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`

Step-by-step 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 =
`(2)/(1)* 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 =
`(1)/(1)* 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`