The approximate $pH$ of $0.01 \ M \ NaHA$ is calculated by: ($K_{a_1} = 10^{-6}$ and $K_{a_2} = 10^{-8}$ are ionization constants of $H_2A$)
- A$pH = 7 + \frac{pK_{a_1}}{2} + \frac{\log C}{2}$
- B$pH = 7 - \frac{pK_{a_1}}{2} - \frac{\log C}{2}$
- C$pH = \frac{pK_{a_1} + pK_{a_2}}{2}$
- D$pH = \frac{pK_{a_1} - pK_{a_2}}{2}$
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