For $H X_{3} PO X_{4}$ : $p K_{a 1} = 2.15$ , $p K_{a 2} = 7.21$ , $p K_{a 3} = 12.35$ . Using the amphiprotic formula, the pH of a solution of disodium hydrogenphosphate, $Na X_{2} HPO X_{4}$ , is approximately

Answer: A

HPO X_{4} X^{2 -}

lies between

H X_{2} PO X_{4} X^{-}

and

PO X_{4} X^{3 -}

, so it uses the neighbouring constants

p K_{a 2}

and

p K_{a 3}

: pH

= \frac{1}{2} (7.21 + 12.35) = 9.78

. Using

p K_{a 1}

and

p K_{a 2}

(4.68) applies to

H X_{2} PO X_{4} X^{-}