Question

What is the silver ion concentration in a solution prepared by mixing 407mL of 0.379 M silver nitrate with 461 mL of 0.565 M sodium phosphate? The Ksp of silverphosphate 2.8x10^-18.

Answer 1

concentration of AgNO3 = 407/(407+461)*0.379 = 0.18 M

concentration of Na3PO4 = 461/(407+461)*0.565 = 0.3 M

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

3mol AgNO3(aq) = 1 mol Na3PO4(aq)

limiting reagent is AgNO3

concentration of Ag3PO4 formed = 0.18/3 = 0.06 M

ksp of silverphosphate = s*(3S)^3

   ksp = 27S^4

(2.8*10^(-18)) = 27*s^4

S = solubility = 1.79*10^(-5) M

so that , amount of Ag3po4soluble = (1.79*10^(-5))*0.06 = 1.074*10^(-6) M

[Ag+] = 1.074*10^(-6)*3 = 3.222*10^-6 M