Answered

Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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)

Sagot :

Lanuel

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:

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

                [tex]AgCl(s) + 2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^{+}(aq)[/tex]   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:

[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.

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

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.