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*.