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.htmlhttp://www.chemteam.info/AcidBase/Hydrolysis.htmlThanks!