Animated Solution for Chemistry - Ionic Equilibrium: The $K_{sp}$ for the following dissociation is $1.6\times 10^{-5}$ $PbC{l}_2(s)\rightleftharpoons P{b}^{2+}(aq)+2C{l}^{-}(aq)$ Which of the following choices is correct for a mixture of 300 mL 0.134 M $Pb(N{O}_3{)}_2$ and 100 mL 0.4 M NaCl ?

Imagine you are in a chemistry lab, holding two beakers. In your left hand, you have $300\text{ mL}$ of a $0.134\text{ M}$ lead nitrate ($Pb(N{O}_3{)}_2$) solution. In your right hand, you have $100\text{ mL}$ of a $0.4\text{ M}$ sodium chloride ($NaCl$) solution.

You are about to pour them both into a larger, empty beaker. The burning question is: Will a solid precipitate of lead chloride ($PbC{l}_2$) form when these two liquids meet?

To answer this, we cannot just guess. We must rely on the principles of ionic equilibrium, specifically the battle between the Ionic Product ($Q$) and the Solubility Product Constant ($K_{sp}$).

The most common trap students fall into is using the initial concentrations directly. But think about it physically: when you mix $300\text{ mL}$ of one liquid with $100\text{ mL}$ of another, the total volume becomes $400\text{ mL}$.

Because the volume has increased, the ions are now swimming in a larger pool. They have been diluted! We must calculate their new concentrations in this mixture before any reaction happens.

For the lead ions ($P{b}^{2+}$), the new concentration is the initial moles divided by the new total volume:

Similarly, for the chloride ions ($C{l}^{-}$):

Calculating Q

Now that we have the true concentrations of the ions in the mixture, we can calculate the Ionic Product, $Q$. The expression for $Q$ takes the exact same form as $K_{sp}$, based on the balanced dissociation equation:

Therefore, the Ionic Product is:

Notice the square on the chloride concentration? That comes directly from the stoichiometric coefficient '2' in the balanced equation. Let's substitute our diluted concentrations:

We have our $Q$ value: $10^{-3}$. The problem gives us the $K_{sp}$ value: $1.6\times 10^{-5}$.

Now, we compare them. Since $10^{-3}$ is significantly larger than $1.6\times 10^{-5}$, we can confidently state that $Q>K_{sp}$.

What does this mean physically? It means the solution is currently holding more dissolved ions than it can stably maintain at equilibrium. It is supersaturated. To relieve this stress and return to equilibrium, the excess ions will crash out of the solution, forming a beautiful white precipitate of solid $PbC{l}_2$.

Thus, the correct choice is indeed (b) $Q>K_{sp}$.