Assuming that the initial pH is pHi.
The initial concentration of [OH-] is
[OH]- = 10^(14 - pHi)
Since half of the Na2HPO4 is neutralized, the concentration of [OH-] will be reduced to half. The new pH will be
[OH-]/2 = 10^(14 - pHi)/2 = 10^(14 - pH)
14 - pHi - log 2 = 14 - pH
In terms of pHi,
pH = pHi + log 2