Question

What is the concentration of each ion in 0.250 M Na2SO4?

Answer 1

Answer: The concentration of $Na^+$ is 0.500 M and $SO_4^{2-}$ is 0.250 M in 0.250 M $Na_2SO_4$

Explanation:

The dissociation of $Na_2SO_4$ is given as:

$Na_2SO_4\rightarrow 2Na^++SO_4^{2-}$

As 1 mole of $Na_2SO_4$ gives  = 2 moles of $Na^+$

0.250 moles of $Na_2SO_4$ gives =$\frac{2}{1}\times 0.250=0.500$ moles of $Na^+$

As 1 mole of $Na_2SO_4$ gives = 1 mole of $SO_4^{2-}$

0.250 moles of $Na_2SO_4$ gives =$\frac{1}{1}\times 0.250=0.250$ moles of $SO_4^{2-}$

Thus the concentration of $Na^+$ is 0.500 M and $SO_4^{2-}$ is 0.250 M in 0.250 M $Na_2SO_4$