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The solubility product constant, known as \(K_{sp}\), is a vital concept in understanding how substances dissolve. \(K_{sp}\) quantifies the solubility of sparingly soluble ionic compounds by indicating the maximum amount of ions that can exist in a solution without precipitating. The dissolution process occurs until the solution reaches a state of equilibrium—meaning no more dissolution happens.
For any slightly soluble compound, such as cerium(III) hydroxide \(\mathrm{Ce(OH)_3}\), it is described by a chemical equation of the form: \[\text{Ce(OH)}_3 \ (s) \rightleftharpoons \text{Ce}^{3+} \ (aq) + 3\text{OH}^- \ (aq)\]This indicates that solid \(\text{Ce(OH)}_3\) dissociates into one cerium ion \(\text{Ce}^{3+}\) and three hydroxide ions \(\text{OH}^-\). Consequently, the \(K_{sp}\) expression for this dissolution is given by: \[K_{sp} = [Ce^{3+}][OH^-]^3\] The precise \(K_{sp}\) value tells us the extent of solubility and is unique to each compound and temperature. For cerium(III) hydroxide, \(K_{sp}\) is \(2.0 \times 10^{-20}\), showing it dissolves only slightly.

Cerium(III) hydroxide \(\text{Ce(OH)}_3\) is an example of a sparingly soluble compound, meaning it dissolves minimally in water. Understanding this compound involves looking at how its ions form when it dissolves.
When \(\text{Ce(OH)}_3\) dissolves, the reaction can be written as: \[\text{Ce(OH)}_3 \ (s) \rightleftharpoons \text{Ce}^{3+} \ (aq) + 3\text{OH}^- \ (aq)\]Here, the solid compound dissociates into free ions: one \(\text{Ce}^{3+}\) ion and three \(\text{OH}^-\) ions per formula unit. The concentration of these ions in solution is determined by the molar solubility \(S\), which reflects how much of the compound actually dissolves. As shown in the solution steps, when \(S\) is the molar solubility:

• \([Ce^{3+}] = S\)
• \([OH^-] = 3S\)

This relationship is crucial for calculating quantities like [] in a saturated solution and for understanding many properties of the solution.

Determining the \(pOH\) of a solution is essential in understanding its alkalinity. It is calculated using the concentration of hydroxide ions \([\text{OH}^-]\) in the solution, following the equation: \[\text{pOH} = -\log [\text{OH}^-]\] In this case, by substituting \([\text{OH}^-]\) which was derived as \(3.87 \times 10^{-5}\) M, the pOH can be calculated:\[\text{pOH} = -\log (3.87 \times 10^{-5}) \approx 4.41\] Steps to perform this calculation:

• Take the concentration of \(\text{OH}^-\) ions in the solution.
• Apply the logarithmic formula to find \(pOH\).

Since \(pH\) and \(pOH\) are related by \(pH + pOH = 14\), knowing \(pOH\) helps you understand the acidic or basic nature of the solution. In the case of cerium(III) hydroxide, the calculated \(pOH\) of 4.41 indicates a basic solution due to the presence of \(OH^-\) ions in water.