Based on the Ksp calculations, the molar solubility of AgCl in 0.500 M of NH₃ is equal to 2.77 × 10⁻² M.
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
Given the following data:
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:
[tex]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}}[/tex]
x = 2.77 × 10⁻² M.
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