In: Chemistry
mass of Na2HPO4 = 0.2498 g
moles of acid = 0.2498 / 141.9588 = 1.76 x 10^-3
moles of conjugate base = 0.1227 / 163.94 = 7.484 x 10^-4
moles of NaOH added = 0.5 x 1 / 1000 = 5 x 10^-4 (C)
pH = pKa + log [conjugate base + C / acid - C]
= 12.32 + log [7.484 x 10^-4 + 5 x 10^-4 / 1.76 x 10^-3 - 5 x 10^-4 ]
= 12.32
pH of the prepared buffer = 12.32