16th Feb, 2023
Emory University
Discussion
Started 11th Jan, 2023
Can someone explain relationship between pH and ammonia/ammonium equilibrium?
I understand that the percent of total ammonia present as un-ionized ammonia (NH3) is dependent on pH and temperature, but I am trying to figure out exactly why and how. I have seen statements that the equilibrium constant of the ammonia/ammonium equilibrium reaction is dependent on pH and temperature, but I believe that this is sloppy terminology. A change in temperature would result in a change in the constant, but changing the pH would not. Changing pH would be akin to a change in concentration, wouldn't it?
I am trying to get a better understanding of equilibrium constants and how they relate to pH in general, and would appreciate any help. Any insights would be appreciated.
Thanks!
Most recent answer
Dear Carlos Araújo Queiroz , I did not say that your calculations are wrong. While you forgot the effect of ionic strength. The numbers in your example are very questionable: "Example: Para CNH4OH = 1.00 M; pOH ≈- ½·{log10(1.80) - 5} =- ½·{0.255 - 5} = ½·{4.745 }= 2.37; pH ≈ 14 - 2.37 = 11.63." pH is not the -Log[H+], but -Log{H+ activity}. pH of 0.1 M NaOH is not equal to 13.
All replies (7)
Hi Eleanor,
If you look at the equations relating these parameters do you have a problem understanding them, or is there another problem?
Regards, Paul.
Paul Milham yes, sorry if my question was not well-phrased. I have quite a basic understanding of equilibria. I guess I am really verifying that changes in concentration don't affect the constants themselves. When we say that changing concentrations results in a shift in the equilibrium, we are talking about what the system does in order to restore equilibrium, correct?
In the case of ammonia/ammonium, at a given pH and the equilibrium constant (known for a certain temperature) the ratio of NH3 to NH4+ can be known. I believe that this is the basis for determining the un-ionized ammonia concentration. So it is not accurate to say that the equilibrium constant itself is pH dependent.
What you wrote looks fine to me. If you are asking about concentration dependence over big concentration ranges that is beyond my knowledge. However, the system will have been well-studied and I'd expect it to be documented in chemical engineering data sources.
1 Recommendation
The pH of not too concentrate pure ammonia (NH4OH) aq. sol. can be predicted as follows:
For NH4OH dissociation: NH4OH ⇌ NH4+ + OH-, Kb = [NH4+]·[OH-]/[NH4OH] = 1.80·10-5 M. Hence: [NH4OH] = [NH4+]·[OH-]/Kb.
Ammonium molar balance writes: CNH4OH = [NH4+] + [NH4OH] + [NH3] ≈ [NH4+] + [NH4OH] =[NH4+]·{1 + [OH-]/Kb}, for enough dil. sol. ([NH3] < CNH4OH/100) and where CNH4OH is the nominal molar concentration of dissolved NH4OH (formality, to be precise). Hence: [NH4+] ≈ CNH4OH/{1 + [OH-]/Kb}.
Charge balance writes: [NH4+] + [H3O+] = [OH-]; or: CNH4OH/{1 + [OH-]/Kb} + [H3O+] = [OH-].
For [H3O+] < [OH-]/100 (pH >8):
CNH4OH/{1 + [OH-]/Kb} ≈ [OH-]
this solves as:
[OH-] ≈ - ½·Kb + √{CNH4OH·Kb + (1/4)·(Kb)2}
or:
[OH-] ≈ ½√Kb·{-√Kb + √{4CNH4OH + Kb}}
for not too dilute sol., 4CNH4OH > 100Kb:
[OH-] ≈ √Kb·{-½·√Kb + √CNH4OH}
for not too dilute sol., √CNH4OH > 100·½·√Kb; CNH4OH > 104·(1/4)·Kb = 4.5·10-2 M:
[OH-] ≈ √(Kb·CNH4OH)
where pOH = -log10{[OH-]/M}; pH = 14 - pOH
Example: Para CNH4OH = 1.00 M; pOH ≈- ½·{log10(1.80) - 5} =- ½·{0.255 - 5} = ½·{4.745 }= 2.37; pH ≈ 14 - 2.37 = 11.63.
Dear Carlos Araújo Queiroz ,
i assume that all your derivations are correct, but they are applicable only for the case, if in solution there are no any other bases or acids.
This is more general text book formula.
pKa of ammonia is 9.25
[NH4+] = [NH4]o x 10^(pka-pH)/(1+ 10^(pKa-pH))
[NH4] = [NH4]o/(1+ 10^(pKa-pH))
Eleanor Carrano the equilibrium constant Ka of the ammonia/ammonium equilibrium reaction
NH4+ = NH3 + H+, Ka
is NOT dependent on pH, but depends on the temperature and ionic strength of solution. pKa = - log10(Ka).
All these are from textbooks.
The pH of not too concentrate ammonia / ammonium chloride (NH4OH / NH4Cl) aq. sol. can be predicted as I have shown elsewhere at this forum:
Dear Carlos Araújo Queiroz , I did not say that your calculations are wrong. While you forgot the effect of ionic strength. The numbers in your example are very questionable: "Example: Para CNH4OH = 1.00 M; pOH ≈- ½·{log10(1.80) - 5} =- ½·{0.255 - 5} = ½·{4.745 }= 2.37; pH ≈ 14 - 2.37 = 11.63." pH is not the -Log[H+], but -Log{H+ activity}. pH of 0.1 M NaOH is not equal to 13.
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