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30.0 mL of 0.10 M Ca(NO3)2 and 15.0 mL of 0.20 M Na3PO4 solutions are mixed. After the...

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lak-86 | Student, Undergraduate | Salutatorian

Posted July 21, 2013 at 7:58 AM via web

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30.0 mL of 0.10 M Ca(NO3)2 and 15.0 mL of 0.20 M Na3PO4 solutions are mixed. After the reaction is complete, which of these ions has the lowest
concentration in the final solution?

(A)  Na+ 
(B)  NO3–
(C)  Ca2+
(D)  PO43–

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

Posted July 21, 2013 at 8:21 AM (Answer #1)

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Amount of `Ca(NO3)_2` mixed `= 0.1xx30/1000 = 0.003`

Amount of `Na_3PO_4^(3-)` mixed`= 0.2xx15/1000 = 0.003 `

 

`3Ca(NO_3)_2+2Na_3PO_4 rarr Ca_3(PO_4)_2+6NaNO_3`

 

`Ca(NO_3)_2:2Na_3PO_4 = 3:2 = 0.003:0.002`

 

So 0.003 moles of `Ca(NO_3)_2` will react with 0.002 moles of `Na_3PO_4^(3-)` .

So in the final mixture we have;

  • `Ca_3(PO_4)_2` which is a precipitate
  • `Na^+` ion
  • `NO_3-` ion
  • `PO_4^(3-)` ions which was excess after reaction.

 


`Na^+` and `NO_3^-` ions remained as it is because it doesn't form any precipitate. All the `Ca^(2+)` was consumed for the precipitation. What was left after precipitation is the rest `PO_4^(3-) ` ions.

`Na_3PO_4 rarr 3Na^++PO_3^(3-)`

 

So it is clear that `PO_4^(3-)` is initially lower than `Na^+` also.

So the lowest concentration would be at `PO_4^(3-)` ions.

Answer is D.

 

 

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