Question

23) What is the molar solubility of AgCl in 0.20 M NaCN if the colorless complex ion Ag(CN)2- forms? 23) Ksp for AgCl is 1.8 × 10-10 and Kf for Ag(CN)2- is 1.0 × 1021.

A) 0.80 M B) 0.10 M C) 0.40 M D) 0.20 M

Answer 1

Ag⁺_(aq) + 2 CN^-_(aq) ===== Ag(CN)₂^-_(aq)                        Kf

AgCl_(s) ===== Ag⁺_(aq) + Cl^-_(aq)                                    Ksp

AgCl_(s) + 2 CN^-_(aq) ====== Ag(CN)₂^-_(aq)   + Cl^-_(aq)        K = Ksp * Kf

K = 1.8 × 10^-10 * 1.0 × 10²¹ = 1.8*10¹¹

As K is very large, all the NaCN (CN-) will react with the AgCl(s), then as we have 0.20 M of CN- an the stoichiometry is 2 mol CN- per 1 mol Cl-, [Cl-] = 0.10 M

Note that the source of Cl- is the AgCl…then that´s the molar solubility of AgCl.

Answer : B) 0.10 M