Chemistry: Post your doubts here!

[gary221]

I requote:

the whole ques?

 

[Ashique]

4 bi) You'll have the KMnO4 in a burrette, and you'll slowly add it to the solution containing Fe2+ ions. So as you keep adding the MnO4- ions, the Fe2+ ions are oxidised to Fe3+ ions, and so the the purple color of the MnO4- solutions keeps on going from purple to colorless. But once all the Fe2+ ions have been oxidised, you've reached the end point where a permanent pale pink color is in the solution. On further addition of the KMnO4, the purple color no longer changes to colorless

c) For a reaction to be feasible, the Er-Eo has to be positive. Since it's reacting with air, it basically means it's reacting with oxygen. The potential is only positive for the Fe2+ ---> Fe3+ reaction.

 

[Ashique]

the whole ques?

No, just part e iv.

gary221 thanks in advance

 

[gary221]

No, just part e iv.

gary221 thanks in advance

okay, first off, find no of moles of NaOH → conc = moles/vol
moles of NaOH = 0.100 * (10 * 10^-3) = 0.001 moles

Moles of CH3CO2H = 0.250 * (10 * 10^-3) = 0.0025 moles
So, clearly, moles of CH3CO2H are in excess.
So, 0.001 moles of NaOH will react with 0.001 moles of CH3CO2H, leaving (0.0025-0.001) = 0.0015 moles of acid in the solution
Now, total volume of solution = 0.02 dm^3

So, conc of acid in soln = moles/vol
conc = 0.0015/0.02 = 0.075 mol/dm^3

Now, substiute values in pH = pKa + log ([salt]/[acid])
conc of salt → moles of salt = moles of NaOH
vol of soln = 0.02 dm^3
conc = 0.001/0.02 = 0.05 mol/dm^3

Hope u gt it!

 

[Ashique]

okay, first off, find no of moles of NaOH → conc = moles/vol
moles of NaOH = 0.100 * (10 * 10^-3) = 0.001 moles

Moles of CH3CO2H = 0.250 * (10 * 10^-3) = 0.0025 moles
So, clearly, moles of CH3CO2H are in excess.
So, 0.001 moles of NaOH will react with 0.001 moles of CH3CO2H, leaving (0.0025-0.001) = 0.0015 moles of acid in the solution
Now, total volume of solution = 0.02 dm^3

So, conc of acid in soln = moles/vol
conc = 0.0015/0.02 = 0.075 mol/dm^3

Now, substiute values in pH = pKa + log ([salt]/[acid])
conc of salt → moles of salt = moles of NaOH
vol of soln = 0.02 dm^3
conc = 0.001/0.02 = 0.05 mol/dm^3

Hope u gt it!

Wow, that was really tricky! :/ Would have never thought of that.

 

[biba]

in 4b (i) we have FeSO4 which has a pale green colour due to Fe^2+ ions... when we start adding KMnO4 (oxidising agent) to FeSO4 the pale green colur starts fading away,the solution turns very pale yellow (almost colourless) because the Fe2+ ions are getting oxidised into Fe3+.. and Fe3+ are yellow coloured when u add more KMnO4 the sloution turns pale pink as the MnO4^- ions are reduced to Mn2+ and Mn2+ ions are pink in colour... but when excess of KMnO4 is added the solution will turn purple because now there are many MnO4^- ions and MnO4- ions are purple in colour!
in 4(c) they said Fe2+ ions (acidified) are stable to oxidation in air.. u open the data booklet and pick ub the equation for Fe2+ in which it is being oxidised to Fe3+ this equation had E* value +0.77.. now u pick up the equation of O2 which is in acidic medium i.e it has H+ ions in it this equation indata booklet has E* value of +1.23.. the reaction occurs between these two equation and E* becums (+1.23)-(0.77)=+0.46V...
in its second part u pick up an equation in which we have Fe(OH)2 and Fe(OH)3.. this equation has E* of -0.56V , now pick up an equation under ALKALINE medium for O2 this one has E* of =0.40.. the reaction occurs between these so the overall E* becums (+0.40)-(-0.56)= +0.96V.. now u comment that the oxidation of Fe(OH)2 is rapid because it has a more positive E* value (remember the more positive the E8 value, the more feasible the reaction is )
​

 

[Oliveme]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf

Its a Ka question:
Q no 3 (e) part iv

Can you please help me out.

I know it has used the formula, pH=pKa + log (salt/acid)
but how the values?

I'll be honest here - this took me a long while to work out, too.

Let's first biuld an equation = CH3COOH + NaOH ------> CH3C00Na + H2O
Moles of acid = 10/1000 x 0.25 = 0.0025 moles
Moles of NaOH = 10/1000 x 0.1 = 0.001 moles
From the equation, the molar ratio should be 1:1 so the acid is clearly in excess.
the total volume of solution is 20 cm^3, right? Because we add 10 and 10.
so moles of salt CH3COOH are 0.001 and so concentration of salt is -----> 0.001/0.02 = 0.05 mol dm^3
The moles of unreacted acid are 0.0025 - 0.001 = 0.0015 moles
so conc. of unreacted acid = 0.0015/0.02 = 0.075 mol dm^3
Plug the values into pH = pKa + log ([acid]/[salt])
this gives ----> pH = 4.76 + log (0.05/0.075) = 4.58

Hope you get it.

