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. Which

of the following is the correct listing of the ions remaining in solution

in 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

Answer:

(A) PO43- < NO3- < Na+

Explanation:

Let us first write the balanced chemical equation

3 AgNO3 + Na3PO4 = Ag3PO4 + 3 NaNO3

Then we calculate the number of moles of each species using the formula

number of moles =  concentration * volume

Remember that sodium phosphate is the limiting reagent and silver phosphate precipitates out of solution and therefore does not contribute to the concentration of ions in solution

Before reaction                                   After reaction

0.01 mol Ag+                                       0 mol of Ag +

0.01 mol NO3-                                     0.01 mol of NO3-

0.03 mol of Na+                                  0.03 mol of Na+

0.01 mol of PO43-                               0.007 mol of PO43-

Answer 2

Answer:

$PO_4^{-3} < NO_3^- < Na^+$

Explanation:

Solutions.

a) 100 ml 1 M of Na3PO4

b) 100 ml 1 M of AgNO3

c) Mixture: 200 ml 0.5 M of Na3PO4 and 0.5 M of AgNO3

Reaction:

$3 Ag^+ + PO_4^{-3} \longrightarrow Ag_3PO_4$

So if silver ion is consumed almost completely:

$[Ag^+]=0 M$

$[PO_4^{-3}]=0.5 M-0.5 M \frac{1 mol PO4}{3 mol Ag}=0.33 M$

$[Na^+]=0.5 M *  \frac{3 mol Na}{mol Na_3PO_4}=1.5 M$

$[NO_3^-]=0.5 M$

In increasing order of concentration:

$PO_4^{-3} < NO_3^- < Na^+$