Since you're dealing with a buffer, you can use the Henderson-Hasselbalch equation to calculate the new pH of the solution when the concentrations of dihydrogen phosphate, #H_2PO_4^(-)#, and hydrogen phosphate, #HPO_4^(2-)#, are equal.
The balanced chemical equations for this buffer are
#H_2PO_4^(-) + H_2O rightleftharpoons HPO_4^(2-) + H_3^(+)O#, #pKa_2 = 7.21#
#HPO_4^(2-) + H_2O rightleftharpoons PO_4^(3-) + H_3^(+)O#, #pKa_3 = 12.7#
Equal concentrations of #H_2PO_4^(-)# and #HPO_4^(2-)# will establish the first equilibrium, which implies that the pH of the solution will now be
#pH_("solution") = pKa_2 + log (([HPO_4^(2-)])/([H_2PO_4^(-)]))#
#pH_("solution") = 7.21 + log ("5.25 mmol/L"/"5.25 mmol/L") = 7.21 + log(1) = 7.21#
You can determine the GIbbs free energy by using the reaction's #pKa_2# by using
#DeltaG = - RT ln(K_(a2))#, where
#K_(a2)# is the acid dissociation constant for the established equilibrium reaction.
You can use the mathematical identity #ln(x) = 2.303 * log(x)# to rewrite the above equation as
#DeltaG = -RT * 2.303 log(K_(a2))#
If you plug #pK_(a2) = - log(K_(a2))# into the equation, you'll get
#DeltaG = 2.303 RT * pK_(a2)#
Therefore,
#DeltaG = 2.303 * 8.3145"J"/("mol" * "K") * (273.15 + 25)"K" * 7.21#
#DeltaG = "41,162.3 J/mol" = "+41.2 kJ/mol"# #-># rounded to three sig figs.
SIDE NOTE. #DeltaG# will be positive because the acid dissociation constant for the established equilibrium is smaller than 1.
