Question

What is the molar solubility of agcl (ksp = 1. 80 × 10⁻¹⁰) in 0. 500 m nh₃? (kf of ag(nh₃)₂⁺ is 1. 7 × 10⁷)

Answer 1

Based on the calculations, the molar solubility of AgCl in 0. 500 M of NH₃ is equal to 2.77 × 10⁻² M.

Given the following data:

• Ksp of mg(oh)₂ = 1.8 × 10⁻¹⁰

• Concentration of NH₃ = 0.500

• Kf of ag(nh₃)₂⁺ = 1.7 × 10⁷

How to determine the molar solubility.

First of al, we would write the properly balanced chemical equation for this chemical reaction:

                          $AgCl(s)\rightleftharpoons Ag^{+} (aq)+ Cl^{-}(aq)$       Ksp = 1.8 × 10⁻¹⁰

                $AgCl(s) + 2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^{+}(aq)$   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:

$K=\frac{ [Ag(NH_3)_2^{+}][Cl]}{[NH_3]^2}\\\\3.06 \times 10^{-3}=\frac{[x][x]}{0.50^{2 }}\\\\3.06 \times 10^{-3}=\frac{x^2}{0.25}\\\\x^2 = 7.65\times 10^{-4}\\\\x=\sqrt{7.65\times 10^{-4}}$

x = 2.77 × 10⁻² M.

Read more on solubility here: brainly.com/question/3006391