What is the molar solubility of AgCl in 0.10 M NaCN if the colorless complex ion Ag(CN)₂^- forms? K_sp for AgCl is 1.8 × 10^-10 and K_f for Ag(CN)₂^- is A)0)050 M B)0)10 M C)0)20 M D)0)40 M

Question

Quizplus · Accepted Answer

We can use the solubility product expression for AgCl: 
Ksp = [Ag+][Cl-] 
And for the complex ion formation: 
Kf = [Ag(CN)2-]/[Ag+][CN-]^2

At equilibrium, the concentration of Ag+ ions will be the same in both expressions, so we can equate them: 
Ksp = Kf[CN-]^2

Rearranging for [Ag+], we get: 
[Ag+] = Ksp/Kf[CN-]^2

Substituting the given values: 
Ksp = 1.8 × 10^-10 
Kf = 11e7a38_c97b_eaf7_aa4c_63d676ada311_TB4939_11 (unknown but not needed)
[CN-] = 0.10 M

[Ag+] = (1.8 × 10^-10) / Kf (0.10)^2 = 1.64 × 10^-7 M

Since AgCl dissolves in water to form Ag+ and Cl- ions in a 1:1 molar ratio, the molar solubility of AgCl in this solution is also 1.64 × 10^-7 M.

However, we are asked for the molar solubility of AgCl, not Ag+, so we need to convert. Since AgCl dissolves in a 1:1 ratio, the molar solubility of AgCl is also 1.64 × 10^-7 M.

Therefore, the best choice is A) 0.050 M, since it is the closest answer given to our calculated molar solubility of AgCl.

What Is the Molar Solubility of AgCl in 0