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What is the Na+ ion concentration in the solution formed by mixing 20. mL of 0.10 M...
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`Na_2SO_4 rarr 2Na^++SO_4^(2-)`
`Na_3PO_4 rarr 3Na^++PO_4^(3-)`
`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` )
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.
Posted by jeew-m on June 11, 2013 at 6:30 PM (Answer #1)
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