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:
![K=\frac{ [Ag(NH_3)_2^{+}][Cl]}{[NH_3]^2}\\\\3.06 \times 10^{-3}=\frac{[x][x]}{0.50^{2 }}\\\\3.06 \times 10^{-3}=\frac{x^2}{0.25}\\\\x^2 = 7.65\times 10^{-4}\\\\x=\sqrt{7.65\times 10^{-4}}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%20%5BAg%28NH_3%29_2%5E%7B%2B%7D%5D%5BCl%5D%7D%7B%5BNH_3%5D%5E2%7D%5C%5C%5C%5C3.06%20%5Ctimes%2010%5E%7B-3%7D%3D%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.50%5E%7B2%20%7D%7D%5C%5C%5C%5C3.06%20%5Ctimes%2010%5E%7B-3%7D%3D%5Cfrac%7Bx%5E2%7D%7B0.25%7D%5C%5C%5C%5Cx%5E2%20%3D%207.65%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5Cx%3D%5Csqrt%7B7.65%5Ctimes%2010%5E%7B-4%7D%7D)
x = 2.77 × 10⁻² M.
Read more on molar solubility here: brainly.com/question/3006391
Answer:
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