Combine the Ksp and Kf equilibria for AgCl and Ag(NH3)2+ respectively and demonstrate Hess's law to determine the equilibrium constant for the dissolution of AgCl in NH3.
AgCl + 2NH3 --> Ag(NH3)2 + Cl-
K= (Ksp)(Kf)
If there is only one mole of both AgCl and Ag(NH3)2, both the Ksp and Kf remain with an exponent of one, correct? The coefficient in front of the NH3 doesn't play a role?
Keq = Kf*Ksp = [Ag(NH3)2]*(Cl^-)/(NH3)^2
From my viewpoint, the 2 for NH3 does play a role.
Dr Bob is 100% right. You must take the NH3 into account in the formation of the silver-amine complex. It is easier to see if simultaneous equations are written.
(1) AgCl <=> Ag^+ + Cl^-
(2) Ag+ + 2NH3 <=> Ag[NH3]2^+
_______________________
(3) AgCl + 2NH3 <=> Ag[NH2]2^+ + Cl-
Ceq:--- (0.10M-2x) x x
(1) Ksp = 1.8E-10
(2) Kf = 1.7E+7
(3) Knet = Ksp x Kf
= (1.8E-10)(1.7E+7)
= 0.0031
Using net equation for analysis of solubility and assuming 0.10M in NH3:
AgCl + 2NH3 <=> (Ag[NH3]2^+)+Cl^-
Ceq ---(0.10M-2x)* x x
__________
The Knet (0.0031) of this equation is too large to allow dropping 2x and avoiding significant error in calculations. However, the Kcplx expression can be reduced to a non-quadratic form by taking square root of both sides of Knet expression.
(Knet)^2
=([Ag(NH2)2^+][Cl^-])/([NH3]^2)
=[(x)(x)/(0.10-2x)^2]=(x^2)/(0.1-2x)^2
[(0.0031)^1/2]=[(x^2)/(0.1-2x)^2]^1/2
0.055 = (x)/(0.10-2x)
0.055(0.10-2x) = (x)
0.0055-0.11x = x
Add 0.11 to both sides & solve for x
1.11x = 0.0055
x = solubility of AgCl in presence of 0.01M NH3 = (0.0055/1.11)M = 0.005M
[AgCl](in HOH) = 1.34e-5M
[AgCl](With Common-ion) ~ 9.0E-9M
[AgCl](with complex form'n) ~ .005M
Yes, you are correct. In this case, since there is one mole of both AgCl and Ag(NH3)2, the coefficients in front of AgCl and Ag(NH3)2 do not affect the equilibrium constant. The equilibrium constant, K, for the overall reaction is given by the product of the individual equilibrium constants, Ksp and Kf:
K = Ksp * Kf
The fact that the coefficient in front of NH3 is 2 does not change the equilibrium constant expression.
Yes, that's correct. The coefficient in front of NH3 doesn't affect the exponents in the equilibrium expressions (Ksp and Kf). Both Ksp and Kf are equilibrium constants, which express the extent of a chemical reaction under specific conditions.
The equilibrium constant, Ksp (solubility product constant), describes the equilibrium between a sparingly soluble salt and its ions in a saturated solution. It is expressed as the product of the concentrations of the dissociated ions raised to the power of their stoichiometric coefficients. For AgCl, the equilibrium expression is:
Ksp = [Ag+][Cl-]
The equilibrium constant, Kf (formation constant), describes the equilibrium between a metal ion and a complex formed from a ligand. In this case, the complex is Ag(NH3)2+. The equilibrium expression is:
Kf = [Ag(NH3)2+]
When we combine both equilibria, we multiply the two equilibrium expressions together:
K = Ksp * Kf
Since both Ksp and Kf have exponents of one, we can write the equilibrium constant as:
K = [Ag+][Cl-] * [Ag(NH3)2+]
The stoichiometric coefficients (1 and 1) are not included in the equilibrium expressions because they do not affect the exponents.
To determine the equilibrium constant for the dissolution of AgCl in NH3, you would need the values of Ksp and Kf. These values can be obtained from experimental data or from tables of equilibrium constants. Once you have the values for Ksp and Kf, you can simply multiply them together to calculate K.