- #1
d.tran103
- 39
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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!
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!