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!
Huh? -log[K_{b}] is pK_{b}. I suppose that's not what you mean, but what you wrote makes no other sense to me.
I still have no idea what you mean by "-log(H3O+) (from the original ice table)". H_{3}O^{+} 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.
Not that you are wrong, but K_{a}K_{b}=K_{w}, so you can give both for any acid/conjugate base pair. People often abbreviate it and say K_{b} of an acid or K_{a} 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: