Chemistry: Post your doubts here!

[Metanoia]

Plz also tell how to solve Q2 (c)(III) using the formula Ka= (H+)(salt) / (acid )
oct 06

Ka= (H+)(salt) / (acid )

log Ka = log [(H+)(salt) / (acid )]

log Ka = log (H+) + log [(salt)/(acid )]

-log (H) = - log Ka + log [(salt)/(acid )]

pH = pKa+ log [(salt)/(acid )]

pH = 7.2 + log (0.002/0.005)

 

[Metanoia]

guys can you work out this question...........................
An excess of sparingly soluble solid, M(OH)3 was added to 200cm3 of 0,1 mol/dm3 KOH. The mixture was shaken, allowed to reach equilibrium at rtp and the filtered.
25.0cm3 of the filtrate was titrated with 0,2mol/dm3 of HCl of which 18,0cm3 were required to reach end point.....
(i) Calculate the number of moles of hydroxide ions in the filtrate...........
(ii)Find the concentration of hydroxide ions in the filtrate..........which are from M(OH)3
(iii)Hence deduce the concentration of M3+ ions
(iv)State what is meant by solubility of a substance and hence give the solubility of M(OH)3 in 0.1 mol/dm3 KOH
(v) Use your answer to calculate Ksp value of M(OH)3........................

i) moles of OH- in filtrate = moles of HCl used in neutralization = 0.2 x 0.018 = 0.0036 mol

ii) moles of OH- from KOH = 0.1 x 0.025 = 0.0025 mol
moles of OH- from M(OH)3 = 0.0036 - 0.0025 = 0.0011 mol
[OH-] from M(OH)3 = 0.0011/0.025 =0.044 mol/dm^3

iii) M(OH)3 <--> M3+ + 3OH-
[M3+] = 0.044/3 = 0.0147 mol/dm^3

iv) solubility of M(OH)3 in 0.1 M of KOH = 0.0147 mol/dm^3

v) Ksp = [M3+] [OH-]^3 = (0.0147)(0.0036/0.025)^3 = 4.39 x 10^-5 (doesn't tally with suggested answer?)

 

[shingiechingz@8]

i) moles of OH- in filtrate = moles of HCl used in neutralization = 0.2 x 0.018 = 0.0036 mol

ii) moles of OH- from KOH = 0.1 x 0.025 = 0.0025 mol
moles of OH- from M(OH)3 = 0.0036 - 0.0025 = 0.0011 mol
[OH-] from M(OH)3 = 0.0011/0.025 =0.044 mol/dm^3

iii) M(OH)3 <--> M3+ + 3OH-
[M3+] = 0.044/3 = 0.0147 mol/dm^3

iv) solubility of M(OH)3 in 0.1 M of KOH = 0.0147 mol/dm^3

v) Ksp = [M3+] [OH-]^3 = (0.0147)(0.044)^3

your answer at (v) is not equal to 1,5*10-4 mol4dm-12

 

[Metanoia]

your answer at (v) is not equal to 1,5*10-4 mol4dm-12

Oh, I spotted my mistake, I used [OH-] from M(OH3) only instead of the total [OH-], a quick calculation still doesn't get me 1,5*10-4 mol4dm-12.

Will relook at it again.

Were the other parts correct?

 

[shingiechingz@8]

Oh, I spotted my mistake, I used [OH-] from M(OH3) only instead of the total [OH-], a quick calculation still doesn't get me 1,5*10-4 mol4dm-12.

Will relook at it again.

Were the other parts correct?

yes all of them......................

 

[Metanoia]

yes all of them......................

Part v) answer is 1,5*10-4?

I've recalculated and got 4.39 x 10^-5.

I'll check later to see if I can spot what went wrong.

 

[Saad the Paki]

Hey is this question for AS or A2 ?
Seems damn tough :/

It's definitely A2....

 

[Mr.Physics]

http://www.alevelchem.com/edexcel_a_level_chemistry/exam/Jan10/6CH01_01_que_20100114.pdf
Q 18 c (ii)

 

[phanttasy]

http://www.alevelchem.com/edexcel_a_level_chemistry/exam/Jan10/6CH01_01_que_20100114.pdf
Q 18 c (ii)

enthalpy change of combustion means the enthalpy change when 1 mole substance is burnt completely in excess oxygen.
Let the answer in c(i) be x
If 0.32g give you x J of energy
Then 1 mole (86g) gives you 86x/0.32

since the enthalpy change of combustion is always exothermic, so the sign is negative
I think is like this >< hope you can understand it. Correct me if I am wrong

 

[Muskan Achhpilia]

Hey,
Can anyone briefly explain to me what are construction lines in paper 5 A level Chemistry?A diagram would really be appreciated.

Thanks.

 

[Muskan Achhpilia]

Can someone please help me out, of how to go about the following question-:




-----------------------------------------------------------------------------------------------------------------------------
Marking scheme



-----------------------------------------------------------------------------------------------------------------------------
Examiner report



2007 summer Paper 5
As such I have never solved paper 5 and my teacher has also not taught me how to go about it, if i could get a sample response or any help it would really be appreciated.

Thanks a lot.

 

[Muskan Achhpilia]

Also is there any link for paper 5 preparation?

 

[Mr.Physics]

enthalpy change of combustion means the enthalpy change when 1 mole substance is burnt completely in excess oxygen.
Let the answer in c(i) be x
If 0.32g give you x J of energy
Then 1 mole (86g) gives you 86x/0.32

since the enthalpy change of combustion is always exothermic, so the sign is negative
I think is like this >< hope you can understand it. Correct me if I am wrong

Yep. Thanks man

 

[Saad the Paki]

The correct answer here is B. Some one explain why

 

[donewithtime80]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_4.pdf

Q 1, b (i) please...

http://studyguide.pk/Past Papers/CI.../9701 - Chemistry/9701_s03_ms_1+2+3+4+5+6.pdf

 

[The Sarcastic Retard]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_4.pdf

Q 1, b (i) please...

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9701 - Chemistry/9701_s03_ms_1+2+3+4+5+6.pdf

Ecell = 0.76V
Electrode potential of (Cu2+/Cu) = 0.34V
Electrode potential of (M2+/M) = ?

There is one rule to find Ecell. Always subtract big value from small. Here we know the big value is of Cu2+/Cu as its mentioned to be positive electrode. Hence,
ECu - EM = Ecell
0.34 - Em = 0.76
Em = -0.42V

 

[The Sarcastic Retard]

The correct answer here is B. Some one explain why View attachment 58111

I always used trial and error method for this kindda questions.
For option B:~

Sum of moles at equilibrium = 2 - 2x + 2x + x = 2 + x

 

[Saad the Paki]

I always used trial and error method for this kindda questions.
For option B:~
View attachment 58116
Sum of moles at equilibrium = 2 - 2x + 2x + x = 2 + x

Niice thanks !

 

[The Sarcastic Retard]

25 a and b
I guess I know the answer, but I am not sure if I am 100% correct.
a) From all three Cr2+ has least positive E(std) value. Hence it needs the strongest reducing agent (more -ve E(std) value agent) to reduce it.
b) From all three Ag has most posivtive E(std) value. Hence it needs the strongest oxidising agent (more +ve E(std) value agent) to oxidise it.
M I corrct? If not then do correct me. Thanks.
Metanoia
DarkEclipse
Dark Destination
awesomaholic101
FranticAmaze
My Name

 

[MYLORD]

I have notes for chemistry unit 3 and physics unit 3, you are doing edexcel IAL correct? If so, just give me your email so i can send you the files

can u kindly give me any notes u have for physics and chemistry ial notes ASAP
Whatever notes u have