Calculate the mass of AgCl that will dissolve in 100ml of 2 M NH3 ksp AgCl = 1.7x10^-10 and Kf of Ag(NH3)2 = 1.6x10^7
AgCl(s)==> Ag^+ + Cl^- k1 = 1.7E-10
Ag^+ + 2NH3 ==> Ag(NH3)2+ k2 = 1.6E7
....AgCl(s) + 2NH3 ==> Ag(NH3)2 + Cl^-
I..............2........0..........0
C.............-2x.......x..........x
E............2-2x.......x..........x
Substitute the E line into K (which is k1k2) and solve for x = (Cl^-) = AgCl.
Then convert M to grams AgCl per L then to 100 mL.
Post your work if you get stuck.
To calculate the mass of AgCl that will dissolve in 100 mL of 2 M NH3, we first need to determine the equilibrium concentrations of Ag+ and Cl- ions in the solution.
The reaction involved in the dissolution of AgCl in NH3 can be represented as follows:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
According to the solubility product constant (Ksp) expression for AgCl,
Ksp = [Ag+][Cl-]
Since NH3 is a complexing agent and can form a complex ion with Ag+, we need to consider the formation constant (Kf) for the complex ion Ag(NH3)2+.
The reaction involved in the formation of the complex ion Ag(NH3)2+ is:
Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq)
The formation constant expression for Ag(NH3)2+ can be written as:
Kf = [Ag(NH3)2+]/[Ag+][NH3]^2
To calculate the mass of AgCl that will dissolve, we will follow these steps:
1. Calculate the concentration of Ag+ ions in the presence of 2 M NH3.
2. Use the formation constant (Kf) to find the concentration of Ag(NH3)2+ complex ions.
3. Determine the concentration of free Cl- ions using the solubility product constant (Ksp) expression.
4. Convert the concentration of Cl- ions to moles.
5. Multiply the moles of Cl- ions by the molar mass of AgCl to obtain the mass of AgCl that will dissolve.
Let's calculate the solution together.
Step 1: Calculate the concentration of Ag+ ions in the presence of 2 M NH3.
Since NH3 reacts with Ag+ to form Ag(NH3)2+, the concentration of Ag+ will decrease by a factor of 2:
[Ag+] = 2 M / 2 = 1 M
Step 2: Use the formation constant (Kf) to find the concentration of Ag(NH3)2+ complex ions.
Using the formation constant expression:
Kf = [Ag(NH3)2+]/[Ag+][NH3]^2
Rearranging the equation:
[Ag(NH3)2+] = Kf * [Ag+][NH3]^2
[Ag(NH3)2+] = (1.6x10^7) * (1 M) * (2 M)^2
[Ag(NH3)2+] = 6.4x10^7 M^3
Step 3: Determine the concentration of free Cl- ions using the solubility product constant (Ksp) expression.
Ksp = [Ag+][Cl-]
[Cl-] = Ksp / [Ag+]
[Cl-] = (1.7x10^-10) / (1 M)
[Cl-] = 1.7x10^-10 M
Step 4: Convert the concentration of Cl- ions to moles.
Moles of Cl- = [Cl-] * (100 mL / 1000) L
Moles of Cl- = (1.7x10^-10 M) * (0.1 L)
Moles of Cl- = 1.7x10^-11 moles
Step 5: Multiply the moles of Cl- ions by the molar mass of AgCl to obtain the mass of AgCl that will dissolve.
Molar mass of AgCl = 107.87 g/mol
Mass of AgCl = Moles of Cl- * Molar mass of AgCl
Mass of AgCl = (1.7x10^-11 moles) * (107.87 g/mol)
Mass of AgCl = 1.83x10^-9 grams
Therefore, the mass of AgCl that will dissolve in 100 mL of 2 M NH3 is approximately 1.83x10^-9 grams.