they come up as q1 well then I'm ****ed for that q

outnumbered

Deleted User wrote: »

I understood completely
Just a quick question, does anyone remember doing redox titrations of group 7 elements? I don't and if they come up as q1 well then I'm ****ed for that q

hm? what you on about? there's vinegar, washing soda, hardness , DO, ammo iron, iron tab, sodimum thio, bleach

outnumbered

outnumbered wrote: »

hm? what you on about? there's vinegar, washing soda, hardness , DO, ammo iron, iron tab, sodimum thio, bleach

Hm sorry, redox reaction, not titration.
(a) Redox reactions of group VII elements (b) Displacement reactions of metals
Still doesn't ring any bells :pac:

ChemHickey

Deleted User wrote: »

I understood completely
Just a quick question, does anyone remember doing redox titrations of group 7 elements? I don't and if they come up as q1 well then I'm ****ed for that q

They aren't titrations, they're just a set of experiments to show reduction/oxidation and how good of oxidising agents the halogens are and to show how they become better oxidising agents as you go up the periodic table (F > Cl as an oxidising agent)

ChemHickey

ChemHickey wrote: »

They aren't titrations, they're just a set of experiments to show reduction/oxidation and how good of oxidising agents the halogens are and to show how they become better oxidising agents as you go up the periodic table (F > Cl as an oxidising agent)

Yeah my friend read it wrong from his book and was talking to me on the phone sorry :P

outnumbered

Deleted User wrote: »

Hm sorry, redox reaction, not titration.
(a) Redox reactions of group VII elements (b) Displacement reactions of metals
Still doesn't ring any bells :pac:

ah yeah get it now! oh i hate them!

outnumbered

I have another few questions on water of crystallisation.
1 why would a standard solution of ethanoic acid not be suitable for this titration?
2 suggest a suitable base for standardisnig H2SO4?
3 why was the titration flask ONLY rinsed with deionised water and not with the sol. it is to contain?

CookieMonster.x

outnumbered wrote: »

I have another few questions on water of crystallisation.
1 why would a standard solution of ethanoic acid not be suitable for this titration?
2 suggest a suitable base for standardisnig H2SO4?
3 why was the titration flask ONLY rinsed with deionised water and not with the sol. it is to contain?

2. Na2co3 I think
3. Because if you rinsed it with the soln you would have an unknown volume on the flask and there would be a change in the number of moles reacting.

outnumbered

CookieMonster.x wrote: »

2. Na2co3 I think
3. Because if you rinsed it with the soln you would have an unknown volume on the flask and there would be a change in the number of moles reacting.

Thanks! you don't kniw when amm fe+2 sulphate is used to standardise do u?
also, yiu dont have the answer to 1 no?

CookieMonster.x

outnumbered wrote: »

Thanks! you don't kniw when amm fe+2 sulphate is used to standardise do u?
also, yiu dont have the answer to 1 no?

It's used to standardise potassium permanganate
I'm not sure about q1 but I would guess that it could be because HCl is a strong acid whereas ethanoic acid is a weak acid and it wouldn't react properly? Not too sure so don't take my word for it! :P are these from exam papers?

outnumbered

CookieMonster.x wrote: »

It's used to standardise potassium permanganate
I'm not sure about q1 but I would guess that it could be because HCl is a strong acid whereas ethanoic acid is a weak acid and it wouldn't react properly? Not too sure so don't take my word for it! :P are these from exam papers?

okay brilliant thanks! they are from the 80's/70's

outnumbered

Okay so I have a Water of crystalisation question from the 1995 paper. I think I have it but as there's no marking scheme, I need to be sure. so if anyone could do it and post the answer it would be brilliant!
7.77g NA2CO3XH20 was made up to 500cm^3. 25cm was titrated agaainst 0.075 molar h2so4. average titre is 20. ratio is 1:1.
a) conc of na2co3xh20 in g dm^-3
b) in mol dm^-3.
c) hence find the % water in sodium carbonate crystals and the value of X.
My answers are: a) 15.54 g/l,,, b) 0.06 and c) 59% with x being 8.5. is this right?

chatterboxxx95

Outnumbered , your q 1 is because ethanoic acid is a weak acid and no indicator would be suitable to show the endpoint :P

outnumbered

chatterboxxx95 wrote: »

Outnumbered , your q 1 is because ethanoic acid is a weak acid and no indicator would be suitable to show the endpoint :P

Why thank you Chatterboxxx95