[Issy123]

does anyone know how to do question 2d) in the specimen paper
i've even looked at the mark scheme but i'm just confused as to why you'd do that :/


As part of an investigation, a student needed to prepare a buffer solution with a pH value of 8.71. From the Ka value of phenol, the student thought that a mixture of phenol and sodium phenoxide could be used to prepare this buffer solution.
The student decided to use a 0.200 mol dm–3 solution of phenol, mixed with an equal volume of sodium phenoxide.
Use your knowledge of buffer solutions to determine the concentration of sodium phenoxide solution that the student would need to mix with the 0.200 mol dm–3 phenol solution.
Ka = 1.3 × 10–10 mol dm–3

please help

[arvin_infinity]

(Original post by Issy123)
does anyone know how to do question 2d) in the specimen paper
i've even looked at the mark scheme but i'm just confused as to why you'd do that :/


As part of an investigation, a student needed to prepare a buffer solution with a pH value of 8.71. From the Ka value of phenol, the student thought that a mixture of phenol and sodium phenoxide could be used to prepare this buffer solution.
The student decided to use a 0.200 mol dm–3 solution of phenol, mixed with an equal volume of sodium phenoxide.
Use your knowledge of buffer solutions to determine the concentration of sodium phenoxide solution that the student would need to mix with the 0.200 mol dm–3 phenol solution.
Ka = 1.3 × 10–10 mol dm–3

please help

MS is wrong the ans. should be 0.013

[Issy123]

(Original post by arvin_infinity)
MS is wrong the ans. should be 0.013

ahhh! phew! i got 0.013 following those steps and couldn't understand how the MS got 0.13

thank you!

[The Illuminati]

I got 0.013 not 0.13
i also got 1.95x10^-10 not 1.99x10^-10

[Issy123]

(Original post by The Illuminati)
I got 0.013 not 0.13
i also got 1.95x10^-10 not 1.99x10^-10

same! apparently the MS is wrong

[arvin_infinity]

(Original post by Issy123)
x

Wondered if you can explain 1aiii to me

[Issy123]

(Original post by arvin_infinity)
Wondered if you can explain 1aiii to me

Hmmm, i've looked at that question a few times, and still don't get it... ive seen a similar q on another past paper but the method to get the answer is completely different.

[The Illuminati]

(Original post by arvin_infinity)
Wondered if you can explain 1aiii to me

(Original post by Issy123)
Hmmm, i've looked at that question a few times, and still don't get it... ive seen a similar q on another past paper but the method to get the answer is completely different.

You work it out using molar ratios the initial rate given is for HCHO which there are 2 mole. You want the formation of O2 which there is half a mole of.
So for every 2 HCHO made half an O2 will be made. the ratio is 2:0.5 or 4:1 therefore to get the initial rate pf the formation of oxygen you take the initial rate of the formation of HCHO in experiment 1 and divide by 4
1x10-12/4 =2.5x10-13

[arvin_infinity]

(Original post by The Illuminati)
You work it out using molar ratios the initial rate given is for HCHO which there are 2 mole. You want the formation of O2 which there is half a mole of.
So for every 2 HCHO made half an O2 will be made. the ratio is 2:0.5 or 4:1 therefore to get the initial rate pf the formation of oxygen you take the initial rate of the formation of HCHO in experiment 1 and divide by 4
1x10-12/4 =2.5x10-13

+rep (owe you from ages ago)

emm..right after I posted this I managed to do it..just had to read the question

[arvin_infinity]

(Original post by Issy123)
Hmmm, i've looked at that question a few times, and still don't get it... ive seen a similar q on another past paper but the method to get the answer is completely different.

Can you pls post the other one

[Issy123]

(Original post by arvin_infinity)
Can you pls post the other one

sure! i'll take a picture of it and upload it

[Issy123]

(Original post by The Illuminati)
You work it out using molar ratios the initial rate given is for HCHO which there are 2 mole. You want the formation of O2 which there is half a mole of.
So for every 2 HCHO made half an O2 will be made. the ratio is 2:0.5 or 4:1 therefore to get the initial rate pf the formation of oxygen you take the initial rate of the formation of HCHO in experiment 1 and divide by 4
1x10-12/4 =2.5x10-13

ahhh! that makes more sense! thanks will hopefully +rep you tomorrow as i've reached my limit today!

[Issy123]

(Original post by arvin_infinity)
Can you pls post the other one

Sorry about the wait! Btw its from a legacy paper

[*Rubina*]

(Original post by Issy123)
does anyone know how to do question 2d) in the specimen paper
i've even looked at the mark scheme but i'm just confused as to why you'd do that :/


As part of an investigation, a student needed to prepare a buffer solution with a pH value of 8.71. From the Ka value of phenol, the student thought that a mixture of phenol and sodium phenoxide could be used to prepare this buffer solution.
The student decided to use a 0.200 mol dm–3 solution of phenol, mixed with an equal volume of sodium phenoxide.
Use your knowledge of buffer solutions to determine the concentration of sodium phenoxide solution that the student would need to mix with the 0.200 mol dm–3 phenol solution.
Ka = 1.3 × 10–10 mol dm–3

please help

Could u please explain how u did this? I got to 1.95 x10-9....so far xx

[arvin_infinity]

(Original post by Issy123)


Sorry about the wait! Btw its from a legacy paper

No worries..

can you tell me what it says at the begining of the question cuz somewhere in there should say "rate is measured from the disapperance/apperance of something"

[Issy123]

(Original post by arvin_infinity)
No worries..

can you tell me what it says at the begining of the question cuz somewhere in there should say "rate is measured from the disapperance/apperance of something"

It says:

In an experiment, maltose, was hydrolysed to form glucose. HCL behaves as a catalyst for the reaction.

maltose + water -> glucose

The reaction was carried out several times using different concentrations of maltose and of HCL. the intitial rate of each experiment run was calculated and the results are shown below. (i think the rest is on the pic?)

[Issy123]

(Original post by *Rubina*)
Could u please explain how u did this? I got to 1.95 x10-9....so far xx

you use [H+] = Ka x [Acid]/[Base]

But swap the [H+] and [Base] around.. so the equation is now [Base] = Ka x [Acid]/[H+]

which is

1.3x10-10 x (0.2/1.95x10-10)

= 0.13



hope that helped!

[arvin_infinity]

(Original post by Issy123)
It says:

In an experiment, maltose, was hydrolysed to form glucose. HCL behaves as a catalyst for the reaction.

maltose + water -> glucose

The reaction was carried out several times using different concentrations of maltose and of HCL. the intitial rate of each experiment run was calculated and the results are shown below. (i think the rest is on the pic?)

+rep
Actually I could work that out from the table -silly me

ermm so the ans. is 0.048

ive seen a similar q on another past paper but the method to get the answer is completely different.

FAILED - (same method)

[The Illuminati]

(Original post by arvin_infinity)
+rep
Actually I could work that out from the table -silly me

ermm so the ans. is 0.048


(same method)

no it's 2.5x10^-13

[Issy123]

(Original post by The Illuminati)
no it's 2.5x10^-13

I just checked the mark scheme & the answer is 0.048