# Acid/base equilbrium problems

by d.tran103
Tags: acid or base, equilbrium
 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!
P: 23,535
 Quote by d.tran103 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.
 P: 39 Oh sorry, I meant to say -log(x) as in taking the -log(H3O+) (from the original ice table)
 Admin P: 23,535 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.
 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.
 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.