- Messages
- 6,440
- Reaction score
- 31,077
- Points
- 698
I requote:
the whole ques?
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
I requote:
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!
I'll be honest here - this took me a long while to work out, too.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?
Bro, in the same paper, how you found the rate constant in Q2 (b) part iii..Wow, that was really tricky! :/ Would have never thought of that.
So Fe3+ ions are basically yellow in colour??? aren't fe3+ red-brown in colourin 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 )
wel its wriiten in chemistry book that in aqueous solutions Fe3+ is yellow..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
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!
yup it is 2 hours!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 revisingumm 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?!
insolubleIs Barium carbonate water soluble or insoluble? :/
OOHHHHHHHHKK!!!!! I did not see that!!!!!! they were asking in the question abt acidified also....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
what is d pH here?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!
4.6, i think.what is d pH here?
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now