Calculate the equilibrium constant K c for the following overall reaction: AgCl(s) + 2CN –(aq) Ag(CN) 2 –(aq) + Cl –(aq) For AgCl, K sp = 1.6 × 10 –10; for Ag(CN) 2 –, K f = 1.0 × 10 21.

1.6x10¹¹ = Kc

Explanation:

For the reaction:

AgCl(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + Cl⁻(aq)

Kc is defined as:

Kc = [Ag(CN)₂⁻] [Cl⁻] / [CN⁻]²

Ksp of AgCl is:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

Where Ksp is:

Ksp = [Ag⁺] [Cl⁻] = 1.6x10⁻¹⁰

In the same way, Kf of Ag(CN)₂⁻ is:

Ag⁺(aq) + 2CN⁻ ⇄ Ag(CN)₂⁻

Kf = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = 1.0x10²¹

The multiplication of Kf with Ksp gives:

[Ag⁺] [Cl⁻] *  [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = Ksp*Kf

[Ag(CN)₂⁻] [Cl⁻] / [CN⁻]² = Ksp*Kf

Obtaining the same expression of the first reaction

That means Ksp*Kf = Kc

1.6x10⁻¹⁰*1.0x10²¹ = Kc

i think the answer is terrae but not 100 percent sure. ▓▬▬ §

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2