What is the Na+ ion concentration in the solution formed by mixing 20. mL of 0.10 M Na2SO4 solution with 50. mL of 0.30 M Na3PO4 solution?

Asked on by lak-86

1 Answer | Add Yours

jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

`Na_2SO_4 rarr 2Na^++SO_4^(2-)`

`Na_3PO_4 rarr 3Na^++PO_4^(3-)`


Mole ratio

`Na_2SO_4:Na^+ = 1:2`

`Na_3PO_4:Na^+ = 1:3`


Amount of `Na_2SO_4` used `= 0.1/1000xx20`

Amount of `Na^+` from `Na_2SO_4 = 0.1/1000xx20xx2`


Amount of `Na_3PO_4` used `= 0.3/1000xx50`

Amount of `Na^+` from `Na_3PO_4 = 0.3/1000xx50xx3`


Total `Na^+`  moles

= (`Na^+` from `Na_2SO_4` )+(`Na^+` from `Na_3PO_4` )

`= 0.1/1000xx20xx2+0.3/1000xx50xx3`

= 0.049


So this 0.049 moles of  `Na^+` is (20+50) = 70ml of solution.


Concentration of `Na^+ = 0.049/70xx1000 M = 0.7M`


So the final concentration of `Na^+` is 0.7M. 

We’ve answered 319,999 questions. We can answer yours, too.

Ask a question