What is the pH of 0.15 M NH4Cl?

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Discussion Overview

The discussion revolves around calculating the pH of a 0.15 M solution of NH4Cl, focusing on the use of an ICE table and the relationship between Ka and Kb. Participants explore the implications of using the acid dissociation constant (Ka) and the base dissociation constant (Kb) in this context.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant sets up an ICE table for the dissociation of NH4 in water and calculates the pH to be 5.04, questioning the use of -log[x] in relation to Kb.
  • Another participant expresses confusion over the meaning of -log(x) in the context of Kb and pH calculations.
  • A different participant clarifies that H3O+ in the ICE table cannot be zero or an unknown before solving for it, which complicates taking the logarithm.
  • Some participants assert that NH4 does not have a Kb because it is an acid, while others note that Kb can be derived from Ka using the relationship KaKb=Kw.
  • There is a discussion about the correct approach to calculating pH from hydroxide concentration if Kb were used, indicating that pH would be derived from pOH in that case.

Areas of Agreement / Disagreement

Participants express differing views on the application of Kb and Ka in the pH calculation, with no consensus reached on the interpretation of the original question or the appropriate method for calculating pH.

Contextual Notes

There are unresolved assumptions regarding the interpretation of Kb and Ka in the context of NH4, as well as the steps involved in calculating pH from the ICE table.

d.tran103
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What is the pH of 0.15 M NH4Cl?
Okay, I set up my ICE table,

NH4 + H2O <---> H3O+ + NH3-
I 0.15 M --- --- ---
Δ -x --- +x +x
E 0.15-x --- x x

KbNH3-=1.8x10-5
So Kw=1.0e-14=Ka*Kb, so KaNH4=5.56e-10

So, Ka=[H3O+][A-]/[HA]

5.56e-10=x^2/0.15-x
[H+]=[H3O+]=[x]=9.13e-6

pH=-log[H+]
pH=5.04

I understand the work, but my question is why can't I take the -log[x] from kb=[OH-][HA]/? Is it because I have the Kb of NH3 and not the Kb of NH4? Thanks!
 
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d.tran103 said:
my question is why can't I take the -log[x] from kb=[OH-][HA]/


Huh? -log[Kb] is pKb. I suppose that's not what you mean, but what you wrote makes no other sense to me.
 
Oh sorry, I meant to say -log(x) as in taking the -log(H3O+) (from the original ice table)
 
I still have no idea what you mean by "-log(H3O+) (from the original ice table)".

H3O+ in your ICE table is either 0 (initial value) - you can't take log of zero, or x - which is an unknown, so you can't calculate logarithm of its value before solving the problem.
 
NH4 has no Kb because it is an acid, not a base. I honestly have no idea what your question is even asking.
 
I'm still not sure what you're asking. You take the -log(x) once you solve for it to get the pH. If you used the Kb, then you would do 14.00 + log(x) because x in that case would equal the hydroxide concentration, so to get pH you must subtract the pOH from 14.00.
 
binomial said:
NH4 has no Kb because it is an acid, not a base

Not that you are wrong, but KaKb=Kw, so you can give both for any acid/conjugate base pair. People often abbreviate it and say Kb of an acid or Ka of a base.

That beiung said, I doubt that's what the OP meant.

binomial said:
I'm still not sure what you're asking. You take the -log(x) once you solve for it to get the pH. If you used the Kb, then you would do 14.00 + log(x) because x in that case would equal the hydroxide concentration, so to get pH you must subtract the pOH from 14.00.

You answered a spammer quoting a random part of the first post :-p
 

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