[Durlag]

Hello everyone,

Could you please help me? I calculated the pH of a solution of 2.5M NH4Cl in 0.1M HCl, and I would like to know if I am right (it took me a lot of attempts, as I find acid-base reactions pretty confusing).

(1) I first calculated the % dissociation of NH4+ at pH=1 (the pH of HCl 0,1M). (2) Then I calculated the concentration of [H3O+] released by the dissociation of our NH4+ at that pH. I added that to the concentration of H3O+ contributed by HCl and I found:

[H3O+] = 0,1 + 1,4125.10^-9 = approx. 0,1. Am I right?
Thus pH = 1 (yeah I know, all this for that...).

P.S.: I found this pretty confusing.
In my first attemp, at step (1), I calculated [H3O+] contributed by NH4+ thanks to the technique of considering [H3O+] = [NH3] = x (from the equation of hydrolysis of NH4Cl and equilibrium with Ka) and calculating x². I got a result, but that would work only in water, and not in HCl 0,1N (right?).
In my second attempt, I took the same equation, and in the equilibrium expression, I replaced [H3O+] by 0,1 and [NH4+] by 2,5.10^-3, and calculated [NH3], and then the % dissociation. But I guess this is wrong, since [H3O+] and [NH3] should have a molar ratio of 1:1.
Explaining this confuses me even more.

I know I'm probably sweating the small stuff, but I like to understand, I don't like to leave a subject with areas that I haven't understood. 
In the beginning I just wanted to understand why I had an amino acid standard (for amino acid analyzer calibration) that contained 2,5.10^-3 M NH4Cl. I called the manufacturer and they couldn't tell me why. It's not a buffer since a buffer is a solution that contains a weak acid (or a base) and the salt of that weak acid (or that base). Or maybe it is a buffer, but only when raising the pH to pH > pKa NH4+/NH3. What do you think?

Don't hesitate to ask me further details about my calculations.

Thank you!!

[Borek]

Do you know how to use ICE tables?

IMHO you can safely ignore presence of NH₄⁺, as it is HCl that dominates the solution, but with ICE table you can pretty easily calculate extent of the NH₄⁺ dissociation - just remember you initially have 2.5 M of NH₄⁺ and 0.1 M of H⁺.

[Durlag]

Hi! Thanks for your answer!

First of all, I'm sorry, it is [NH4Cl]=2,5.10^-3 M and not [NH4Cl]=2,5M.

I've done what you said. Here is what I did:

Question: calculate the exent of dissociation of NH4Cl in HCl 0.1M.

We know:
[NH4Cl]=[NH4+]=2,5.10^-3 M
[HCl]=[H+]=[H3O+]=0,1 M

The reaction of hydrolysis of NH4Cl (the salt of a weak base) in H2O is:
NH4+ + H2O  NH3 + H3O+

ICE table:

Reaction          NH4+                     NH3          H3O+ Initial             2,5.10^-3               0              0,1 Change             -x                       +x             +x Equilib.          (2,5.10^-3)-x          0+x           0,1+x

NH4+ acts as an acid (donating a proton to H2O), so we can write, at equilibrium, the acid dissociation constant:

Ka = ([NH3][H3O+])/[NH4+]
Ka = (x.(0,1+x))/((2,5.10^-3)-x)

Here we can assume x is negligible, and thus make the following approximations:
0,1+x = 0,1
(2,5.10^-3)-x = 2,5.10^-3

So:
Ka = (0,1.x)/(2,5.10^-3)
x = (Ka.(2,5.10^-3))/0,1
We know that Ka = 5,65.10^-10 (can be caluclated from Kb, known from tables, and Kw).
So, x = 1,4125.10^-11
% dissoc. = (x/[NH4+]₀)*100 and [NH4+]₀ = 2,5.10^-3 M
Answer: So % dissoc. = 5,65.10^-7 %, which is a lot inferior to 5%, so our approximation was correct and x is negligible.

So, [NH3] = x = 1.4125.10^-11
And [H3O+] = 0,1 + 1,4125.10^-11 = 0,1
Thus pH = 1.

Note:
I applied the theory about hydrolysis of a salt of a weak base, but I don't really understand why the salt of a weak base (e.g. NH4Cl) acts like an acid (donating an H+ to water which creates H3O+ which makes an acidic solution). If you have NH4+ in water, the pH<pKa of NH4+/NH3, so at pH7, NH4+ should stay in its protonated state. It's contrary to what I've learnt before about how the protonated form of a base behaves in water.
Similarly, I don't understand why the salt of a weak acid (e.g. NaAc) acts as a base (taking an H+ from water which creates OH- which makes a basic solution). In water, pH>pKa of HAc/Ac-, so Ac- should stay unprotonated! Why does it protonate?

I did the calculations, and indeed, a 2,5.10^-3 M solution of NH4Cl in water gives a pH of 5.73 (if I'm not wrong).

I got the theory from there:
http://www.chemteam.info/AcidBase/Salts.html
http://www.chemteam.info/AcidBase/Hydrolysis.html

Thanks!

[Borek]

So, [NH3] = x = 1.4125.10^-11

Yes.

If you have NH4+ in water, the pH<pKa of NH4+/NH3, so at pH7, NH4+ should stay in its protonated state.

No, it doesn't work this way.

Or, more precisely - it is only an approximation. Yes, in neutral solution MOST of the ammonia is protonated. Doesn't mean all of it, doesn't mean NH₄⁺ can't dissociate acidifying solution.

[Durlag]

Thanks!

I find that NH4+ being able to lower a pH of 7 by more than 1 pH unit is suprisingly a lot.

Is that why you have to place the pH at minimum 2 units from the pKa if you want to be sure that most of your acid / base is protonated /unprotonated? I think I've read that somewhere.

[Borek]

See the formula derived here: http://www.titrations.info/acid-base-titration-indicators

While it is applied to indicators there, it holds for every weak acid.

[Durlag]

Ok, I think I got it. Thanks to the modified Henderson-Hasselbalch equation, you can calculate the % of each form of the acid-base couple, knowing the pH and the pKa.

At 1 pH unit below pKa of a weak acid (HA), you get 90,9% HA and 9,1% A-(% dissociation = 9,1%).
At 2 pH units below pKa of a weak acid (HA), you get 99% HA and 1% A-(% dissociation = 1%).
At 3 pH units below pKa of a weak acid (HA), you get 99,9% HA and 0,1% A-(% dissociation = 0,1%).

It's the same of you go above the pKa by 1, 2 or 3 pH units, except the numbers are reversed between the to forms.

However, this works if you have only your weak acid in solution, but what if the pH variation is caused by HCl, for example? I guess it still works as long as [A-] is negligible, but what if it is not negligible? I'm going to post about this in another thread.

[Durlag]

However, this works if you have only your weak acid in solution, but what if the pH variation is caused by HCl, for example? I guess it still works as long as [A-] is negligible, but what if it is not negligible? I'm going to post about this in another thread.

I've checked the maths and come to the conclusion that it works in all cases.