Based on the Ksp calculations, the molar solubility of AgCl in 0.500 M of NH₃ is equal to 2.77 × 10⁻² M.
<h3>How to determine the molar solubility.</h3>First of al, we would write the properly balanced chemical equation for this chemical reaction:
AgCl(s) ⇆ Ag⁺(aq)+ Cl⁻(aq) Ksp = 1.8 × 10⁻¹⁰
AgCl(s) + 2NH₃(aq) ⇆ Ag(NH₃)₂⁺(aq) Kf = 1. 7 × 10⁷
<u>Given the following data:</u>
- Ksp of AgCl = 1.8 × 10⁻¹⁰
- Concentration of NH₃ = 0.500
- Kf of ag(nh₃)₂⁺ = 1.7 × 10⁷
K = Ksp × Kf
K = 1. 80 × 10⁻¹⁰ × 1.7 × 10⁷
K = 3.06 × 10⁻³
Mathematically, the Ksp for the above chemical reaction is given by:
x = 2.77 × 10⁻² M.
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