Answer
$Na_2A$ and $NaHA$ are the best choice for $pH = 5.80$.
Work Step by Step
1. Calculate the $pK_{a1}$ Value
$pK_{a1} = -log(K_{a1})$
$pK_{a1} = -log( 1.1 \times 10^{- 3})$
$pK_{a1} = 2.959$
2. Calculate the $pK_{a2}$ Value
$pK_{a2} = -log(K_{a2})$
$pK_{a2} = -log( 2.5 \times 10^{- 6})$
$pK_{a2} = 5.602$
Therefore, the compounds in the second ionization are more appropriate, because the $pK_{a2}$ is closer to 5.80;
Second ionization of a diprotic acid:
$HA^-(aq) + H_2O(l) \lt -- \gt A^{2-}(aq) + H_3O^+(aq)$
Since we are using sodium salts:
$NaHA$ and $Na_2A$
