Question

Calculate the solubility of silver chloride in a solution that is 0.150 M in NH3.

Answer 1

th ereactions

AgCl(s) ↔ Ag + + Cl- Ksp = 1.8 ×10-10

Ag + + 2NH3(aq) → Ag(NH3)2 + Kf = 1.7 ×107

add both so you get:

AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ + Cl- whic hhas a K overall:

calcualte K overall as follows:

K = Ksp × Kf = 1.8 ×10-10 ×1.7 ×107 = 3.1 ×10-3

K = [Ag(NH3)+2][Cl-] / [NH3]2

K = = S^2/(M-2s)^2

M = 0.15 so

K = = S^2/(0.15 -2s)^2

solve for S

sqrt(3.1*10^-3) = S/(0.15-2s)

0.0556(0.15-2s) = s

0.00834 - 0.111s = s

1.11s = 0.00834

s = 0.00834 /1.1 = 0.00758181818 M