Answer
So $0.10M$ $NaHCO_3$ is more acidic.
Work Step by Step
** You can find the $K_a$ values on table: 14.2 (Page 624)
Since both solutions have the same concentration, we just have to determine which compound is more acidic:
-$NaHCO_3$:
It is a salt formed by $Na^+$ and $HC{O_3}^-$.
$Na^+$ has negligible acidity, so, we don't need to consider it.
$HC{O_3}^-$ is an acid with $K_a = 4.7 \times 10^{-11}$
- $Na_2HPO_4$:
It is a salt formed by $2Na^+$ and $HP{O_4}^{2-}$.
$Na^+$ has negligible acidity, so, we don't need to consider it.
$HP{O_4}^{2-}$ is an acid with $K_a = 4.6 \times 10^{-13}$
$HC{O_3}^-$ has a greater $K_a$ value, therefore, it is more acidic.
So, $0.10M$ $NaHCO_3$ is more acidic.