 

[messi10]

Wow, that was really tricky! :/ Would have never thought of that.

Bro, in the same paper, how you found the rate constant in Q2 (b) part iii..


Thank You so much guys for helping.. You are saviour!

 

[D0cEngi]

Plz tell me how to balance an ionic equation.

 

[Nab900]

in 4b (i) we have FeSO4 which has a pale green colour due to Fe^2+ ions... when we start adding KMnO4 (oxidising agent) to FeSO4 the pale green colur starts fading away,the solution turns very pale yellow (almost colourless) because the Fe2+ ions are getting oxidised into Fe3+.. and Fe3+ are yellow coloured when u add more KMnO4 the sloution turns pale pink as the MnO4^- ions are reduced to Mn2+ and Mn2+ ions are pink in colour... but when excess of KMnO4 is added the solution will turn purple because now there are many MnO4^- ions and MnO4- ions are purple in colour!​

in 4(c) they said Fe2+ ions (acidified) are stable to oxidation in air.. u open the data booklet and pick ub the equation for Fe2+ in which it is being oxidised to Fe3+ this equation had E* value +0.77.. now u pick up the equation of O2 which is in acidic medium i.e it has H+ ions in it this equation indata booklet has E* value of +1.23.. the reaction occurs between these two equation and E* becums (+1.23)-(0.77)=+0.46V...​

in its second part u pick up an equation in which we have Fe(OH)2 and Fe(OH)3.. this equation has E* of -0.56V , now pick up an equation under ALKALINE medium for O2 this one has E* of =0.40.. the reaction occurs between these so the overall E* becums (+0.40)-(-0.56)= +0.96V.. now u comment that the oxidation of Fe(OH)2 is rapid because it has a more positive E* value (remember the more positive the E8 value, the more feasible the reaction is )​

​

So Fe3+ ions are basically yellow in colour??? aren't fe3+ red-brown in colour

and for part(c), can you please explain why are we taking acidic and alkaline medium into consideration :/

Anyway, THNKS for ur time

 

[biba]

So Fe3+ ions are basically yellow in colour??? aren't fe3+ red-brown in colour

and for part(c), can you please explain why are we taking acidic and alkaline medium into consideration :/

Anyway, THNKS for ur time

wel its wriiten in chemistry book that in aqueous solutions Fe3+ is yellow..
we have to take the acidic and alakine conditions into consideration because they have specified it in the question, if u donot take the conditions into consideration u wil get stuck in the question because there are 2 different equations for O2 in th data booklet

 

[nidanas]

umm hello ppl!
I wanna ask is it true dat the duration of d chem paper for tomorrow will be 2 hours?!
I mean I jst noticed dat 2011 and 2012 ques paper wz 2 hrs long... so any idea abt 2mrw' paper?!

 

[Ashique]

Bro, in the same paper, how you found the rate constant in Q2 (b) part iii..


Thank You so much guys for helping.. You are saviour!

k= rate/[Cr(CO)6]
to find the k
Half life= In2/k
k=In2/half life

 

[biba]

umm hello ppl!
I wanna ask is it true dat the duration of d chem paper for tomorrow will be 2 hours?!
I mean I jst noticed dat 2011 and 2012 ques paper wz 2 hrs long... so any idea abt 2mrw' paper?!

yup it is 2 hours!

 

[aleezay]

Is Barium carbonate water soluble or insoluble? :/

 

[KurayamiKimmi]

umm hello ppl!
I wanna ask is it true dat the duration of d chem paper for tomorrow will be 2 hours?!
I mean I jst noticed dat 2011 and 2012 ques paper wz 2 hrs long... so any idea abt 2mrw' paper?!

best to practice and keep yourself in check for 1 hr and 45 mins , the 15 minutes, if they are given will come in handy for revising

 

[KurayamiKimmi]

Is Barium carbonate water soluble or insoluble? :/

insoluble

 

[Nab900]

wel its wriiten in chemistry book that in aqueous solutions Fe3+ is yellow..
we have to take the acidic and alakine conditions into consideration because they have specified it in the question, if u donot take the conditions into consideration u wil get stuck in the question because there are 2 different equations for O2 in th data booklet

OOHHHHHHHHKK!!!!! I did not see that!!!!!! they were asking in the question abt acidified also....

THANK YOU SO MUCH

 

[sumaiyarox:)]

okay, first off, find no of moles of NaOH → conc = moles/vol
moles of NaOH = 0.100 * (10 * 10^-3) = 0.001 moles

Moles of CH3CO2H = 0.250 * (10 * 10^-3) = 0.0025 moles
So, clearly, moles of CH3CO2H are in excess.
So, 0.001 moles of NaOH will react with 0.001 moles of CH3CO2H, leaving (0.0025-0.001) = 0.0015 moles of acid in the solution
Now, total volume of solution = 0.02 dm^3

So, conc of acid in soln = moles/vol
conc = 0.0015/0.02 = 0.075 mol/dm^3

Now, substiute values in pH = pKa + log ([salt]/[acid])
conc of salt → moles of salt = moles of NaOH
vol of soln = 0.02 dm^3
conc = 0.001/0.02 = 0.05 mol/dm^3

Hope u gt it!

what is d pH here?

 

[gary221]

what is d pH here?

4.6, i think.