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A study of the reaction ²⁴Mg(p, ⁶He)¹⁹Na at E_p = 54.7 MeV reveals a group of ⁶He particles corresponding to a state in ¹⁹Na with M - A = 12.974 ± 0.070 MeV. It is presumed to be the ground state of ¹⁹Na, although the close proximity of the second T = 3/2 state in ¹⁹O from the first (96 keV), does not permit a definite assignment. If it is assumed that ¹⁹Na(0) has M - A = 12.974 ± 0.070 MeV, then ¹⁹Na is unbound with respect to decay into ¹⁸Ne + p by 0.366 ± 0.070 MeV (CE69). The mass excess of ¹⁹Na is calculated to be 12.965 ± 0.025 MeV, using a seniority scheme, and 12.968 ± 0.025 MeV, using a supermultiplet scheme [both these values were calculated assuming that the lowest J^π = 3/2⁺; T = 3/2 state in ¹⁹Na is 0.096 MeV above the J^π = 5/2⁺; T = 3/2 state, as in ¹⁹O] (HA69H). See also (BA60Q, KE66C, MU68G, CE71).