Answer to Question #137727 in General Chemistry for hi

Question

Calculate solubility of AgI in 0.1 M NH3 ?

Answer 1

Ks (AgI) = 8,3 ∙ 10-¹⁷

AgI = > Ag⁺ + I^- Ksp = [Ag+ ] [ I-] = 8.3 * 10 ^ ^{- 17}

A^g+ + 2NH3 <=> Ag(NH3)²⁺

Kf = [Ag(NH3)²⁺] / [Ag+] [ NH3-] ^{^2} = 1.7 * 10 ^{^ 7}

this is complexation

one Ag ⁺ that dissolves one I ^- ion is released

I^- = [Ag(NH3)²⁺] = c

solving for [Ag+]

c^2 = Ksp * Kf * [NH3] ^{^2}

c^2 = 8.3 * 10 ^{^-17} * 1.7*10 ^{^7} * (0.1-2c) ^{^2}

c = -5.644 * 10 ^{^ -10} moles per liter