100 mL of 0.10 M $H X_{3} PO X_{4}$ is mixed with 180 mL of 0.10 M NaOH at 298 K. $p K_{a 1} = 2.12$ , $p K_{a 2} = 7.21$ , $p K_{a 3} = 12.32$ ; use $lo g 2 = 0.301$ . The pH of the mixture is closest to

Answer: B

10 mmol

H X_{3} PO X_{4}

reacts with the first 10 mmol of NaOH to give 10 mmol

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

. The remaining 8 mmol NaOH converts 8 mmol of it to

HPO X_{4} X^{2 -}

, leaving 2 mmol

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

. pH

= p K_{a 2} + lo g (8/2) = 7.21 + 2 (0.301) = 7.81