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–
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...
(The entire section contains 130 words.)
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