Chemistry: Post your doubts here!

[***amd***]

see attachment for the explanation
hope you can understand

thanks

 

[ashcull14]

 

[qwertypoiu]

View attachment 57782

Let CH3CO2H = x cm3
Let CH3CO2Na = (100-x) cm3

pH = pKa + log(salt/acid)

5.5 = -lg(1.79E-5) + log(salt/acid)
5.5 = 4.75 + log(salt/acid)
log(salt/acid) = 0.75
Salt/acid = 0.1778
(100-x)/x = 0.1778
100-x = 0.1778x
100 = 1.1778x
x = 100/1.1778 = 85
100-x = 15

 

[ashcull14]

Let CH3CO2H = x cm3
Let CH3CO2Na = (100-x) cm3

pH = pKa + log(salt/acid)

5.5 = -lg(1.79E-5) + log(salt/acid)
5.5 = 4.75 + log(salt/acid)
log(salt/acid) = 0.75
Salt/acid = 0.1778
(100-x)/x = 0.1778
100-x = 0.1778x
100 = 1.1778x
x = 100/1.1778 = 85
100-x = 15

thanks

 

[My Name]

http://maxpapers.com/wp-content/uploads/2012/11/9701_s13_qp_23.pdf
question 5 part a iv

 

[The Sarcastic Retard]

http://maxpapers.com/wp-content/uploads/2012/11/9701_s13_qp_23.pdf
question 5 part a iv

2 products. One will be organic and other is CO2.
Organic product will be HO2CCO2H (Break the alkene, you get CHO it further oxidises to COOH and OH is oxidised to COOH as well)

 

[My Name]

2 products. One will be organic and other is CO2.
Organic product will be HO2CCO2H (Break the alkene, you get CHO it further oxidises to COOH and OH is oxidised to COOH as well)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_ms_22.pdf
question 4 part a
product of
C if I write CH3(CH2)2CH2OH instead of CH3CH2CH2CH2OH
and product of
E if I write CH3CH=CHCOOH instead of CH3CH=CHCO2H

Fine?

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9701 - Chemistry/9701_w09_ms_22.pdf
question5 part b
for equation if I CH3CH2CH2CH2OH instead of C4H9OH

Fine?

 

[Saad the Paki]

Help with this question please

 

[My Name]

Help with this question please View attachment 57985

I think it is A.
What is your doubt?
About whether it is A or B?

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_ms_22.pdf
question 4 part a
product of
C if I write CH3(CH2)2CH2OH instead of CH3CH2CH2CH2OH
and product of
E if I write CH3CH=CHCOOH instead of CH3CH=CHCO2H

Fine?

http://studyguide.pk/Past Papers/CIE/International A And AS Level/9701 - Chemistry/9701_w09_ms_22.pdf
question5 part b
for equation if I CH3CH2CH2CH2OH instead of C4H9OH

Fine?

Yes. What you've written is fine.

 

[Saad the Paki]

I think it is A.
What is your doubt?
About whether it is A or B?

The answer is A
But yes I'm confused between A or B... even D seems likely
Can u point out some reasons why the Ans is A and not B or D

 

[My Name]

The answer is A
But yes I'm confused between A or B... even D seems likely
Can u point out some reasons why the Ans is A and not B or D

It can't be D since I don't think that an ester can dissolve in water completely.
Then you have A and B.
It can't be B since alcohols are not acidic enough to react with aqueous solutions of strong alkalis.( However,they can react with the recatice metals.)

 

[Saad the Paki]

It can't be D since I don't think that an ester can dissolve in water completely.
Then you have A and B.
It can't be B since alcohols are not acidic enough to react with aqueous solutions of strong alkalis.( However,they can react with the recatice metals.)

Ooh okay..! Thanks

 

[My Name]

Ooh okay..! Thanks

Welcome

 

[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 :/