If the K_spof AgCl is equal to 3.89×10^-10 and the K_f of $\mathrm { AgCl } _ { 2 } ^ { - }$ is 2.8×10⁵, what is the equilibrium constant for the reaction?

Question

Quizplus · Accepted Answer

The equilibrium constant for the reaction is equal to the K_f of $\mathrm { AgCl } _ { 2 } ^ { - }$ divided by the K_sp of AgCl. Therefore, the equilibrium constant is equal to 2.8×10⁵ / 3.89×10^-10 = 1.1×10^-4.

If the Ksp of AgCl Is Equal to 3 AgCl2−\mathrm { AgCl } _ { 2 } ^ { - }AgCl2−​

If the K_spof AgCl Is Equal to 3 $\mathrm { AgCl } _ { 2 } ^ { - }$