Answer
(b) $H_2P{O_4}^-$ and $H{PO_4}^{2-}$
This pair is the best choice because the $pKa$ for its acid "$H_2P{O_4}^-$" is the closest to 7.00.
Work Step by Step
Calculate the $pK_a$ for each acid pairs:
(a)
$pKa = -log(Ka)$
$pKa = -log( 7.5 \times 10^{- 3})$
$pKa = 2.13$
(b)
$pKa = -log( 6.2 \times 10^{- 8})$
$pKa = 7.21$
(c)
$pKa = -log( 4.8 \times 10^{- 13})$
$pKa = 12.32$
** You can find these values in the page 1045 of this book.
Identify the closest to the wanted pH (7.00)
- It is the (b) pair.
