Question

A solution contains $2.0\times {10}^{-3}M{\mathrm{Ce}}^{3+}$ and $1.0\times {10}^{-2}M$ IO ${}_3^{3-}$ Will $\mathrm{Ce}{\left({\mathrm{IO}}_3\right)}_3(s)$ $[K_{\mathrm{sp}}\text{ for }\mathrm{Ce}{\left({\mathrm{IO}}_3\right)}_3\text{ is }3.2\times {10}^{-10}.]$

Short answer

The ion product (Q) for the given concentrations of $2.0\times {10}^{-3}M{\mathrm{Ce}}^{3+}$ and $1.0\times {10}^{-2}M{\mathrm{IO}}_3^{3-}$ is $2.0\times {10}^{-9}$, which is greater than the solubility product constant (Ksp) of $\mathrm{Ce}{\left({\mathrm{IO}}_3\right)}_3$, $3.2\times {10}^{-10}$. Since Q > Ksp, the solution is supersaturated and precipitation of $\mathrm{Ce}{\left({\mathrm{IO}}_3\right)}_3$ will occur.

Step-by-step solution

Write the balanced equation and expression for Ksp

First, write down the balanced equation for the dissolution of the solid in the solution: $$\mathrm{Ce}{\left({\mathrm{IO}}_3\right)}_3(s)⟷{\mathrm{Ce}}^{3+}(aq)+3{\mathrm{IO}}_3^{3-}(aq)$$ Now, write down the expression for the solubility product constant (Ksp): $$K_{\mathrm{sp}}=[{\mathrm{Ce}}^{3+}][{\mathrm{IO}}_3^{3-}{]}^3$$

Calculate the ion product (Q) for the given concentrations

We are given the concentrations of the ions: $$[{\mathrm{Ce}}^{3+}]=2.0\times {10}^{-3}M$$ $$[{\mathrm{IO}}_3^{3-}]=1.0\times {10}^{-2}M$$ Now, find the ion product (Q) for the given concentrations by substituting the values into the expression for Q: $$Q=(2.0\times {10}^{-3})(1.0\times {10}^{-2}{)}^3$$

Calculate the value of Q

Now, calculate the value of Q: $$Q=(2.0\times {10}^{-3})(1.0\times {10}^{-6})$$ $$Q=2.0\times {10}^{-9}$$

Compare Q with Ksp to determine if precipitation will occur

Now, we have: $$K_{\mathrm{sp}}=3.2\times {10}^{-10}$$ $$Q=2.0\times {10}^{-9}$$ Since the ion product (Q) is greater than Ksp, we can conclude that the solution is supersaturated, and precipitation of $\mathrm{Ce}{\left({\mathrm{IO}}_3\right)}_3$ will occur.

Key concepts

Chemical Equilibrium

In the world of chemistry, chemical equilibrium refers to the state at which the reactants and products of a reversible reaction are formed at the same rate, resulting in no net change in their concentrations over time. It's like a tug-of-war where both teams are equally strong, and the rope doesn't move. Chemical equilibrium occurs in reversible reactions, which can proceed in both forward and backward directions.

For instance, the dissolution of a salt into its constituent ions in a solution can reach a point at which the rate of the salt dissolving is equal to the rate of the ions recombining to form the salt. This balanced state can be mathematically described by the equilibrium constant $K_c$ for reactions with concentration terms or $K_p$ when dealing with gases and pressure terms.

One particular type of equilibrium constant is the solubility product constant $K_{\mathrm{sp}}$, which is critical for predicting the solubility of ionic compounds. The $K_{\mathrm{sp}}$ specifically pertains to the equilibrium that exists between a solid and its ions in a saturated solution. The expression for $K_{\mathrm{sp}}$ is derived from the stoichiometry of the dissolved species and involves the concentrations of the ionic species raised to the power corresponding to their coefficients in the balanced chemical equation.

Saturation and Supersaturation

The concepts of saturation and supersaturation pertain to how much solute a solution can hold. Saturation occurs when a solution contains the maximum amount of solute that can be dissolved at a given temperature. This is when the solution is at equilibrium with the undissolved solute, and any additional solute will not dissolve.

Suppose you've added so much sugar to your tea that it can't dissolve any more—your tea is now saturated with sugar. If you've ever made rock candy, you're familiar with supersaturation. This happens when a solution temporarily contains more dissolved solute than it would under normal equilibrium conditions—like dissolving extra sugar in hot tea and then cooling it down.

The critical factor here is that supersaturation is a non-equilibrium state; it's unstable. Given time or a trigger, such as a seed crystal or a disturbance, the excess solute will typically precipitate out. In the context of our exercise, by calculating the ion product (Q) and comparing it with the solubility product (K_sp), we can tell whether the solution is saturated, unsaturated, or supersaturated with respect to a particular ionic compound. When $Q>K_{\mathrm{sp}}$ as in our scenario, it indicates a supersaturated solution, suggesting that the likelihood of precipitation is high.

Precipitation Reactions

A precipitation reaction takes place when ions in solution combine to form an insoluble compound, resulting in the formation of a solid, also known as a precipitate. We often witness such reactions in the mixing of two salt solutions where the product of their interaction is not soluble in water.

Let's use a party analogy: imagine ions in solution as guests at a dance party. When they 'dance' freely in the solution, they're solvated by the solvent (usually water). A precipitation reaction is like a couple (ion pair) deciding to leave the dance floor (solution) to sit down (form a solid).

For any given precipitation reaction, the solubility product constant $K_{\mathrm{sp}}$ allows us to predict whether a precipitate will form when two ionic species are combined in solution. In comparing K_sp with the reaction quotient Q, which is calculated using the initial concentrations of the ions, we can determine the direction in which the reaction will proceed. If $Q<K_{\mathrm{sp}}$ the solution is unsaturated, and no precipitate will form. If $Q=K_{\mathrm{sp}}$ the solution is at the brink of forming a precipitate and is at equilibrium. If $Q>K_{\mathrm{sp}}$ as seen in our original exercise, this indicates that the solution is supersaturated and a precipitate is expected to form.