Question

Consider a beaker containing a saturated solution of \(\mathrm{Pbl}_{2}\) in equilibrium with undissolved \(\mathrm{Pbl}_{2}(s)\). Now solid KI is added to this solution. (a) Will the amount of solid \(\mathrm{Pbl}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Pb}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(1^{-}\)ions in solution increase or decrease?

Short answer

In summary, when solid KI is added to a saturated solution of \(\mathrm{PbI}_2\) in equilibrium with undissolved \(\mathrm{PbI}_{2}(s)\): a) The amount of solid \(\mathrm{PbI}_2\) at the bottom of the beaker will increase. b) The concentration of \(\mathrm{Pb}^{2+}\) ions in the solution will decrease. c) The concentration of \(\mathrm{I}^{-}\) ions in the solution will increase.

Step-by-step solution

Write the balanced equation

The balanced equation for the dissolution of \(\mathrm{PbI}_{2}\) is: \[ \mathrm{PbI}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{I}^-(aq) \] This reaction is in equilibrium in the beaker.

Apply Le Chatelier's Principle

According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change, it will shift its equilibrium position to counteract the change. In this exercise, KI is added to the beaker, which increases the concentration of \(\mathrm{I}^{-}\) ions in the solution. To counteract this change, the equilibrium will shift to the left, to consume the added \(\mathrm{I}^{-}\) ions.

Identify the effect on the amount of solid PbI2

As the equilibrium shifts to the left, more \(\mathrm{PbI}_2\) will be formed (to consume the added \(\mathrm{I}^{-}\) ions). Therefore, the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker will increase.

Identify the effect on the concentration of Pb2+

Since the equilibrium shifts towards the left and \(\mathrm{PbI}_2\) is formed, the concentration of \(\mathrm{Pb}^{2+}\) ions in the solution will decrease as they combine with \(\mathrm{I}^{-}\) ions to form more solid \(\mathrm{PbI}_{2}\).

Identify the effect on the concentration of I-

As the concentration of \(\mathrm{I}^{-}\) ions is initially increased by adding KI, the equilibrium shifts to the left to consume the added \(\mathrm{I}^{-}\) ions; however, as KI keeps dissolving, the concentration of \(\mathrm{I}^{-}\) ions will continue to increase as it doesn't fully combine with the \(\mathrm{Pb}^{2+}\) ions. So, overall, the concentration of \(\mathrm{I}^{-}\) ions in the solution will increase. In conclusion: a) The amount of solid \(\mathrm{PbI}_2\) at the bottom of the beaker will increase. b) The concentration of \(\mathrm{Pb}^{2+}\) ions in the solution will decrease. c) The concentration of \(\mathrm{I}^{-}\) ions in the solution will increase.

Key concepts

Chemical Equilibrium

Chemical equilibrium represents a state of balance in a reversible chemical reaction, where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, but this does not imply that the reactions have ceased. Rather, the processes continue to occur simultaneously at an equal pace.

Different factors can disturb the equilibrium of a chemical system. These include changes in concentration, temperature, and pressure. Le Chatelier's Principle explains how systems at equilibrium react to these disturbances. For example, increasing the concentration of one substance typically causes the equilibrium to shift away from the added substance, thus consuming it and attempting to restore equilibrium. Understanding this dynamic is essential for predicting the outcomes of changes in a given chemical system.

Solubility Product

The solubility product (\(K_{sp}\) is a constant that provides crucial information about the solubility of sparingly soluble salts. It is defined for a solid that dissolves to form its constituent ions in a saturated solution. The solubility product is the product of the concentrations of the ions, each raised to the power of its coefficient in the balanced chemical equation for dissolution.

The value of the solubility product can indicate whether a precipitate will form when solutions of two soluble salts are mixed. A higher concentration of one of the ions can lead to exceeding the product's solubility, resulting in precipitation. The calculation of the solubility product is fundamental when dealing with saturated solutions, and it helps in predicting the extent to which a solid can dissolve.

Common Ion Effect

The common ion effect refers to the decrease in solubility of an ionic compound when a common ion is added to the solution. It is a direct application of Le Chatelier's Principle. When an ionic compound is dissolved in water, it separates into its constituent ions. If one of these ions is added externally, the system will respond by reducing the solubility of the ionic compound to re-establish equilibrium.

For instance, adding a common ion to a solution of lead iodide (\(PbI_2\) affects its solubility. This is a result of the increased concentration of the common ion pushing the equilibrium towards the solid, which leads to more of the solid forming. Understanding the common ion effect is crucial for controlling the solubility of compounds in various chemical processes, including precipitation reactions and analytical chemistry.