The reaction quotient is defined identically to the equilibrium constant but input initial concentrations into it. Let's see an example with silver iodide:

$AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)$

$$K_{ps}=[Ag^+][I^-]=8.5x10^{-17}$$

• We mixed two solutions of silver nitrate and potassium iodide to prepare a solution of silver iodide with concentrations $[I^-]=0.01\;M$ and $[Ag^+]=0.02\;M$. To find out if the resulting solution is unsaturated, saturated or supersaturated, we calculate the reaction quotient:

$Q_{ps}=[Ag^+][I^-]=0.01x0.02=2x10^{-4}$

Comparing the reaction quotient with the equilibrium constant, we observe that: $Q_{ps}>K_{ps}$. This means that the concentrations of $I^-$ and $Ag^+$ in the solution are higher than those that a saturated solution would present, producing the precipitation of silver iodide, the solution is supersaturated.

• In a second experiment we mixed silver nitrate and potassium iodide to prepare a solution with concentrations $[Ag^+]=10^{-8}\;M$ and $[I^-]=10^{-10}\; M$. How is this new dissolution?

We write the reaction quotient and compare it with the solubility product constant:

$Q_{ps}=[Ag^+][I^-]=10^{-8}x10^{-10}=10^{-18}$

Since $Q_{ps} • The third case is obtained when the reaction quotient and the solubility product constant are the same,$Q_{ps}=K_{ps}\$, the solution is saturated. Any addition of iodide or silver ions will cause precipitation of silver iodide.