Question

Determine the molar solubility of AgI (Ksp = 8.50e-17) in 2.577 M S2O32- if the complex ion [Ag(S2O3)2]3- forms with a Kf = 2.90e13. Answers 0.09193/ 0.11637/ 0.04516/ or 0.14730

Answer 1

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
note that [Ag⁺] = [I⁻]

[Ag⁺][I⁻] = [Ag⁺]² = 8.50X10^-17 ⇒ [Ag⁺] = 2.91X10^-8⇒ solubility is therefore 2.91 × 10⁻⁸ mol/liter (quite insoluble)

PART - B

Ag⁺(aq) + 2 (S2O3)⁻2(aq) ⇌ [Ag(S2O3)2]3- . . . ; . . . Kf = [Ag(S2O3)2]3-/[Ag⁺][S2O3⁻2]² = 2.90 × 10¹³

But to get the solubility of AgI in this solution, you need the equilibrium constant for the reaction
AgI(s) + 2 S2O3²⁻(aq) ⇌ Ag(S2O3)₂⁻ + I⁻ ; K = [Ag(S2O3)₂⁻][I⁻]/[S2O3²⁻]² = ?

K = [Ag(S2O3)₂⁻][I⁻]/[S2O3^2⁻]² = {[Ag(S2O3)³₂⁻]/[Ag⁺][S2O3⁻]²} • {[Ag⁺][I⁻]}
K = Kf × Ksp = 2.90 × 10¹³ × 2.91X10^-8 = 8.45 × 10⁵

Thus,

in 2.577M [Ag(S2O3)2]3-, S2O32- = 2.577 M (initially)
In general, [S2O3⁻2] = 2.577 – 2x
where, in this case, [Ag(S2O3)2]3- = [I⁻] = x, so x² = K × [S203⁻]² = (4.5 × 10⁴)(2.577– 2x)²

Solve for x ⇒ x = 0.09193/0.11637 = [Ag(S2O3)2]3- = molar solubility

Basically, this means that AgI will dissolve until almost all the [S2O3⁻2] ion is used up.