CHEMISTRY TITRATION HELP

[zahraahmed]

In case a titration is coming this thursday in paper 33 i need help in the following :
please
1. How many times do we perform the titration? 2, 3 ,or 4
2. to how many significant figures is the burette reading to be quoted?
3. What does the examiner mean by a rough titre?
4. do we have to fill the burette again till the zero mark before carrying a titration
5. What do we put in the burette and what in the conical flask?

PLEASE help im damn sure that a titration is coming this time and it would help many if these questions are answered.
many thanks :%) :%) :crazy: :crazy: :shock: :bad: :sorry:

 

[libra94]

1. until u get the same reading!! if u r not getting that, do it atleast 3 times and find the average
2. 4 s.f. !! it must be to 2decimal places!!!
3. ur first titration value on the burette...the first one is the rough one
4. in the first one, yes...but in the 2nd one its not necessary! u can just subtract the final volume from the initial volume
5.well tht depends on the ques...sometimes acid is in the burette and sometimes base..so that depends on the question, it tells u!!

 

[ihatephysics]

@zahra
i think two titrations are more than enough!! because of time constraints
besides the examiner reports say u mention titres to 3 d.p hope libra checks it again!!
and u fill the titration depending on ur own convenience!!

 

[libra94]

@zahra
i think two titrations are more than enough!! because of time constraints
besides the examiner reports say u mention titres to 3 d.p hope libra checks it again!!
and u fill the titration depending on ur own convenience!!

how come 3d.p?? u can write like 34.555?!!?
can u plz send me the link of that marksheme please...

 

[libra94]

@ihatephysics http://www.xtremepapers.me/CIE/Internat ... _ms_31.pdf in this markscheme it says to the nearest 0.05cm^3 which is 2 d.p!!

 

[filza94]

FOR Chemistry PAPER 33!!!!!!!!!!

hydrated ammonim salt titrated with KMnO4
usual salt analysis
one thermal question

We have to titrate ammonium nitrate with KMn04 or would we need to oxidize potassium iodide in the

presence of ammonium nitrate and then titrate the resulting solution with sodium thiosulphate? :|

see question no.1 in june2008-31
a similar question is expected!

 

[zahraahmed]

this is what the examiners say abt titrations:
Titrations were generally performed well. Burette readings for “accurate” titrations were recorded
to 2 decimal places (nearest 0.05 cm as required by the syllabus). The Examiners were pleased
to see that few candidates recorded “impossible” burette readings such as 27.43 cm
The majority of candidates produced consistent titres as described in the syllabus (2 titres within 0.10 cm3
. Many candidates, having obtained two titres within 0.10 cm wasted time by performing
further titrations: 3, or even 4, identical titres or titres within 0.10 cm3 was not unusual
The selection of titres for calculation of the “average” was less successfully performed. Many
candidates ticked only one titre. In this case Examiners accepted the candidate’s chosen value
when assessing accuracy. The difference between the chosen value and the next nearest was
used to calculate spread, and a penalty applied if necessary.

 

[filza94]

hey guyz recent info i got for u ppl for chemistry ppr 33

my class fellow get information that FA1 is ammonium iron salt for heating, Q2 is reaction KMnO4 with KI in the presence of H2SO4, then titration with Na2S2O3...and Q3 contain Ammonium Bromide and zincsulfate salts.....

 

[filza94]

Anyone please explain it in a simple, understandable way. Thanks...

 

[kshumaila52]

hey guyz recent info i got for u ppl for chemistry ppr 33

my class fellow get information that FA1 is ammonium iron salt for heating, Q2 is reaction KMnO4 with KI in the presence of H2SO4, then titration with Na2S2O3...and Q3 contain Ammonium Bromide and zincsulfate salts.....

filza94
You just uploaded my informations......
plz dnt confuse ppl...i told you that these are not sure.....

 

[zahraahmed]

"Concentrated sulfuric acid reacts with KMnO4 to give Mn2O7, which can be explosive.[9] Similarly concentrated hydrochloric acid gives chlorine. The Mn-containing products from redox reactions depend on the pH. Acidic solutions of permanganate are reduced to the faintly pink manganese(II) ion (Mn2+) and water. In neutral solution, permanganate is only reduced by 3e− to give MnO2, wherein Mn is in a +4 oxidation state. This is the material that stains one's skin when handling KMnO4. KMnO4 spontaneously reduces in an alkaline solution to green K2MnO4, wherein manganese is in the +6 oxidation state."
2 KMnO4 + 10 FeSO4 + 8 H2SO4 = K2SO4 + 2 MnSO4 + 5 Fe2(SO4)3 + 8 H2O
2 KMnO4 + 8 H2SO4 + 10 KI = 2 MnSO4 + 8 H2O + 5 I2 + 6 K2SO4

 

[filza94]

it confirmed i called my teacher n asked n i evn tolf u on fb dont shout at me here .....

 

[filza94]

http://notezone.net/cambridgechem/chemi ... sis%5D.pdf plz go dix link pg 11 n c answers below of it

First of all you need the mass of FA1. In the first table you have the mass of the empty test tube and the

mass of the test tube + FA1. All you do is subtract the values 12.80 - 10.50 giving you 2.30g of FA1.

In the second table you need the mass of FA1 after heating. Again, you just subtract the values 11.76 -

10.50 = 1.26g FA1 after heating.

1 d. (i) The mass of anhydrous XS04 present in the crystals is what is left over after heating = 1.26g

1 d. (ii) The mass of water driven from the crystals is the loss in mass that has occurred when the FA1 was

heated. Before heating there were 2.30g of FA1 and after heating there were 1.26g. So the mass of water

lost is 2.30 - 1.26 = 1.04g

1 e. To calculate the number of moles of water in the sample of FA1 you use the formula no. of moles =

mass/mass of 1 mole which gives you 1.04/18 = 0.058 mol

1 f. The number of moles in XSO4
7 moles of H20 gives 1 mole XSO4 (from the formula XSO4.7H20)
So 0.058 moles of water give 0.058/7 = 8.29 x 10 ^ -3 moles of XSO4

1 g. To calculate the relative formula mass of XSO4 you use the formula mass of 1 mole = mass/no. of

moles which gives you 1.26/8.29 x 10 ^ -3 = 152.7

1 h. To calculate the Ar of element X you first find the Mr of SO4 which is 32 + (4 x 16) = 96
You subtract this from the RFM from part g giving you 152.7 - 96 = 56.7