Question

When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whichof the following is the correct listing of the ions remaining in solutionin order of increasing concentration?(A) PO43- < NO3- < Na+(B) PO43- < Na+ < NO3-(C) NO3- < PO43- < Na+(D) Na+ < NO3- < PO43-(E) Na+ < PO43- < NO3-

Answer 1

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Answer:

A. PO43- < NO3- < Na+

Explanation:

Here’s the reaction equation:

Na3PO4 + 3AgNO3 = Ag3PO4 + 3Na+ + 3NO3-

All of the Ag and PO43- in the solution reacted to form the Ag3PO4 which is the yellow precipitate, leaving little or no Ag or PO43- left in the solution. This narrows down the answer to options A and B.

Since the concentrations and the volumes of the reactants are the same, the number of moles will also be the same.

Na3PO4 will release 3 moles of Na+ while AgNO3 releases 1 mole of NO3-. Hence, the Na+ would be more in solution than the NO3-, narrowing the answer further down to option A.

PO43- < NO3- < Na+

Answer 2

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be  PO43- < NO3- < Na+