When does the equation pH=(1/2)(pKa1+pKa2) used?

1 Answer
Aug 12, 2018

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)