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# urgent help needed find pka watch

1. hi guys ,

i need help, i need to find the pKa of citric acid

pH =2.3

concentration = 1 mol dm-3
volume = 50 cm3 = 0.05
i find the number of moles
1 *0.05 = .05 moles

[H+] = 10^-2.3
[H+} = 5.01*10^-2.3
because its a one proton acid it dissociates as shown below
HA<--> H+ + A-

the concentration of the [A-] = H+

how do i calculate the concentration of non dissociated ions ?

then i know that i insert the data

ka= [A-][H+]
--------
[HA]

then pKa=-log Ka

thanks a lot
i really appreciate it
2. What level is that?
3. what do you mean ? higher level?
4. What is the question? I see a lot of working out but I don't follow it.
5. how do i calculate the concentration of non dissociated ions ?

i need help to get to
ka= [A-][H+]
--------
[HA]

cheers
6. I don't follow your working. Remember this is about concentrations, so amounts are irrelevant-

[H+] = 10^-2.3 looks correct

[H+} = 5.01*10^-2.3 looks incorrect

Citric acid is a tribasic acid (3 acidic protons) but assume you are only dissociating the first one, ie the reaction is H3A ----> H+ + H2A-

If you have generated 10^-2.3 M H+, then you must have also generated 10^-2.3 M H2A-

Original conc of acid was 1 M, so undissociated acid conc must be (1 - 10^-2.3)

So now you have top and bottom lines of the Ka calculation

Hope this helps

C
7. Charco and Eier FTW
8. i meant with [H+} = 5.01*10^-2.3 is that
that pH = 10^ -pH
10^-2.3 gives the result of 5.01*10^-3

is therefore dissociated acid conc 1- 5.01*10^-3
Are there any numbers squared ?

ka= (5.01*10^-3)^2 / 1- 5.01*10^-3 ??

thanks a lot and thumbs up for you

sorry but [H+} = 5.01*10^-2.3 was a typo
9. (Original post by INDIGO123456789)
hi guys ,

i need help, i need to find the pKa of citric acid

pH =2.3

concentration = 1 mol dm-3
volume = 50 cm3 = 0.05
i find the number of moles
1 *0.05 = .05 moles

[H+] = 10^-2.3
[H+} = 5.01*10^-2.3
because its a one proton acid it dissociates as shown below
HA<--> H+ + A-

the concentration of the [A-] = H+

how do i calculate the concentration of non dissociated ions ?

then i know that i insert the data

ka= [A-][H+]
--------
[HA]

then pKa=-log Ka

thanks a lot
i really appreciate it
Furrowing through the chaff..

if the pH = 2.3 then

[H+] = 0.00501

Assuming you are dealing with the first dissociation

HA <==> H+ + A-

so that [H+] = [A-]

Then:

Ka = [H+]2/[HA]

You are told that [HA] = 1.0 mol dm-3

Therefore: Ka = [H+]2 = 2.5 x 10-5

so pKa = 4.6

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

NOTE: This is well off the accepted pKa value of either the first or second dissociation of citric acid (3.13 & 4.76)
10. iif [HA] equals the mol dm^-3 why do we find the undissociated acid conc? which would be 1- 10^-2.3 ?

Ka=2.5 x 10^-5 / 1- 10^-2.3?
11. (Original post by INDIGO123456789)
iif [HA] equals the mol dm^-3

why do we find the undissociated acid conc??
These two phrases are contradictory.

[HA] is the undissociated acid concentration.

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