Question

Calculate the solubility of AgCl (Ksp=1.77×10-10)

a. In 0.100M NaCl

b. Added to 0.050M ammonia. The Kform of Ag(NH3)2+ is 1.6×107.

Please show all work! Thank you!

Answer 1

a) 0.100M NaCl

Ksp=1.77×10^-10

AgCl -----------> Ag+ + Cl-

s s s = solubility of AgCl

Given that 0.1 M NaCl is added i.e [Cl-] = 0.1 M

Then,

Ksp = [Ag+] [Cl-] = s ( s+ 0.1 )

1.77×10^-10 = s² + 0.1s

Since s is very small, s2 is eliminated.

Then,

1.77×10^-10 = 0.1s

s = 1.77 x 10^-9 M

Therefore,

solubility of AgCl in 0.100M NaCl = 1.77 x 10^-9 M

b) AgCl + 2NH3 -------------> Ag(NH3)2+ + Cl-

Kf of Ag(NH₃)₂+ = [AgCl] [NH3]²

1.6×10⁷ = [AgCl] (0.05)²

[AgCl] = 640 x 10⁷ M