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Buffer solution and PH change question.

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phen
How did you arrive at [H+] = Ka * (number of moles propanoic acid / number of moles of the conjugate base)?



it's a buffer solution and in my book it says that
if someone confirms i am right i can PM the method with information but it would be a bit pointless if i am wrong^^
Ka =

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

therefore

[H+] =

Ka x [CH3CH2COOH]
_________________
[CH3CH2COOH]

I think this is the right answer.

:crazy:
Reply 23
Avatar for xSkyFire
xSkyFire
19
Ok this is it, these are my answers :
moles of acid originally present = 0.0165
NaOH present = 0.0023

CH3CH2COOH + NaOH = CH3CH2COONa +H2O
0.0023 NaOH reacts with 0.0165 leaving 0.0142 moles of acid
Sodium propanoate (salt) is 0.0023 moles.

Ok, now concentration time, divide all through by (40/1000)
Acid = 0.355 mol dm^-3
Salt = 0.0575 mol dm^-3

1.35 x 10^-5 = [0.0575] x [H+] / [0.355]
[H+] = 8.33 X 10^-5

PH = 4.08
xSkyFire
Ok this is it, these are my answers :
moles of acid originally present = 0.0165
NaOH present = 0.0023

CH3CH2COOH + NaOH = CH3CH2COONa +H2O
0.0023 NaOH reacts with 0.0165 leaving 0.0142 moles of acid
Sodium propanoate (salt) is 0.0023 moles.

Ok, now concentration time, divide all through by (40/1000)
Acid = 0.355 mol dm^-3
Salt = 0.0575 mol dm^-3

1.35 x 10^-5 = [0.0575] x [H+] / [0.355]
[H+] = 8.33 X 10^-5

PH = 4.08



yep to the last bits
pH = 4.08
Mr. Tizzy XII
Ka =

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

therefore

[H+] =

Ka x [CH3CH2COOH]
_________________
[CH3CH2COOH]

I think this is the right answer.

:crazy:




phen
How did you arrive at [H+] = Ka * (number of moles propanoic acid / number of moles of the conjugate base)?



for further elaboration

this is basically your buffer equilibrium
CH3CH2COOH + NaOH --> CH3CH2COONa + H20
in other words
CH3CH2COOH = H+ + CH3CH2COO-
and
Ka = [H+][CH3CH2COO-]
-------------------
CH3CH2COOH

rearranging this you get

[H+] = Ka x [CH3CH2COOH] / [CH3CH2COO-]


remember it's acid / salt and the CH3CH2COO- is your salt as it's really CH3CH2COONa
:smile:

working out concentration of acid from what theyve given us = 0.015
working out concentration of NaOH base = 0.0023
(first two questioins)
moles of conjugate base = 0.0023 due to the CH3CH2COONa

and moles of ACID remaining

=0.0165 - 0.0023
and

H + = Ka x 0.0165/0.0023
= 8.3 10^-5
log this
= pH = 4.08
Reply 26
Avatar for xSkyFire
xSkyFire
19
I've done these sort of equations hundreds of times, I guarantee the PH is 4.08 so yeah @ equation above
I have a tendency to omit random bits of calculation since I speed through things in my head + I'm lazy, just incase you didn't catch on from the beginning OP lol :p:
Original post by Mr. Tizzy XII
Here is my attempt -

a) [H+] = sq. root Ka

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

-log(3.67x10-3) = 2.43

Don't you have to multiply Ka by the concentration of propanoic acid.
[H+]= sq.root (ka x propanoic acid)

sq.root (1.35x10^-5 x 0.125) = 1.2990x10^-3
-log(1.2990x10^-3) = 2.89
I did it like this, is this not right?
[QUOTE="Ishtiaq;60262843" Kiran="Kiran"]
Original post by Mr. Tizzy XII
Here is my attempt -

a) [H+] = sq. root Ka

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

-log(3.67x10-3) = 2.43

Don't you have to multiply Ka by the concentration of propanoic acid.
[H+]= sq.root (ka x propanoic acid)

sq.root (1.35x10^-5 x 0.125) = 1.2990x10^-3
-log(1.2990x10^-3) = 2.89
I did it like this, is this not right?


No. The people who got 4.08 are correct.

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