Question

What is the molar solubility of AgCl in 0.40 M NaCN if the colorless complex ion Ag(CN)2 - forms? Ksp for AgCl is 1.8

Answer 1

let the molar solubility of AgCl be 'x'

the reactions are

   AgCl (s)------->Ag⁺    +   Cl^-       K_sp=1.8*10^-10

Ag⁺   +   2CN^------>Ag(CN)₂^-        K_f=1.0*10²¹

--------------------------------------------------------

AgCl (s) +2CN^-------> Cl^- +Ag(CN)₂^-

initial:         0.10M           0          0

change:          -2x             x          x

equilibrium:  (0.10-2x)       x          x

the equilibrium constant K for this reaction is

K= K_sp* K_f =(1.8*10^-10)*(1.0*10²¹) = 1.8*10¹¹

?K=[Cl^-][Ag(CN)₂^- ]/[CN^-]²=(x*x)/(0.10-2x)²= 1.8*10¹¹

==>x²=1.8*10¹¹*(0.10-2x)²

==>720x²-72x+1.8=0

on solving we get

x=0.050M

therefore the molar solubility of AgCl = 0.050 M