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]

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..

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⁺ + SO₄^2-

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]

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⁺ + SO₄^2-

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⁺.

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]

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

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

[Deadly_king]

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+?

Yupz........it usually depends on that

Basically this is the principle!

I can see that I have been a good tutor

[Deadly_king]

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

You're welcome pal

You are a very good student..........

Yeah it's alright! But it would be best to calculate a solution using the minimum mass possible 

[Greed444]

need helps on May/June 2010 paper52 Q1(c) and (d)

Q1(c)
what do we have to do draw here? i simply drew a solution-filled beaker on top of a stand with a bunsen burner underneath it for boiling. and also a thermometer inserted into the beaker. Is this a good approach or the other way around? and are there better ways to answer this?

Q1(d)
can someone explain why the limited supply of 100g of d.water?
do we have to divide them into beakers
(e.g 10g in one beaker, 20g in the 2nd beaker while 30g and 40g in the 3rd and 4th one)?


what do we have to write to gain a mark of three?
can someone guide me to answer this type of question? i can't really answer this very well.
Thanks

[Deadly_king]

need helps on May/June 2010 paper52 Q1(c) and (d)

Q1(c)
what do we have to do draw here? i simply drew a solution-filled beaker on top of a stand with a bunsen burner underneath it for boiling. and also a thermometer inserted into the beaker. Is this a good approach or the other way around? and are there better ways to answer this?

Q1(d)
can someone explain why the limited supply of 100g of d.water?
do we have to divide them into beakers
(e.g 10g in one beaker, 20g in the 2nd beaker while 30g and 40g in the 3rd and 4th one)?


what do we have to write to gain a mark of three?
can someone guide me to answer this type of question? i can't really answer this very well.
Thanks

(c) I guess your diagram should be ok. But you need to add some details.

The question ask to label all the apparatus used and its capacity.

Hence you should state the capacity of the thermometer ( Ex : 100^oC) and also that of the beaker used (Ex : 500cm³).

For the measurement of temperature, I would suggest you to make use of two thermometers.
One which is just immersed in the solution, that is its bulb is just in the water. Another one which is just on the surface of the liquid without touching it though. Make use of retort stands and clamps to hold your thermometers in position.

NOTE : Both thermometers should have same range and same sensitivity. This also should be stated in your diagram.

If you add all these, all the three marks should be yours 

(d) Yeah, you should divide the water accurately so that you will not need more than 100g. You should be preparing solutions of different concentrations each time using some water so as to keep total mass used constant.

But you should not in any way exceed the use of 100g of deionised water. This is a limitation which you must precisely make sure that you cope with.

Then you need to devise a way such that you obtain at least 5 solutions of different concentration in an appropriate table.

The last mark is for correct calculation of the molality of the different solutions that you need to calculate.

[moon]

For diprotic acids number of moles should still not exceed 0.06/2 since 1 mole of acod already provides 2 moles of H⁺.
[/quote]
Why do we have to divide by 2 and not multiplying??  1mole NaOH reacts with 1mole of sulphuric acid and hence reacts with 2H⁺
so 0.06moles NaOH reacts with 0.06*2 of H⁺ Isn't it? I still didn't understand this point and how can we keep the total final volumes the same in each experiment??? I really need help urgently cuz the markschemes and the examiner's reports are not very useful............PLZ HELP!!!! thanx in advance

[moon]

F0r nov 2009 ppr52 I really need an explanation for Q1(f) & Q3(d) the markscheme is not giving an answer.I would really appreciate ur help.

[ashish]

F0r nov 2009 ppr52 I really need an explanation for Q1(f) & Q3(d) the markscheme is not giving an answer.I would really appreciate ur help.

q1(f)

H = Mc deltaT

you have to take M as the total volume of the solution
H= V_T*4.3 * delta T

NaOH is the reagent in excess and HCl is the limiting reagent

therefore you should use the limiting reagent for calculation the heat change for 1 mole

suppose you use 20cm³  of acid of concentration 2.0 mol/dm³

1000cm³ of HCl contains 2.0 mol HCl
20cm³-----------  (2.0/1000)*20
                                       =0.04

then 0.04 mole of HCl gives H J of heat (V_T*4.3 * delta T)
           1 mole----------- (H/0.04)*1 J



Q3(d)

you have to work out the molecular formula let say your mean value for Cu was 2.00

element  mass    Ar          mole         /simplest
O           1.00    16.0        0.0625        1.98  =2
Cu          2.00    63.5        0.0315       1

Copper (I) oxide
CuO₂

[Deadly_king]

For diprotic acids number of moles should still not exceed 0.06/2 since 1 mole of acod already provides 2 moles of H⁺.

Why do we have to divide by 2 and not multiplying??  1mole NaOH reacts with 1mole of sulphuric acid and hence reacts with 2H⁺
so 0.06moles NaOH reacts with 0.06*2 of H⁺ Isn't it? I still didn't understand this point and how can we keep the total final volumes the same in each experiment??? I really need help urgently cuz the markschemes and the examiner's reports are not very useful............PLZ HELP!!!! thanx in advance

Nan.........1 mole of sulfuric acid provides 2 moles of H⁺
2NaOH + H₂SO₄ ----> Na₂SO₄ + 2H₂O
1 mole of sulfuric acid reacts with 2 moles of NaOH

In your case 0.03 moles of sulfuric acid is required to neutralise 0.006 moles of NaOH because sulfuric acid provides 2H⁺

Total volume is kept constant so that volume can be used as a measure of the concentration of the solutions. You keep total volume constant by adding water.

If you have any problem, let us know and we'll try to help you as much as you can.
Relax dude

[moon]

Q3(d)

you have to work out the molecular formula let say your mean value for Cu was 2.00

element  mass    Ar          mole         /simplest
O           1.00    16.0        0.0625        1.98  =2
Cu          2.00    63.5        0.0315       1

Copper (I) oxide
CuO₂



How did u get an average 2??? my average is 3.983 so the formula of copper oxide is CuO PLZ help..... thanks 4 ur help

[moon]

Total volume is kept constant so that volume can be used as a measure of the concentration of the solutions. You keep total volume constant by adding water.

If you have any problem, let us know and we'll try to help you as much as you can.
Relax dude

what is the volume of water do we have to use?

Thanx a lot for ur help..