Chemistry: Post your doubts here!

[Sarosh Jameel]

can u plz tell why the moles for A is halved !! ??? OCT/06 q2 part iv

 

[Metanoia]

can u plz tell why the moles for A is halved !! ??? OCT/06 q2 part iv

Dibasic acid means it releases 2 moles of H+ per mole of acid.

So ratio of dibasic acid : NaOH is 1: 2

 

[Sarosh Jameel]

Dibasic acid means it releases 2 moles of H+ per mole of acid.

So ratio of dibasic acid : NaOH is 1: 2

thanks !

 

[Sarosh Jameel]

Dibasic acid means it releases 2 moles of H+ per mole of acid.

So ratio of dibasic acid : NaOH is 1: 2

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

 

[shingiechingz@8]

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........................

 

[Mr.Physics]

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........................

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

 

[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-:




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Marking scheme



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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