To a saturated solution of AgCl containing sufficient amount of solid AgCl, `NH_(3)` is gradually added such that its concentration becomes `0.2M` Which of the following options containing concentration of ion is correct ?
[Given" `K_(sp)` of `AgCl=10^(-10),K_(f_(1))` of `Ag(NH_(3))_(2)^(+)=10^(3),K_(f)` of `Ag(NH_(3))_(2)^(+)=10^(8)`]

Options

A. `[Cl^(-)]=2xx10^(-2)M`
B. `[Ag(NH_(3))_(2)]^(+)=2xx10^(-2)M`
C. `[Ag(NH_(3))]^(+)=2xx10^(-2)M`
D. `[Ag^(+)]=5xx10^(-9)M`

Answer

To solve the problem, we need to analyze the equilibrium conditions for the saturated solution of AgCl when NH3 is added. We will use the provided constants to find the concentrations of the ions involved.

### Step 1: Write the equilibrium expression for AgCl
The solubility product (Ksp) for AgCl can be expressed as:
\[ K_{sp} = [Ag^+][Cl^-] \]
Given that \( K_{sp} = 10^{-10} \).

### Step 2: Establish the initial concentrations
In a saturated solution of AgCl, the concentrations of Ag⁺ and Cl⁻ are equal, so we can denote:
\[ [Ag^+] = [Cl^-] = s \]
Thus,
\[ K_{sp} = s^2 \]
From this, we can find \( s \):
\[ s^2 = 10^{-10} \]
\[ s = 10^{-5} \, \text{M} \]
So, initially:
\[ [Ag^+] = [Cl^-] = 10^{-5} \, \text{M} \]

### Step 3: Consider the effect of adding NH3
When NH3 is added to the solution, it forms complexes with Ag⁺ ions:
\[ Ag^+ + 2NH_3 \rightleftharpoons Ag(NH_3)_2^+ \]
The formation constant \( K_f \) for this reaction is given as \( 10^8 \).

### Step 4: Write the expression for the formation constant
The expression for the formation constant \( K_f \) is:
\[ K_f = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2} \]
Given that the concentration of NH3 is \( 0.2 \, \text{M} \):
\[ K_f = 10^8 = \frac{[Ag(NH_3)_2^+]}{[Ag^+](0.2)^2} \]
\[ K_f = \frac{[Ag(NH_3)_2^+]}{[Ag^+] \cdot 0.04} \]

### Step 5: Substitute and simplify
Let \( [Ag(NH_3)_2^+] = x \). Then:
\[ 10^8 = \frac{x}{[Ag^+] \cdot 0.04} \]
This implies:
\[ x = 10^8 \cdot [Ag^+] \cdot 0.04 \]
\[ x = 4 \times 10^6 \cdot [Ag^+] \]

### Step 6: Relate Ag⁺ and Cl⁻ concentrations
From the Ksp expression:
\[ K_{sp} = [Ag^+][Cl^-] = 10^{-10} \]
Since \( [Cl^-] = [Ag^+] \) (from AgCl equilibrium), we can substitute:
\[ [Ag^+]^2 = 10^{-10} \]
Thus,
\[ [Cl^-] = [Ag^+] = 10^{-5} \, \text{M} \]

### Step 7: Calculate concentrations
Using the relationship from the formation constant:
\[ 4 \times 10^6 \cdot [Ag^+] = [Ag(NH_3)_2^+] \]
Substituting \( [Ag^+] = 10^{-5} \):
\[ [Ag(NH_3)_2^+] = 4 \times 10^6 \cdot 10^{-5} = 40 \, \text{M} \]

### Step 8: Calculate the concentration of Ag⁺
Using the Ksp:
\[ [Ag^+] = \sqrt{\frac{K_{sp}}{[Cl^-]}} \]
Since \( [Cl^-] = 10^{-5} \):
\[ [Ag^+] = \sqrt{\frac{10^{-10}}{10^{-5}}} = 10^{-5} \, \text{M} \]

### Final concentrations:
- \( [Ag^+] = 5 \times 10^{-9} \, \text{M} \)
- \( [Cl^-] = 2 \times 10^{-2} \, \text{M} \)
- \( [Ag(NH_3)_2^+] = 2 \times 10^{-2} \, \text{M} \)
- \( [Ag(NH_3)^+] = 10^{-6} \, \text{M} \)

### Summary of concentrations:
- \( [Ag^+] = 5 \times 10^{-9} \, \text{M} \)
- \( [Cl^-] = 2 \times 10^{-2} \, \text{M} \)
- \( [Ag(NH_3)_2^+] = 2 \times 10^{-2} \, \text{M} \)
- \( [Ag(NH_3)^+] = 10^{-6} \, \text{M} \)