What is the solubility of AgCl in 10 M NH3? The Kf of Ag(NH3)2+ from AgCl(s) in aqueous NH3 is 0.0031, which is for the balanced reaction shown below.
AgCl(s) + 2 NH3(aq) → Ag(NH3)2+(aq) + Cl-(aq)
Work:
Ksp = x^2 /(2x+10)^2 = 1.8 x 10^-10
Now I am confused.......
At first I got 0.557 an now.. I don't think this is correct... :(
I used the ICE chart to calculate, but it seems that I am doing this incorrectly...
The first thing wrong is you are using Ksp. You still haven't answered my question about the Kf. Kf is the formation constant for the complex.
Ag^+ + 2NH3 ==> [Ag(NH3)2]^+
The literature shows that as about 1.7E7 and your question gives it as 0.0031. I think that 0.0031 is the equilibrium constant for the reaction
AgCl(s) + 2 NH3(aq) → Ag(NH3)2+(aq) + Cl-(aq) and that is NOT Kf.
In addition, if I calculate the value of Keq for that reaction it is Keq = Kf*Ksp = 0.0031 which adds some credibility to what I've said above. Here is the way I would set up the ICE chart.
.AgCl + 2NH3 =>[Ag(NH3)2]^+ + Cl^-
I..solid..10.....0.............0
C....s....-2x....x.............x
E..s....10-2x....x.............x
K = 0.0031 = (x)(x)/(10-2x)^2
Solve for x.
To solve this problem, you need to use the concept of the solubility product constant (Ksp) and the formation constant (Kf). The equilibrium expression for the dissociation of AgCl is given by:
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
Let's begin by setting up the Ksp expression for AgCl:
Ksp = [Ag+][Cl-]
Now, we need to express the concentration of Ag+ and Cl- in terms of x, which represents the solubility of AgCl in moles per liter:
[Ag+] = x
[Cl-] = x
We also need to consider the formation of Ag(NH3)2+ from AgCl in aqueous NH3:
AgCl(s) + 2 NH3(aq) ↔ Ag(NH3)2+(aq) + Cl-(aq)
The formation constant (Kf) of Ag(NH3)2+ is given as 0.0031. This means that for every Ag(NH3)2+ formed, the concentration of Ag+ and NH3 will be multiplied by Kf. So, we can express the concentration of Ag+ and NH3 in terms of x:
[Ag+] = Kf * x
[NH3] = 2 * Kf * x
Next, we substitute these expressions into the Ksp equation and solve for x:
Ksp = [Ag+][Cl-]
1.8 x 10^-10 = (Kf * x)(x)
Simplifying this equation, we have:
1.8 x 10^-10 = Kf * x^2
Now, you can solve for x by rearranging the equation and taking the square root:
x^2 = (1.8 x 10^-10) / Kf
x = sqrt((1.8 x 10^-10) / Kf)
Substituting the given value of Kf (0.0031), we have:
x = sqrt((1.8 x 10^-10) / 0.0031)
Calculating this value will give you the solubility of AgCl in 10 M NH3.