Buffer solution and PH change question.

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#1
"When answering this question assume that the temperature is 298K and give all pH values to 2 decimal places."

"The acid dissociation constant, Ka, of propanoic acid, CH3CH2COOH, has the value 1.35x10-5 moldm-3."

"Ka = [H+][CH3CH2COO-]"
________________
[CH3CH2COOH]

a) Calculate the pH of a 0.550 moldm-3 solution of propanoic acid.

b) A buffer solution is formed when 10.0 cm3 of 0.230 moldm-3 aqueous sodium hydroxide are added to 30.0 cm3 of 0.550 moldm-3 aqueous propanoic acid

(i) Calculate the number of moles of propanoic acid originally present.
(ii) Calculate the number of moles of sodium hydroxide added.
(iii) Hence, calculate the number of moles of propanoic acid present in the buffer solution.
(iv) Hence, calculate the pH of the buffer solution.

----------------------------------------------------------------

Thank you very much if you have read this far, this is an A2 question that i am stuck on, i have answered most of the questions but the figures definitely look wrong. If anyone could help me out with the answers and how to get to them properly i would greatly appreciate it. Thanks in advance, Tizzy.
0
12 years ago
#2
(Original post by Mr. Tizzy XII)
"When answering this question assume that the temperature is 298K and give all pH values to 2 decimal places."

"The acid dissociation constant, Ka, of propanoic acid, CH3CH2COOH, has the value 1.35x10-5 moldm-3."

"Ka = [H+][CH3CH2COO-]"
________________
[CH3CH2COOH]

a) Calculate the pH of a 0.550 moldm-3 solution of propanoic acid.

b) A buffer solution is formed when 10.0 cm3 of 0.230 moldm-3 aqueous sodium hydroxide are added to 30.0 cm3 of 0.550 moldm-3 aqueous propanoic acid

(i) Calculate the number of moles of propanoic acid originally present.
(ii) Calculate the number of moles of sodium hydroxide added.
(iii) Hence, calculate the number of moles of propanoic acid present in the buffer solution.
(iv) Hence, calculate the pH of the buffer solution.

----------------------------------------------------------------

Thank you very much if you have read this far, this is an A2 question that i am stuck on, i have answered most of the questions but the figures definitely look wrong. If anyone could help me out with the answers and how to get to them properly i would greatly appreciate it. Thanks in advance, Tizzy.
Why not show us how far you've got then?
0
12 years ago
#3
(Original post by Mr. Tizzy XII)
"When answering this question assume that the temperature is 298K and give all pH values to 2 decimal places."

"The acid dissociation constant, Ka, of propanoic acid, CH3CH2COOH, has the value 1.35x10-5 moldm-3."

"Ka = [H+][CH3CH2COO-]"
________________
[CH3CH2COOH]

a) Calculate the pH of a 0.550 moldm-3 solution of propanoic acid.

b) A buffer solution is formed when 10.0 cm3 of 0.230 moldm-3 aqueous sodium hydroxide are added to 30.0 cm3 of 0.550 moldm-3 aqueous propanoic acid

(i) Calculate the number of moles of propanoic acid originally present.
(ii) Calculate the number of moles of sodium hydroxide added.
(iii) Hence, calculate the number of moles of propanoic acid present in the buffer solution.
(iv) Hence, calculate the pH of the buffer solution.

----------------------------------------------------------------

Thank you very much if you have read this far, this is an A2 question that i am stuck on, i have answered most of the questions but the figures definitely look wrong. If anyone could help me out with the answers and how to get to them properly i would greatly appreciate it. Thanks in advance, Tizzy.
1)square root (ka x propanoic acid conc) then take -log{10} of that
2) M = C X V
II) Same
III) 1 : 1 reaction, pretty easy
iv) Use the added volumes for the new volumes, take C = M/V, use that to get H+ conc, then -log{10} that.
0
#4
(Original post by charco)
Why not show us how far you've got then?
Here is my attempt -

a) [H+] = sq. root Ka

sq.root 1.35x10-5 = 3.67x10-3

-log(3.67x10-3) = 2.43

b)

(i) 0.550 x (30/1000) = 0.0165 moles
(ii) 0.230 x (10/1000) = 2.3x10-3 moles
(iii) 0.0165 - 2.3x10-3 = 0.0142 moles.

(0.0142 x (40/1000) = 5.68x10-4 < moles of excess acid converted to moldm-3>

(iv)

1.35x10-5 =

[H+]x[2.3x10-3]
_____________
5.68x10-4

[H+] =

1.35x10-5 x 5.68x10-4
___________________
2.3x10-3

= 3.33x10-6

-log(3.33x10-6) = pH of 5.48
0
#5
(Original post by xSkyFire)
1)square root (ka x propanoic acid conc) then take -log{10} of that
2) M = C X V
II) Same
III) 1 : 1 reaction, pretty easy
iv) Use the added volumes for the new volumes, take C = M/V, use that to get H+ conc, then -log{10} that.
III) 1 : 1 reaction, pretty easy

I was under the impression that in a buffer solution 1 : 1 moles of the acid / base would react to leave in this case an excess of acid (0.0142 moles) Is this what you mean?

IV) Please can you explain that further?

Thanks for the help.
0
12 years ago
#6
(Original post by Mr. Tizzy XII)
III) 1 : 1 reaction, pretty easy

I was under the impression that in a buffer solution 1 : 1 moles of the acid / base would react to leave in this case an excess of acid (0.0142 moles) Is this what you mean?

