Based on the calculations, the molar solubility of AgCl in 0. 500 M of NH₃ is equal to 2.77 × 10⁻² M.
<u>Given the following data:</u>
- Ksp of mg(oh)₂ = 1.8 × 10⁻¹⁰
- Concentration of NH₃ = 0.500
- Kf of ag(nh₃)₂⁺ = 1.7 × 10⁷
<h3>How to determine the molar solubility.</h3>
First of al, we would write the properly balanced chemical equation for this chemical reaction:
Ksp = 1.8 × 10⁻¹⁰
Kf = 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.
Read more on solubility here: brainly.com/question/3006391