J.R. S. answered 02/06/23
Ph.D. University Professor with 10+ years Tutoring Experience
First, you have typo, and the solution should contain 0.10 M NH3 and 0.10 M NH4Cl.
Now, let's look at the reactions taking place when a small volume of NaOH is added. This is a buffer and so the added NaOH being a base will react with the acid present (NH4Cl).
NaOH(aq) + NH4Cl(aq) ==> NaCl(aq) + NH4OH(aq) .. molecular equation
Na+(aq) + OH-(aq) + NH4+(aq) + Cl-(aq) ==> Na+(aq) + Cl-(aq) + NH4OH(aq) .. ionic equation
NH4+(aq) + OH-(aq) ==> NH4OH(aq) .. net ionic
So, in this reaction the spectator ions are Na+ and Cl-
H2O + H2O ==> H3O+ + OH-
In this reaction, the spectator ion is H3O+