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. Whi

The correct ordering for the ions in concentration is: A. PO43- < NO3- < Na+. The reaction can be described as follows: Na3PO4 + 3AgNO3 → Ag3PO4 + 3Na+ + 3NO3-. During this process, all the Ag and PO43- in the solution have reacted to produce the yellow precipitate, Ag3PO4, leaving minimal amounts of Ag or PO43- in the solution. Given that both the concentrations and volumes of reactants are identical, the resulting moles would also remain equal. The dissociation of Na3PO4 generates 3 moles of Na+, while AgNO3 provides 1 mole of NO3-, leading us to conclude that Na+ will dominate the solution, pointing towards option A.

The correct answer is Option A. The calculation goes as follows: Number of millimoles of Na3PO4 = 1 × 100 = 100 Number of millimoles of AgNO3 = 1 × 100 = 100 Dissociating 1 mole of Na3PO4 yields 3 moles of sodium ions and 1 mole of phosphate ions, whereas 1 mole of AgNO3 releases 1 mole of Ag+ and 1 mole of NO3-. The Ag+ ion concentration becomes negligible since it forms a precipitate with the phosphate ion, indicating that the concentration of phosphate ions is also low. With 100 millimoles of Na3PO4, we get 300 millimoles of Na+ and 100 millimoles of PO43-, and with 100 millimoles of AgNO3 we have 100 millimoles of Ag+ and 100 millimoles of NO3-. Thus, the order of increasing concentration is: PO43- < NO3- < Na+.