Question

375 mL of a .150 M aqueous solution of silver (1) nitrate is mixed with 125 mL of .125 M aqueous solution of sodium phosphate. Calculate the mass of precipitate that forms and the final concentration of each ion in the mixed solution. Volumes are additive and the precipitation reaction goes to completion.

Answer 1

concentration of AgNO3 = 375 x 0.150 / 375 + 125 = 0.1125

concentration of Na3PO4 = 125 x 0.125 / 375 + 125 = 0.03125

3 AgNO3 (aq)   + Na3PO4 (aq)   -----------------> Ag3PO4 (s) + 3 NaNO3 (aq)

     3                           1                                                1

0.1125                0.03125                                          ?

here limiting reagent is Na3PO4 .

remaining AgNO3 = 0.1125 - (0.03125 x 3) = 0.01875 M

concnetation of Ag+ = 0.01875 M

moles of Ag3PO4 = 0.0156

mass of Ag3PO4 = 0.0156 x 418.576 = 6.54 g

mass of Ag3PO4 = 6.54 g

[Na+] = 0.938 M

[NO3-] = 0.1125 M

Ag3PO4 -----------------> 3 Ag+ + PO43-

Ksp = [Ag+]^3[PO43-]

8.89×10^–17 = (0.01875)^3 x [PO43-]

[PO43-] = 1.35 x 10^-11 M