# Acid/base equilbrium problems

1. Apr 11, 2012

### d.tran103

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!

2. Apr 11, 2012

### Staff: Mentor

Huh? -log[Kb] is pKb. I suppose that's not what you mean, but what you wrote makes no other sense to me.

3. Apr 11, 2012

### d.tran103

Oh sorry, I meant to say -log(x) as in taking the -log(H3O+) (from the original ice table)

4. Apr 11, 2012

### Staff: Mentor

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.

5. Apr 19, 2012

### binomial

NH4 has no Kb because it is an acid, not a base. I honestly have no idea what your question is even asking.

6. Apr 19, 2012

### binomial

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.

7. Apr 20, 2012

### Staff: Mentor

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.

You answered a spammer quoting a random part of the first post :tongue2: