Sufficient strong acid is added to a solution containing na2hpo4 to neutralize one-half of it. what will be the ph of this solution?

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