Q.x. Explain the relation between ionic product and solubility product to predict whether a precipitate will form when two solutions are mixed?

Step 1. Define Ksp. For a sparingly soluble salt $B_xA_y(s)\rightleftharpoons xB^{y+}(aq)+yA^{x-}(aq)$, $K_{sp}=[B^{y+}]^x[A^{x-}]^y$, using strictly the EQUILIBRIUM (saturated) ion concentrations.

Step 2. Define IP. The ionic product (IP) is calculated using the exact same expression, $IP=[B^{y+}]^x[A^{x-}]^y$, but evaluated with whatever ion concentrations are ACTUALLY present at a given moment -- which need not be equilibrium/saturated values at all, e.g. immediately after two solutions are mixed together.

Step 3. State the three-way comparison. (i) If $IP=K_{sp}$, the ion concentrations present happen to exactly match the saturation condition -- the solution is exactly saturated and at equilibrium, with no net further dissolution or precipitation.

(ii) If $IP>K_{sp}$, more of the ions are present than the equilibrium (saturated) state can support -- the solution is supersaturated, and the sparingly soluble salt precipitates out until IP falls back down to equal Ksp.

(iii) If $IP<K_{sp}$, fewer ions are present than the saturation limit -- the solution is unsaturated, and no precipitate forms (more of the salt could still dissolve, if any solid were present). …

Define IP as the same mathematical expression as Ksp but evaluated at arbitrary (non-equilibrium) concentrations, state the three-way IP-vs-Ksp comparison, and connec …

• Treating IP and Ksp as different formulas -- they use the IDENTICAL expression; the only difference is whether the concentrations plugged in are equilibrium/saturated values or just whatever is currently present. …