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Acid/base equilbrium problems

by d.tran103
Tags: acid or base, equilbrium
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d.tran103
#1
Apr11-12, 12:29 PM
P: 39
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]/[B]? Is it because I have the Kb of NH3 and not the Kb of NH4? Thanks!
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Borek
#2
Apr11-12, 01:03 PM
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Quote Quote by d.tran103 View Post
my question is why can't I take the -log[x] from kb=[OH-][HA]/[B]
Huh? -log[Kb] is pKb. I suppose that's not what you mean, but what you wrote makes no other sense to me.
d.tran103
#3
Apr11-12, 02:14 PM
P: 39
Oh sorry, I meant to say -log(x) as in taking the -log(H3O+) (from the original ice table)

Borek
#4
Apr11-12, 03:30 PM
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Acid/base equilbrium problems

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.
binomial
#5
Apr19-12, 09:06 PM
P: 7
NH4 has no Kb because it is an acid, not a base. I honestly have no idea what your question is even asking.
binomial
#6
Apr19-12, 10:02 PM
P: 7
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.
Borek
#7
Apr20-12, 01:43 AM
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Quote Quote by binomial View Post
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.

Quote Quote by binomial View Post
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.
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