As the previous answerer said, an acid base titration wouldn't be suitable for NaNO_3NaNO3 if water is the solvent, as NO_3^-NO−3 is too weak a base to change the pH to make a titration suitable. The same goes for Na^+Na+ but being too weak an acid.
NO_3^-NO−3 has no insoluble salts, or at least no common insoluble salts, so we can't do a precipitation titration either. Na^+Na+ has this same trouble.
NO_3^-NO−3 isn't a good coordinating agent so a complexation titration isn't viable either. There is a complexating agent for Na^+Na+ and other alkali metals, but it's rarely used and expensive so it isn't worth considering.
Na^+Na+ won't reduce to Na^0Na0 easily, and if it does it'll react with water violently so it isn't a good idea to do it either.
NO_3^-NO−3 does have a redox reaction though,
NO_3^(-) + 3H^+ +2e^(-) harr HNO_2 + H_2ONO−3+3H++2e−↔HNO2+H2O E_(red)^0 = 0,94E0red=0,94
Which could theoretically be coupled with I^-I− to produce I_3^-I−3 which can then be titrated with S_2O_3^-2S2O−23.
3I^(-) harr I_3^(-) + 2e^(-)3I−↔I−3+2e− E_(o.x)^0 = -0,536E0⊙x=−0,536
That being said, I'm not sure if the DeltaE would be big enough outside of standard conditions.
E^0 values copied from Skoog's fundamentals of Analytic Chemistry
