What is the molar solubility of AgCl in 1.0 M K₂S₂O₃ if the complex ion Ag(S₂O₃)₂^3- forms? The Ksp for AgCl is 1.8 × 10^-10 and the K_f for Ag(S₂O₃)₂^3- is 2.9 × 10¹³. A)0.50 M B)1.0 M C)1.5 M D)2.0 M

Question

Quizplus · Accepted Answer

Using the given Ksp for AgCl, we can determine that the concentration of Ag+ and Cl- ions in solution at equilibrium are both equal to √(1.8 × 10^-10 M), which is approximately 1.34 × 10^-5 M.

Next, we can use the given Kf for the complex ion to determine the concentration of Ag+ ions that are complexed with SS12O13- ligands to form Ag(SS12O13)S12P13-. Let's denote the molar solubility of AgCl as x. Then, the concentration of Ag+ in solution is also x. The concentration of SS12O13- ligands can be assumed to be negligible compared to x.

The Kf expression for the complex ion is:
Kf = ([Ag(SS12O13)S12P13-]/[Ag+][SS12O13]^2) = 2.9 × 10^13

Substituting in the known values, we have:
2.9 × 10^13 = ([Ag(SS12O13)S12P13-]/(x)(0))^2
[Ag(SS12O13)S12P13-]/x^2 = 2.9 × 10^13

Solving for [Ag(SS12O13)S12P13-], we have:
[Ag(SS12O13)S12P13-] = (x^2)(2.9 × 10^-13)

Now, we can use the fact that Ag+ ions that are complexed with SS12O13- ligands do not contribute to the equilibrium concentration of Ag+ ions. Therefore, the total concentration of Ag+ ions at equilibrium is:
Total [Ag+] = x - [Ag(SS12O13)S12P13-]

Substituting in the previously obtained expressions, we have:
Total [Ag+] = x - (x^2)(2.9 × 10^-13)

The Ksp expression for AgCl at equilibrium is:
Ksp = [Ag+][Cl-]^2 = (x)(1.34 × 10^-5)^2

Setting the Ksp expression equal to the total [Ag+] expression, we have:
(x)(1.34 × 10^-5)^2 = x - (x^2)(2.9 × 10^-13)

Solving for x using the quadratic equation, we obtain:
x = 0.50 M

Therefore, the molar solubility of AgCl in 1.0 M KF is 0.50 M.

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