Calculate the molar solubility of AgCl in a 0.15 M solution of NH3(aq).
(Ksp(AgCl) = 1.6 x 10^-10; Kf(Ag(NH3)2+) = 1.5 x 10^7)
To calculate the molar solubility of AgCl in a 0.15 M solution of NH3(aq), we need to consider the complex formation between Ag+ ions and NH3 molecules.
First, let's write the balanced equation for the dissolution of AgCl in water:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
The solubility product constant (Ksp) expression for AgCl is given as:
Ksp = [Ag+][Cl-] = 1.6 x 10^-10
However, in this case, NH3 is present in solution, which can form a complex with the Ag+ ions. The formation constant (Kf) expression for the complex Ag(NH3)2+ is given as:
Kf = [Ag(NH3)2+]/[Ag+][NH3]^2 = 1.5 x 10^7
Let's assume that the molar solubility of AgCl is represented by "x". Since AgCl dissociates into Ag+ and Cl- ions, we can write:
[Ag+] = x
[Cl-] = x
For the complex formation, we need to consider the concentration of Ag(NH3)2+:
[Ag(NH3)2+] = 2x (since two NH3 molecules are involved in the complex)
Since NH3 is present in excess (0.15 M) compared to the solubility of AgCl, we can assume that the concentration of NH3 is not significantly changed before and after the complex formation.
Now, let's rewrite the Ksp and Kf expressions using the concentrations:
Ksp = [Ag+][Cl-] = x * x = x^2 = 1.6 x 10^-10
Kf = [Ag(NH3)2+]/[Ag+][NH3]^2 = (2x)/((x)(0.15)^2) = (2x)/(0.0225*x^2) = 1.5 x 10^7
By solving the equations simultaneously, we can find the value of x, which represents the molar solubility of AgCl:
1.6 x 10^-10 = x^2
1.5 x 10^7 = (2x)/(0.0225*x^2)
Solving these equations, we find that x ≈ 1.3 x 10^-5 M
Therefore, the molar solubility of AgCl in a 0.15 M solution of NH3(aq) is approximately 1.3 x 10^-5 M.
To calculate the molar solubility of AgCl in the NH3(aq) solution, we need to consider the complex ion formation of Ag(NH3)2+.
Let's break down the process step by step:
Step 1: Write the balanced chemical equation for the dissociation of AgCl:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Step 2: Write the equilibrium expression for the dissolution of AgCl:
Ksp = [Ag+][Cl-]
Step 3: Write the balanced chemical equation for the complex ion formation of Ag(NH3)2+:
Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq)
Step 4: Write the equilibrium expression for the complex ion formation:
Kf = [Ag(NH3)2+]/([Ag+][NH3]^2)
Step 5: Since AgCl partially dissolves, let's assume x moles of AgCl dissolve in the 0.15 M NH3 solution. This results in x moles of Ag+ and Cl- in the solution.
Step 6: Using the assumption from the previous step, we can set up the equilibrium expressions. For the dissolution of AgCl, the concentrations would be:
[Ag+] = x
[Cl-] = x
For the complex ion formation, the concentrations would be:
[Ag(NH3)2+] = x
[Ag+] = x
[NH3] = 0.15 M
Step 7: Apply the equilibrium expressions and solve for x using the given values of Ksp and Kf:
Ksp = [Ag+][Cl-] = x * x = x^2
Kf = [Ag(NH3)2+]/([Ag+][NH3]^2) = x/(x * (0.15)^2) = 1.5 x 10^7
Step 8: Solve the equation derived from the Kf expression for x:
1.5 x 10^7 = x/(x * (0.15)^2)
1.5 x 10^7 = 1/(0.15)^2
1.5 x 10^7 = 1/0.0225
x = 0.0225 mol/L
Therefore, the molar solubility of AgCl in the 0.15 M NH3 solution is 0.0225 mol/L.
.......AgCl(s) ==> Ag^+ + Cl^-....................Ksp = (Ag^+)(Cl^-)
.......Ag^+ + 2NH3 ==> [Ag(NH3)2]^+....Kf = [Ag(NH3)2]^+/(Ag^+)(NH3)^2
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.......AgCl(s) + 2NH3 ==> Cl^- + [Ag(NH3)2]^+.... Keq = Ksp*Kf
I........solid........0.15...........0............0
C......solid..........-2x............x.............x
E......solid.........0.15-x........x..............x
Plug the E line into the Keq expression and solve for x = solubility of AgCl in mols/L
Post your work if you get stuck.