Answer
(b) $NaH_2PO_4$ and $Na_2HPO_4$ are the best choice for this situation.
Work Step by Step
1. First, determine the $pK_a$ for the acid in each pair:
(a)
1. Calculate the pKa value $(H_3P{O_4}):$
$pKa = -log(Ka)$
$pKa = -log( 7.5 \times 10^{- 3})$
$pKa = 2.12$
(b)
1. Calculate the pKa value $H_2P{O_4}^-:$
$pKa = -log(Ka)$
$pKa = -log( 6.2 \times 10^{- 8})$
$pKa = 7.21$
(c)
1. Calculate the pKa value $(H{PO_4}^{2-})$:
$pKa = -log(Ka)$
$pKa = -log( 3.6 \times 10^{- 13})$
$pKa = 12.44$
2. Identify the acid with the closest $pK_a$ to the desired pH (7):
- It is the $H_2P{O_4}^-$.
Therefore: Correct answer = (b)
