What is the molar solubility of AgCl in 0.50 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 1.0 × 10²¹. A)0.25 M B)0.50 M C)1.0 M D)2.0 M

Question

Quizplus · Accepted Answer

Using the equilibrium reaction:

AgCl(s) ↔ Ag+(aq) + Cl-(aq)

The equilibrium constant expression for this reaction is:

Ksp = [Ag+][Cl-]

Using the equilibrium reaction:

Ag(CN)2-(aq) + Cl-(aq) ↔ AgCl(s) + 2CN-(aq)

The equilibrium constant expression for this reaction is:

Kf = [AgCl][CN-]2 / [Ag(CN)2-]

We are given the values of Ksp and Kf, but we need to find the molar concentration of Ag+ and Cl- in solution in order to calculate the molar solubility of AgCl. Let x be the molar solubility of AgCl in mol/L.

Then [Ag+] = x mol/L and [Cl-] = x mol/L

Using the equilibrium constant expressions and substituting in the given values, we get:

Ksp = x*x = 1.8 × 10^-10

x = sqrt(1.8 × 10^-10) = 1.34 × 10^-5 M

Next, we can use the equilibrium constant expression for the formation of Ag(CN)2- to find the molar concentration of Ag+ in the presence of CN-:

Kf = [AgCl][CN-]^2 / [Ag(CN)2-]

1.0 × 10^21 = [(1.34 × 10^-5)^2][0.50] / [Ag(CN)2-]

[Ag(CN)2-] = 9.49 × 10^-14 M

Since [Ag(CN)2-] = [Ag+], we have [Ag+] = 9.49 × 10^-14 M

Finally, we can use the Ksp expression to find the molar solubility of AgCl in the presence of CN-:

Ksp = [Ag+][Cl-]

1.8 × 10^-10 = (9.49 × 10^-14)(x)

x = 1.89 × 10^-4 M

However, we should keep in mind that AgCl forms a complex ion with CN-, which decreases the molar solubility of AgCl.

Thus, the molar solubility of AgCl in 0.50 M NaCN is 1.89 × 10^-4 M, which is closest to choice A (0.25 M).

What Is the Molar Solubility of AgCl in 0