Answer to Question #134999 in General Chemistry for hi

Question

Calculate the concentration of NH3 required dissolving 0.01 mol of AgCl in a liter of
solution. Ksp(AgCl) =1.56x10-10 Kdiss (Ag (NH3)2+) =6.8x10-8 ?

Answer 1

C(AgCl) = 0.01 mol/1.0L = 0.01 M

AgCl(s) + 2NH₃ <=> [Ag(NH₃)₂]⁺ + Cl^-

0.01 M 2×0.01 M 0.01 M 0.01 M

AgCl <=> Ag⁺ + Cl^-

[Ag⁺] = [Cl^-]

Ag + 2NH₃ (aq) → [Ag(NH₃)₂⁺] (aq)

$k_{dis}= \frac{[NH_3]^2[Ag^+] }{ [[Ag(NH_3)_2^+]]} = 6.8×10^{-8}$

K_sp = [Ag⁺][Cl^-] = [Ag⁺]² = 1.56×10^-10

[Ag⁺] = 1.25×10^-5

$6.8×10^{-8} = \frac{[NH_3]^2(1.25×10^{-5}) }{ 0.01}$

[NH₃]² = 5.44×10^-5

[NH₃] = 7.37×10^-3 M

[NH₃]_total = 7.37×10^-3 + (2×0.01) M = 0.0273 M