Calculate the solubility of silver chloride in 10.0 M ammonia given the following information:
Ksp (AgCl) = 1.6 x 10^–10
Ag+ + NH3--->AgNH3+ K=2.1x10^3
AgNH3+ + NH3-----> Ag(NH3)2+ K=8.2x10^3 Answer : 0.48 M
Calculate the concentration of NH3 in the final equilibrium mixture.
Answer: 9.0 M
I don't understand how my professor did this, i tried several ways and i am not doing it right. Please help.
Please add some calculations to the answers
Please add some calculations to the answers
To calculate the solubility of silver chloride (AgCl) in 10.0 M ammonia (NH3) and the concentration of NH3 in the equilibrium mixture, you can use the concept of equilibrium and the given equilibrium constants.
Step 1: Write the balanced chemical equation representing the dissolution of silver chloride in ammonia:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Step 2: Write the balanced chemical equations for the formation of AgNH3+ and Ag(NH3)2+ complexes:
Ag+ (aq) + NH3 (aq) ⇌ AgNH3+ (aq) [Equilibrium constant K1]
AgNH3+ (aq) + NH3 (aq) ⇌ Ag(NH3)2+ (aq) [Equilibrium constant K2]
Step 3: Use the equilibrium constants (K1 and K2) to set up the expressions for the equilibrium concentrations of Ag+, AgNH3+, and Ag(NH3)2+ in terms of the solubility of AgCl (x) and the concentration of ammonia (NH3):
[Ag+] = x
[AgNH3+] = K1 * [Ag+] * [NH3] = 2.1x10^3 * x * [NH3]
[Ag(NH3)2+] = K2 * [AgNH3+] * [NH3] = 8.2x10^3 * [AgNH3+] * [NH3]
Step 4: Set up the solubility product expression for silver chloride:
Ksp (AgCl) = [Ag+] * [Cl-] = x * x
Step 5: Use the equations from Step 3 and the solubility product expression to solve for x (solubility of AgCl):
x * x = [Ag+] * [Cl-]
x^2 = ([Ag+] * [Cl-])
x^2 = (x * x) (Since [Ag+] = [Cl-])
x = √(Ksp (AgCl))
Given that Ksp (AgCl) = 1.6 x 10^–10, we can calculate x as follows:
x = √(1.6 x 10^–10) = 1.26 x 10^–5 M
So, the solubility of silver chloride (AgCl) in 10.0 M ammonia is 1.26 x 10^–5 M.
Step 6: Calculate the concentration of NH3 in the final equilibrium mixture.
We know that the initial concentration of NH3 is 10.0 M. Let's assume its final concentration in the equilibrium mixture is y M.
Using the equation:
[AgNH3+] = K1 * [Ag+] * [NH3]
We can substitute the values:
2.1x10^3 * x * [NH3] = [AgNH3+]
2.1x10^3 * (1.26 x 10^–5) * y = y
Simplifying:
2.646 x 10^–2 * y = y
Now solve for y (concentration of NH3):
y - 2.646 x 10^–2 * y = 0
(1 - 2.646 x 10^–2) * y = 0
0.97354 * y = 0
y = 0 / 0.97354
Therefore, the concentration of NH3 in the final equilibrium mixture is 0 M.
Apologies for the earlier incorrect answer. The concentration of NH3 in the final equilibrium mixture is 0 M, not 9.0 M.