Question

Write the \(K_{\mathrm{sp}}\) expressions for each of the following com- pounds: (a) \(\mathrm{AgI},(\mathbf{b}) \mathrm{CuI},(\mathbf{c}) \mathrm{AuCl}_{3},(\mathbf{d}) \mathrm{Hg}_{2} \mathrm{I}_{2},(\mathrm{e}) \mathrm{PbI}_{2}\).

Short answer

Ksp expressions: (a) AgI: \( K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{I}^{-}] \), (b) CuI: \( K_{\mathrm{sp}} = [\mathrm{Cu}^{+}][\mathrm{I}^{-}] \), (c) AuCl3: \( K_{\mathrm{sp}} = [\mathrm{Au}^{3+}][\mathrm{Cl}^{-}]^3 \), (d) Hg2I2: \( K_{\mathrm{sp}} = [\mathrm{Hg}_{2}^{2+}][\mathrm{I}^{-}]^2 \), (e) PbI2: \( K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{I}^{-}]^2 \)

Step-by-step solution

Identify the Dissociation Products for each Compound

The first step in writing the solubility product constant expression, or Ksp, is to identify how each ionic compound dissociates in water. In each case, you'll end up with cations and anions.

Write the Balanced Dissociation Equations

Write the balanced chemical equation for the dissociation of each compound into its ions in water. For example, \(\mathrm{AgI}\) dissociates into \(\mathrm{Ag}^{+} \) and \(\mathrm{I}^{-} \), \(\mathrm{CuI}\) dissociates into \(\mathrm{Cu}^{+} \) and \(\mathrm{I}^{-} \), and so forth.

Write the Ksp Expression

For each balanced equation, write the Ksp expression by taking the product of the concentrations of the ions, each raised to the power of its coefficient in the balanced equation.

Expression for \(\mathrm{AgI}\)

The dissociation of \(\mathrm{AgI}\) is given by: \( \mathrm{AgI} \rightarrow \mathrm{Ag}^{+} + \mathrm{I}^{-} \). The Ksp expression for \(\mathrm{AgI}\) is: \( K_{\mathrm{sp}} = [\mathrm{Ag}^{+}][\mathrm{I}^{-}] \)

Expression for \(\mathrm{CuI}\)

The dissociation of \(\mathrm{CuI}\) is given by: \( \mathrm{CuI} \rightarrow \mathrm{Cu}^{+} + \mathrm{I}^{-} \). The Ksp expression for \(\mathrm{CuI}\) is: \( K_{\mathrm{sp}} = [\mathrm{Cu}^{+}][\mathrm{I}^{-}] \)

Expression for \(\mathrm{AuCl}_{3}\)

The dissociation of \(\mathrm{AuCl}_{3}\) is given by: \( \mathrm{AuCl}_{3} \rightarrow \mathrm{Au}^{3+} + 3\mathrm{Cl}^{-} \). The Ksp expression for \(\mathrm{AuCl}_{3}\) is: \( K_{\mathrm{sp}} = [\mathrm{Au}^{3+}][\mathrm{Cl}^{-}]^{3} \)

Expression for \(\mathrm{Hg}_{2}\mathrm{I}_{2}\)

The dissociation of \(\mathrm{Hg}_{2}\mathrm{I}_{2}\) is given by: \( \mathrm{Hg}_{2}\mathrm{I}_{2} \rightarrow \mathrm{Hg}_{2}^{2+} + 2\mathrm{I}^{-} \). The Ksp expression for \(\mathrm{Hg}_{2}\mathrm{I}_{2}\) is: \( K_{\mathrm{sp}} = [\mathrm{Hg}_{2}^{2+}][\mathrm{I}^{-}]^{2} \)

Expression for \(\mathrm{PbI}_{2}\)

The dissociation of \(\mathrm{PbI}_{2}\) is given by: \( \mathrm{PbI}_{2} \rightarrow \mathrm{Pb}^{2+} + 2\mathrm{I}^{-} \). The Ksp expression for \(\mathrm{PbI}_{2}\) is: \( K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{I}^{-}]^{2} \)

Key concepts

Chemical Equilibrium

In our journey to understand how different substances behave in solutions, mastering the concept of chemical equilibrium is key. It's basically the state where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of the reactants and products. Think of it as a busy street where the number of cars entering equals the number of cars leaving, keeping the traffic constant.

In the case of ionic compounds dissolving in water, equilibrium is reached when the solubility product constant, or Ksp, comes into play. Ksp is a special type of equilibrium constant that tells us the extent to which a compound will dissolve. Higher Ksp values indicate a more soluble compound. For example, if Silver iodide (AgI) has a low Ksp, it means that it won't dissociate much in water, whereas something with a higher Ksp would be more solubility-friendly.

Ionic Compound Dissociation

When ionic compounds take the dive into water, they start to dissociate, which is just a fancy way of saying that they break apart into ions. These ions, cations (positive charge) and anions (negative charge), are the life of the party in aqueous solutions. Understanding dissociation is like learning how dancers split off into pairs at a ball.

The process is guided by the nature of the ionic compound and the properties of water. For instance, when CuI (Copper(I) iodide) is placed in water, it dissociates into Cu+ and I- ions. But not all of it will dissociate—only until equilibrium is reached. The way these ions relate in concentration forms the basis for the Ksp expression we write, which is critical for predicting solubility and analyzing reactions in a solution.

Balancing Chemical Equations

Balancing chemical equations is not just an exercise in counting—it's an indispensable skill for understanding chemical reactions. Imagine having a see-saw where you need to have an equal weight on both sides to maintain balance. Similarly, in chemical equations, we must have the same number of each type of atom on both sides of the reaction.

Why is this balance crucial? Because it respects the law of conservation of mass: matter is neither created nor destroyed. When you see a balanced equation, such as the dissociation of Hg2I2 into Hg22+ and 2I- ions, it follows this rule, ensuring that the atoms involved in reactants are accounted for in the products. And for Ksp, we use these balanced equations to write expressions that predict how much of an ionic compound will dissolve in water. It's this balance that allows us to tackle problems of solubility and reactivity with confidence.