Question

a. The solubility product, Ksp, for AgI is 8.3 x 10-17. What is the concentration of iodide ion (I- ) in solution saturated in silver iodide.

b. What would the concentration of silver(I) ion (Ag+ ) be for a saturate solution of AgI which is also 0.20 M in NaI (soluble, of course)

Answer 1

(a.) The equilibrium reaction in saturated solution of AgI is :

AgI (s) Ag⁺ (aq) + I^- (aq)

K_sp AgI = [Ag⁺][I^-]

K_sp AgI = (s) * (s)

where s is the molar solubility of AgI

8.3 x 10^-17 = s²

s = (8.3 x 10^-17)^1/2

s = 9.1 x 10^-9 M

Concentration of iodide, [I^-] = s = 9.1 x 10^-9 M

(b.) initial concentration of I^- = [NaI] = 0.20 M

K_sp AgI = [Ag⁺][I^-]

K_sp AgI = (s) * (s + 0.20 M)

where s is the molar solubility of AgI

8.3 x 10^-17 = s² + 0.20s

s² + 0.20s - 8.3 x 10^-17 = 0

Solving for s, s = 4.2 x 10^-16 M

Concentration of Ag⁺ = [Ag⁺] = 4.2 x 10^-16 M