$50 mL$ of $0.10 M$ $H X_{3} PO X_{4}$ is mixed with $80 mL$ of $0.10 M$ $NaOH$ . Given $p K_{a 1} = 2.15$ , $p K_{a 2} = 7.21$ , $p K_{a 3} = 12.32$ , find the pH (two decimals). Take $lo g 2 = 0.301$ , $lo g 3 = 0.477$ and treat the final mixture as a simple $H X_{2} PO X_{4} X^{-} / HPO X_{4} X^{2 -}$ buffer.

Numerical answer type, tolerance 0.01.

Answer: 7.39

H X_{3} PO X_{4}

5.0 mmol

;

NaOH

8.0 mmol

. The first

5.0

mmol convert

H X_{3} PO X_{4}

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

; the remaining

3.0

mmol convert

3.0

mmol of it to

HPO X_{4} X^{2 -}

, leaving

H X_{2} PO X_{4} X^{-} = 2.0

and

HPO X_{4} X^{2 -} = 3.0

mmol.

pH = 7.21 + lo g (3/2) = 7.21 + 0.477 - 0.301 = 7.39

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