A solution is prepared that has [NH3]=1.00M and [Cl-]=0.100M. How many grams of AgNO3 can be dissolved in 1.00L of this solution without a precipitate of AgCl(s) forming?
Ksp of AgCl = 1.8x10^-10
Kf of [(Ag(NH3)2)+] = 1.6x10^-7
To determine the maximum amount of AgNO3 that can be dissolved without a precipitate of AgCl forming, we need to compare the solubility product constant (Ksp) of AgCl with the concentrations of the ions in the solution.
First, let's write the balanced equation for the dissolution of AgCl:
AgCl ⇌ Ag+ + Cl-
From the equation, we can see that the concentrations of Ag+ and Cl- are equal. So, we need to find the concentration of Ag+ in the solution.
Step 1: Write the balanced equation for the formation of the complex ion Ag(NH3)2+:
Ag+ + 2NH3 ⇌ Ag(NH3)2+
From the equation, we can see that the concentration of Ag+ is the concentration of the complex ion Ag(NH3)2+.
Step 2: Use the formation constant (Kf) to find the concentration of Ag(NH3)2+:
Kf = [Ag(NH3)2+]/([Ag+][NH3]^2)
Given Kf = 1.6x10^-7 and [NH3] = 1.00 M, we can rearrange the equation to find [Ag(NH3)2+]:
[Ag(NH3)2+] = Kf * [Ag+][NH3]^2
[Ag(NH3)2+] = (1.6x10^-7) * (1.00)^2
[Ag(NH3)2+] = 1.6x10^-7 M
Step 3: Since [Ag(NH3)2+] = [Ag+], the concentration of Ag+ in the solution is 1.6x10^-7 M.
Step 4: Compare the concentration of Ag+ with the concentration of Cl- to determine if AgCl will precipitate. If [Ag+] < [Cl-], AgCl will not precipitate.
Given [Cl-] = 0.100 M, we see that [Ag+] < [Cl-], so AgCl will not precipitate.
Step 5: Now, we can use the concentration of Ag+ to calculate the maximum amount of AgNO3 that can be dissolved in the solution.
The molar mass of AgNO3 is 169.87 g/mol.
We have 1.6x10^-7 moles of Ag+ in 1.00 L of solution.
Therefore, the maximum amount of AgNO3 that can be dissolved is:
Mass of AgNO3 = moles of Ag+ * molar mass of AgNO3
Mass of AgNO3 = (1.6x10^-7) * (169.87)
Mass of AgNO3 = 2.718x10^-5 grams
Therefore, approximately 2.718x10^-5 grams of AgNO3 can be dissolved in 1.00L of this solution without a precipitate of AgCl(s) forming.