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

Aug 12, 2018

The given relation is obtained in case of an aqueous solution where amphiprotic species $H {A}^{-}$is produced on ionization of acid salt like $N a H C {O}_{3}$ .

Here the spices acts both as acceptor or donor of proton.

Dissociation of original di-basic acid ${H}_{2} A$ occurs in two steps as follows

${H}_{2} A + {H}_{2} O \stackrel{{K}_{1}}{r} i g h t \le f t h a r p \infty n s H {A}^{-} + {H}_{3} {O}^{+}$

${K}_{1} = \frac{\left[H {A}^{-}\right] \left[{H}_{3} {O}^{+}\right]}{\left[{H}_{2} A\right]} \ldots \ldots . \left[1\right]$

$H {A}^{-} + {H}_{2} O \stackrel{{K}_{2}}{r} i g h t \le f t h a r p \infty n s {A}^{2 -} + {H}_{3} {O}^{+}$

${K}_{2} = \frac{\left[{A}^{2 -}\right] \left[{H}_{3} {O}^{+}\right]}{\left[H A\right]} \ldots \ldots \left[2\right]$

From [1] and [2] we get

${K}_{1} \times {K}_{2} = \frac{\left[H {A}^{-}\right] \left[{H}_{3} {O}^{+}\right]}{\left[{H}_{2} A\right]} \times \frac{\left[{A}^{2 -}\right] \left[{H}_{3} {O}^{+}\right]}{\left[H A\right]}$

$\implies {K}_{1} \times {K}_{2} = \frac{{\left[{H}_{3} {O}^{+}\right]}^{2} \times \left[{A}^{2 -}\right]}{\left[{H}_{2} A\right]} \ldots . . \left[3\right]$

Now if we consider the following equilibrium

$2 H {A}^{-} r i g h t \le f t h a r p \infty n s {H}_{2} A + {A}^{2 -}$

then we can say that $\left[{H}_{2} A\right] = \left[{A}^{2 -}\right]$

Applying this to [3] we get

${K}_{1} \times {K}_{2} = {\left[{H}_{3} {O}^{+}\right]}^{2}$

Taking ${\log}_{10}$ on both sides

${\log}_{10} \left({K}_{1} \times {K}_{2}\right) = {\log}_{10} {\left[{H}_{3} {O}^{+}\right]}^{2}$

$\implies - {\log}_{10} {K}_{1} - {\log}_{10} {K}_{2} = - 2 {\log}_{10} \left[{H}_{3} {O}^{+}\right]$

$\implies p {K}_{1} + p {K}_{2} = 2 p H$

$\implies p H = \frac{1}{2} \left(p {K}_{1} + p {K}_{2}\right)$