Find the solubility of AgBr in 3.9 M NH3 [Ksp of AgBr = 5.0 10-13 and Kf of Ag(NH3)2+ = 1.7 107]
AgBr==> Ag^+ + Br^-
Ag^+ + 2NH3 ==> Ag(NH3^+
eqn 1..Ksp = (Ag^+)(Br^-)
Kf = [Ag(NH3)2^+]/(Ag^+)(NH3)^2
Let S = solubility, then
eqn 2..S = (Br^-) = (Ag^+) + [Ag(NH3)2^+]
eqn 3..3.9 = (NH3) + (NH4^+) + 2[Ag(MH3)2^+]
Three equations as above.
1..as is
2..(Ag^+)<< [Ag(NH3)2^+]
3..3.9 = (NH3) as (NH4^+) is small and 2[Ag(NH3)2^+] is much smaller than (NH3)
Three equation and three unknowns.
To find the solubility of AgBr in 3.9 M NH3, we need to consider the formation of the complex ion Ag(NH3)2+ and its effect on the solubility of AgBr.
First, let's write the chemical equation for the dissolution of AgBr in water:
AgBr(s) ⇌ Ag+(aq) + Br-(aq)
The solubility product constant (Ksp) expression for AgBr is:
Ksp = [Ag+][Br-]
Since the dissolution of AgBr can also be affected by the complex ion Ag(NH3)2+, we need to consider its formation equation:
Ag+(aq) + 2 NH3(aq) ⇌ Ag(NH3)2+(aq)
The formation constant (Kf) expression for Ag(NH3)2+ is:
Kf = [Ag(NH3)2+]/([Ag+][NH3]^2)
Now, let's assume that "x" is the solubility of AgBr in moles per liter (M), which is the same as the concentration of Ag+ and Br- ions in the solution, since AgBr dissociates completely.
Using the balanced equation for the dissolution of AgBr, we can express the concentrations of Ag+ and Br- ions in terms of "x":
[Ag+] = x
[Br-] = x
Next, we can write the concentration of Ag(NH3)2+ in terms of "x" using the Kf expression:
[Ag(NH3)2+] = Kf * [Ag+][NH3]^2
Substituting the values of [Ag+] and [NH3] in terms of "x", we have:
[Ag(NH3)2+] = Kf * x * (3.9)^2
Now, let's substitute the concentrations of Ag+ and Br- ions as well as the concentration of Ag(NH3)2+ into the Ksp expression for AgBr:
Ksp = [Ag+][Br-] = (x)(x) = x^2
Since we know the value of Ksp for AgBr is 5.0 x 10^-13, we can set up the equation:
x^2 = 5.0 x 10^-13
Take the square root of both sides to solve for "x":
x = √(5.0 x 10^-13)
Calculating this expression gives the final solubility value (x) of AgBr in 3.9 M NH3.
To find the solubility of AgBr in 3.9 M NH3, we need to consider the dissolution and complex formation reactions.
The dissolution of AgBr can be represented as:
AgBr(s) ⇌ Ag+(aq) + Br-(aq)
The complex formation reaction between Ag+ and NH3 can be represented as:
Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq)
Given the equilibrium constants, Ksp of AgBr = 5.0 * 10^-13 and Kf of Ag(NH3)2+ = 1.7 * 10^7, we can write the equations for the equilibrium concentrations:
Ksp = [Ag+][Br-]
Kf = [Ag(NH3)2+]/[Ag+][NH3]^2
Let's assume 'x' represents the molar solubility of AgBr in NH3.
At equilibrium, the concentrations can be expressed as:
[Ag+] = [Ag(NH3)2+] = x
[Br-] = x (since AgBr dissociates completely)
Using the given equilibrium constants, we can set up the equations:
Ksp = [Ag+][Br-]
5.0 * 10^-13 = x * x
Kf = [Ag(NH3)2+]/[Ag+][NH3]^2
1.7 * 10^7 = x/x * (3.9)^2
Simplifying the equations, we get:
x^2 = 5.0 * 10^-13
x * (3.9)^2 = 1.7 * 10^7
Solving for 'x', we find:
x ≈ 7.07 * 10^-7 M
Therefore, the solubility of AgBr in 3.9 M NH3 is approximately 7.07 * 10^-7 M.