A0.100 l solution of 0.300 m agno3 is combined with a 0.100 l solution of 1.00 m na3po4. calculate the concentration of ag and po43– at equilibrium after the precipitation of ag3po4 (ksp = 8.89 × 10–17).

Moles of Ag+ = 0.1 L * 0.3 m = 0.03 Moles

moles of PO4-3 = 0.1 L * 1 = 0.1 moles

According to the reaction equation which can be written as:

                 3Ag +    +   PO4 3-   ↔   Ag3PO4(s)
initial        0.03                0.1
change    -3X                    -X                        -

Equ         (0.03-3X)      (0.1-X)

when Ksp = [Ag]^3 [PO43-]
               
                 = (0.03-3X)^3 ( 0.1-X)

∴8.89 x 10^-17 = (0.03 - 3X)^3 (0.1-X)

8.89 x 10^-17 = 27 * (X-0.1)^4

∴ X = 2.9 x 10^-6

∴[Ag+] = 0.03 - (3 * (2.9 X 10^-6))
  
             = 0.02998 M

[PO43-] = 0.1 - X = 0.1 - (4.3 x 10^-5)
                         
                             = 0.0999 M