What mass of Agl will dissolve in 1.0 L of 1.0 M `NH_(3)`? Neglect change in conc. Of `NH_(3)`.
[Given: `K_(sp)(AgI)=1.5xx10^(-16)),` `K_(f)[Ag(NH_(3))_(2)^(+)]=1.6xx10^(7)],` (At. Mass Ag=108,1=127)

Options

A. `4.9xx10^(-5)` g
B. `0.0056` g
C. `0.035` g
D. `0.011` g

Answer

To solve the problem of how much AgI will dissolve in 1.0 L of 1.0 M NH₃, we will follow these steps:

### Step 1: Write the Dissolution Reaction
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

### Step 2: Write the Formation Reaction
Ag⁺(aq) + 2 NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq)

### Step 3: Define the Solubility Product (Ksp) and Formation Constant (Kf)
Given:
- Ksp(AgI) = 1.5 × 10⁻¹⁶
- Kf([Ag(NH₃)₂]⁺) = 1.6 × 10⁷

### Step 4: Set Up the Equilibrium Expressions
Let the solubility of AgI be 's' mol/L. The equilibrium concentrations will be:
- [Ag⁺] = s
- [I⁻] = s
- [Ag(NH₃)₂]⁺ = 2s (since each Ag⁺ reacts with 2 NH₃)

The Ksp expression is:
Ksp = [Ag⁺][I⁻] = s²

The Kf expression is:
Kf = [Ag(NH₃)₂⁺] / [Ag⁺][NH₃]² = (2s) / (s)(1.0)² = 2s

### Step 5: Combine Ksp and Kf
The overall equilibrium constant (K) for the dissolution of AgI in the presence of NH₃ can be expressed as:
K = Ksp × Kf = (1.5 × 10⁻¹⁶)(1.6 × 10⁷)

### Step 6: Calculate K
K = 1.5 × 10⁻¹⁶ × 1.6 × 10⁷ = 2.4 × 10⁻⁹

### Step 7: Set Up the Equation
From the K expression:
K = (2s)² / (1 - 2s)
Assuming that 2s is much less than 1, we can simplify this to:
K ≈ (2s)²

### Step 8: Solve for 's'
2s = √(K)
2s = √(2.4 × 10⁻⁹)
s = (1/2)√(2.4 × 10⁻⁹)

Calculating s:
s ≈ 0.5 × 1.55 × 10⁻⁴ = 7.75 × 10⁻⁵ mol/L

### Step 9: Calculate the Moles of AgI
Since we have 1.0 L of solution, the moles of AgI that dissolve is:
Moles of AgI = s × volume = (7.75 × 10⁻⁵ mol/L) × (1 L) = 7.75 × 10⁻⁵ mol

### Step 10: Calculate the Mass of AgI
Using the molar mass of AgI (Ag = 108, I = 127):
Molar mass of AgI = 108 + 127 = 235 g/mol

Mass of AgI = moles × molar mass = (7.75 × 10⁻⁵ mol) × (235 g/mol) = 0.0182 g

### Final Answer
The mass of AgI that will dissolve in 1.0 L of 1.0 M NH₃ is approximately **0.0182 g**.

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