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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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- `Ca_3(PO_4)_2` which is a precipitate
- `Na^+` ion
- `NO_3-` ion
- `PO_4^(3-)` ions which was excess after reaction.
Best answer as selected by question asker.
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;
`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.
Posted by jeew-m on July 21, 2013 at 8:21 AM (Answer #1)
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