What is the molar solubility of AgCl(s) in 0.1 M `NH_(3)(aq)?K_(sp)(AgCl)=1.8xx10^(-10), K_(f)[Ag(NH_(3))_(2)]^(+)=1.6xx10^(7).`

Explanation

To find the molar solubility of AgCl in a 0.1 M NH3 solution, we will follow these steps:

### Step 1: Write the dissociation equation for AgCl
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Let the molar solubility of AgCl be \( S \) mol/L. At equilibrium, the concentration of Ag⁺ will be \( S \) and the concentration of Cl⁻ will also be \( S \).

### Step 2: Write the expression for Ksp
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