Qualification > Sciences
Need help on CIE Chemistry Paper 5!
Greed444:
on OCT/NOV 2009 paper 52, Q1(d) and (e)
Q1(d)
can someone explain how to fill in the tables?? do we need to guess evrything?
and the Q says that there must be the SAME total final volumes.
Im more confused with the marking scheme saying to use < 6x10^2 mol for monoprotic acid,
how is that related in filling the tables?
another thing is, why [diprotic acid]= 0.5 x [monoprotic acid] ?
Q1(e)
can someone outline a simple answer for this? im VERY weak and lost in this.. if you could
help me step-by-step..
Deadly_king:
--- Quote from: Greed444 on October 02, 2010, 07:30:13 am ---on OCT/NOV 2009 paper 52, Q1(d) and (e)
Q1(d)
can someone explain how to fill in the tables?? do we need to guess evrything?
and the Q says that there must be the SAME total final volumes.
Im more confused with the marking scheme saying to use < 6x10^2 mol for monoprotic acid,
how is that related in filling the tables?
another thing is, why [diprotic acid]= 0.5 x [monoprotic acid] ?
Q1(e)
can someone outline a simple answer for this? im VERY weak and lost in this.. if you could
help me step-by-step..
--- End quote ---
Monoprotic acid refers to acids which donate only 1 mole of H+ upon dissociation of 1 mole of acid. Apart from H2SO4 and ethanedioic acid all the other 3 acids are monoprotic.
examples : HCl ---> H+ + Cl-
Only one mole of H+ is formed.
Diprotic acids are those which can donate 2 moles of H+ upon dissociation of 1 mole of acid.
Example : H2SO4 ---> 2H+ + SO42-
To fill the tables now :
Number of moles of NaOH present : 2/1000*30 = 0.06
Since NaOH should be in excess number of moles of monoprotic acids should not exceed 0.06.
So you can state any concentration less than that in your table.
For diprotic acids number of moles should still not exceed 0.06/2 since 1 mole of acod already provides 2 moles of H+.
Deadly_king:
Oops I did not notice part (e)
Anyway am sure by now you must have realised that ethanedioic acid is a diprotic acid.
Let's say you stated 10.0cm3 of ethanedioic acid with concentration 2.0 mol/dm3
As you can see, number of moles of H+ used is 0.04 moles which is less than 0.06 moles. Therefore it's good.
1. Find the Mr of the salt.
Mr : 2(12) +4(16) + 2 +2(18) = 126
Number of moles = Mass/Mr
You need to produce a solution of concentration 2.0mol/dm3
In 250cm3 you'll have only (2/1000 * 250) = 0.5 moles
0.5 = Mass/126 ----> Mass to be used = 63g
Measure 63g of the salt in the graduated flask and add water upto the 250 mark.
Your solution is done.
NOTE : That is just an example to show you the method. In reality you can't really use 63g. It's way too much. You should calculate it for a reasonable mass of about 5-10g.
Hope you understand. If not, let me know your difficulties and i'll try to elaborate more :)
Greed444:
--- Quote from: Deadly_king on October 02, 2010, 07:56:52 am ---Monoprotic acid refers to acids which donate only 1 mole of H+ upon dissociation of 1 mole of acid. Apart from H2SO4 and ethanedioic acid all the other 3 acids are monoprotic.
examples : HCl ---> H+ + Cl-
Only one mole of H+ is formed.
Diprotic acids are those which can donate 2 moles of H+ upon dissociation of 1 mole of acid.
Example : H2SO4 ---> 2H+ + SO42-
To fill the tables now :
Number of moles of NaOH present : 2/1000*30 = 0.06
Since NaOH should be in excess number of moles of monoprotic acids should not exceed 0.06.
So you can state any concentration less than that in your table.
For diprotic acids number of moles should still not exceed 0.06/2 since 1 mole of acod already provides 2 moles of H+.
--- End quote ---
i understand now why we have to use <0.06 moles.
does the [acid] depends mainly only on 1 mole of H+ but not 2 moles of H+?
is that why we have to divide by 2?
how bout for triprotic acid like phosphoric acid?
H3PO4 --> 3H+ + PO4^3- (am i wrong? correct me if im wrong plz ^_^'')
so does that mean we must use less than 0.06/3? since 1mole of the acid provides 3 moles of H+?
Greed444:
--- Quote from: Deadly_king on October 02, 2010, 08:09:00 am ---Oops I did not notice part (e)
Anyway am sure by now you must have realised that ethanedioic acid is a diprotic acid.
Let's say you stated 10.0cm3 of ethanedioic acid with concentration 2.0 mol/dm3
As you can see, number of moles of H+ used is 0.04 moles which is less than 0.06 moles. Therefore it's good.
1. Find the Mr of the salt.
Mr : 2(12) +4(16) + 2 +2(18) = 126
Number of moles = Mass/Mr
You need to produce a solution of concentration 2.0mol/dm3
In 250cm3 you'll have only (2/1000 * 250) = 0.5 moles
0.5 = Mass/126 ----> Mass to be used = 63g
Measure 63g of the salt in the graduated flask and add water upto the 250 mark.
Your solution is done.
NOTE : That is just an example to show you the method. In reality you can't really use 63g. It's way too much. You should calculate it for a reasonable mass of about 5-10g.
Hope you understand. If not, let me know your difficulties and i'll try to elaborate more :)
--- End quote ---
Thanks Bro!!
I finally get it!
so it doesn't matter if i used diffrent volumes and conc and get half of 63g or twice the 63g?
when i use 20cm^3 and 1.0 mol/dm^3, n=0.02 which is <0.06/2,
so to produce 1,0mol/dm3 in 250cm3, 0.25moles needed
and i get mass=31.5g
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