The given relation is obtained in case of an aqueous solution where amphiprotic species HA^(-)is produced on ionization of acid salt like NaHCO_3 .
Here the spices acts both as acceptor or donor of proton.
Dissociation of original di-basic acid H_2A occurs in two steps as follows
H_2A+H_2Ostackrel(K_1)rightleftharpoonsHA^(-)+H_3O^(+)
K_1=([HA^(-)][H_3O^(+)])/([H_2A]).......[1]
HA^(-)+H_2Ostackrel(K_2)rightleftharpoonsA^(2-)+H_3O^(+)
K_2=([A^(2-)][H_3O^(+)])/([HA])......[2]
From [1] and [2] we get
K_1xxK_2=([HA^(-)][H_3O^(+)])/([H_2A])xx([A^(2-)][H_3O^(+)])/([HA])
=>K_1xxK_2=([H_3O^(+)]^2xx[A^(2-)])/([H_2A]).....[3]
Now if we consider the following equilibrium
2HA^(-)rightleftharpoonsH_2A+A^(2-)
then we can say that [H_2A]=[A^(2-)]
Applying this to [3] we get
K_1xxK_2=[H_3O^(+)]^2
Taking log_10 on both sides
log_10(K_1xxK_2)=log_10[H_3O^(+)]^2
=>-log_10K_1-log_10K_2=-2log_10[H_3O^(+)]
=>pK_1+pK_2=2pH
=>pH=1/2(pK_1+pK_2)