IV) Please can you explain that further?

Thanks for the help.
That's the point, they react 1:1, that's why you have an "excess" amount of acid, so you have some left over.

IV) By mixing your acid and and alkali, you change the volume, therefore changing the concentration. You take the salt formed,
CH3CH2COO-Na+ and work out the concentration, same with the acid, plug in the values and rearrange to get = [H+]

-log{10} [H+] to get the PH
0
#7
(Original post by xSkyFire)
That's the point, they react 1:1, that's why you have an "excess" amount of acid, so you have some left over.

IV) By mixing your acid and and alkali, you change the volume, therefore changing the concentration. You take the salt formed,
CH3CH2COO-Na+ and work out the concentration, same with the acid, plug in the values and rearrange to get = [H+]

-log{10} [H+] to get the PH
(iv)

1.35x10-5 =

[H+]x[2.3x10-3]
_____________
5.68x10-4

[H+] =

1.35x10-5 x 5.68x10-4
___________________
2.3x10-3

= 3.33x10-6

-log(3.33x10-6) = pH of 5.48

This is what i managed to get but isn't a pH change from 2.43 - 5.48 too big for a buffer solution that is meant to 'resist' pH changes?

Thanks again for the help.
0
12 years ago
#8
(Original post by Mr. Tizzy XII)
(iv)

1.35x10-5 =

[H+]x[2.3x10-3]
_____________
5.68x10-4

[H+] =

1.35x10-5 x 5.68x10-4
___________________
2.3x10-3

= 3.33x10-6

-log(3.33x10-6) = pH of 5.48

This is what i managed to get but isn't a pH change from 2.43 - 5.48 too big for a buffer solution that is meant to 'resist' pH changes?

Thanks again for the help.
The PH of 2.43 was your acid, not your buffer. Your buffer has PH of 5.48, if you add small amounts acid or alkali to that solution now, the PH won't change much e.g. go from 5.48 to 5.40 on addition of [H+] instead of going to a PH 2.00 (random figures I made up) lol
0
12 years ago
#9
THIS is A2? :s
0
#10
(Original post by xSkyFire)
The PH of 2.43 was your acid, not your buffer. Your buffer has PH of 5.48, if you add small amounts acid or alkali to that solution now, the PH won't change much e.g. go from 5.48 to 5.40 on addition of [H+] instead of going to a PH 2.00 (random figures I made up) lol
Thanks alot for the help, i just found it hard doing the question with the equations but not fully understanding the logic It's all good now though 0
12 years ago
#11
(Original post by phen)
THIS is A2? :s
Yeah, this is the easier stuff you do in A2, along with the enthalpy stuff from earlier on the Edexcel syllabus (early unit 4 stuff). It looks hard but with a bit of practice, background knowledge and setting your work out really well, it's really easy 0
12 years ago
#12
(Original post by Mr. Tizzy XII)
Thanks alot for the help, i just found it hard doing the question with the equations but not fully understanding the logic It's all good now though awesome 0
12 years ago
#13
(Original post by xSkyFire)
Yeah, this is the easier stuff you do in A2, along with the enthalpy stuff from earlier on the Edexcel syllabus (early unit 4 stuff). It looks hard but with a bit of practice, background knowledge and setting your work out really well, it's really easy Sorry, I meant it differently: I was disappointed to see a question like that would be used in the curriculum. D;
But, you're saying it's the easier stuff, so I guess it's okay.

(We had buffers first thing this year (same year as A2), and a question like that would be pretty much a give-away ): )
In fact, I have a test on this stuff tomorrow.
0
12 years ago
#14
i get a pH of 4 for the buffer solution?
where have i gone wrong 0
12 years ago
#15
(Original post by Dior-perfume)
i get a pH of 4 for the buffer solution?
where have i gone wrong Pretty hard to tell when you don't show how you've gotten that answer, wouldn't you say? ):

You should have two decimal places anyway.
0
12 years ago
#16
(Original post by Mr. Tizzy XII)
b)

(i) 0.550 x (30/1000) = 0.0165 moles
(ii) 0.230 x (10/1000) = 2.3x10-3 moles
(iii) 0.0165 - 2.3x10-3 = 0.0142 moles.

i got this and then I did H+ = Ka x (0.0142 / 0.0023)
= 8.3 x 10-5
then log
= pH 4
0
12 years ago
#17
(Original post by phen)
Pretty hard to tell when you don't show how you've gotten that answer, wouldn't you say? ):

You should have two decimal places anyway.

0
#18
(Original post by Dior-perfume)
i get a pH of 4 for the buffer solution?
where have i gone wrong If you showed the working out i might be able to see Who knows, you may be the one with the right answer EDIT:

I used

Ka =

[H+][CH3CH2COO-]
________________
[CH3CH2COOH]

therefore

[H+] =

Ka x [CH3CH2COOH]
_________________
[CH3CH2COOH]

Is this wrong and what are you using?
0
12 years ago
#19
(Original post by Mr. Tizzy XII)
If you showed the working out i might be able to see Who knows, you may be the one with the right answer see above 0
12 years ago
#20
(Original post by Dior-perfume)
i got this and then I did H+ = Ka x (0.0142 / 0.0023)
= 8.3 x 10-5
then log
= pH 4
How did you arrive at [H+] = Ka * (number of moles propanoic acid / number of moles of the conjugate base)?
0
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