Answer
$[Na^+] = 0.508M$
$[CO{_3}^{2-}] = 0.254M$
Work Step by Step
As we calculated in the last exercise, the molarity of $Na_2CO_3$ is equal to 0.254M.
There are 2 $Na^+$ in each $Na_2CO_3$, so:
$[Na^+] = 2 \times 0.254M = 0.508M$
There is 1 $C{O_3}^{2-}$ in each $Na_2CO_3$, so:
$[C{O_3}^{2-}] = 1 \times 0.254M = 0.254M$
