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Understanding the solubility product constant, often denoted as \(K_{sp}\), is crucial when dealing with substances that dissolve sparingly, like cerium(III) hydroxide. The solubility product constant is a measure of the extent to which a compound can dissolve in a solvent to form ions. It's calculated from the concentrations of the ions in a saturated solution of a sparingly soluble salt.

For the compound \(Ce(OH)_3\), its dissolution can be expressed as an equilibrium reaction:

• \[ Ce(OH)_3 (s) \rightleftharpoons Ce^{3+} (aq) + 3OH^- (aq) \]

The solubility product, \(K_{sp}\), is then given by the expression:

• \[ K_{sp} = [Ce^{3+}][OH^-]^3 \]

The measured value of \(K_{sp}\) provides insights into how much of the compound can exist in solution as distinct ions at equilibrium. Smaller \(K_{sp}\) values typically indicate that little dissolves, indicating low solubility.

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution to form a saturated solution. To calculate molar solubility from \(K_{sp}\), especially for compounds like \(Ce(OH)_3\), it's important to set up the correct expression.

• Write the dissolution equation: where one mole of \(Ce(OH)_3\) produces one mole of \(Ce^{3+}\) and three moles of \(OH^-\).
• Set the molar solubility of \(Ce(OH)_3\) equal to \(s\), where \([Ce^{3+}] = s\) and \([OH^-] = 3s\).
• Substitute \(s\) into the \(K_{sp}\) expression: \[ K_{sp} = s(3s)^3 = 27s^4 \]

Solving \(27s^4 = 2.0 \times 10^{-20}\) gives you the molar solubility, \(s = (2.0 \times 10^{-20}/27)^{1/4} \), revealing how much \(Ce(OH)_3\) can dissolve under these conditions.

The pOH of a solution is a measure of its hydroxide ion concentration. It can be calculated using the expression:

• \[ \text{pOH} = -\log_{10}([OH^-]) \]

For the saturated solution of cerium(III) hydroxide, the concentration of hydroxide ions, \([OH^-]\), comes from the molar solubility of the compound. Since three moles of \(OH^-\) ions result from each mole of \(Ce(OH)_3\) that dissolves:

• \[ [OH^-] = 3s \]

Substitute the calculated molar solubility, \(s\), into this relationship to find \([OH^-]\), and then use the formula for pOH. In this exercise, \([OH^-] = 4.74 \times 10^{-5} M\) leads to a pOH:

• \[ \text{pOH} \approx 4.32 \]

This value helps determine the basicity of the solution.

Writing a dissolution equation helps us understand how a solid compound dissociates into ions in a solution. For sparingly soluble salts like cerium(III) hydroxide, it's crucial for calculating related quantities like \(K_{sp}\) and solubility.

Consider the dissolution of \(Ce(OH)_3\):

• \[ Ce(OH)_3 (s) \rightleftharpoons Ce^{3+} (aq) + 3OH^- (aq) \]

This balanced equation tells us:

• One mole of \(Ce(OH)_3\) dissociates to give one mole of \(Ce^{3+}\) ions.
• The same process releases three moles of \(OH^-\) ions.

Such equations show the stoichiometry of the dissolution process and become the foundation for other calculations, including those for \(K_{sp}\), molar solubility and related ion concentrations.