Question

What is the silver ion concentration in a solution prepared by mixing 423 mL of 0.386 M silver nitrate with 459 mL of 0.509 M sodium phosphate? The Ksp of silver phosphate is 2.8X10^-18

Answer 1

Ag3 PO4 == > 3Ag+ + PO4 ^2-

^for every 1 mole of phosphate, 3 moles of Ag+ is required

let x= Po4-2

Ksp = [Ag+]³ [PO4-2] = (3x)3*x

27x4= 2.8*10-18

x=1.6*10-5 mole/L of silver phosphate

moles of AgNO3 in 0.386M of 423ml =0.386*423/1000=0.163278

Moles of Sodium phosphate = 0.509*459/1000= 0.233631

the reaction between silver nitrate and sodium phosphate is

3 AgNO3 + Na3PO4 = Ag3PO4 + 3 NaNO3

3 moles requires 1 mole of of sodium phoshate molar ratio of AgNo3: Na3PO4= 3:1

Given moalr ratio = 0.163278: 0.233631 = 0.69871 :1

all the silver reacts and concentration of silver= 3* 1.68*10-5mol/L =0.0000504 mole/L