16.10
Solubility Equilibria
Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal rates. These equilibria underlie many…
Sodium chloride is considered soluble because large quantities of it will dissolve in water, but when lead chloride is added to water, only a small amount dissolves, while the rest remains insoluble.
The undissolved solid coexists with the lead and chloride ions that are in solution. Some of the solid lead chloride continues to dissolve, while some of the ions in solution recombine to form a precipitate.
When the rate of dissolution equals the rate of precipitation, a solubility equilibrium is established.
The equilibrium constant can be calculated from the equilibrium concentrations of the ions according to the dissolution reaction—where lead chloride dissociates into one lead and two chloride ions.
So, the equilibrium constant is given by the molar concentration of lead ions multiplied by the square of the molar concentration of chloride ions. Because the concentration of the solid lead chloride remains constant, it is excluded from the calculation.
This equilibrium constant is called the solubility product, denoted by Ksp. At 25 °C, the Ksp of lead chloride is 1.17 × 10−5.
The value of Ksp represents the extent to which a compound can dissolve to form a saturated aqueous solution. At a given temperature, the Ksp of a compound is constant.
The solubility of a compound in moles per liter, known as the molar solubility, is often used to express the concentration of the dissolved solid in a saturated solution. The solubility of a compound can vary depending on factors, such as the pH of the solution and if there are other ions present.
The molar solubility of a compound, x, can be calculated from its Ksp using an ICE table.
The initial concentrations of lead ions and chloride ions in the solution are zero.
At equilibrium, the molar concentration of lead ions is represented by x, while that of chloride ions is 2x.
Substituting into the equilibrium expression, the solubility product for lead chloride is equal to x times 2x2, which equals 4x3.
As the Ksp for lead chloride is 1.17 × 10−5, x is solved to be 1.43 × 10−2 molar.
For compounds that have the same dissociation stoichiometry, such as lead chloride and calcium fluoride, where 1 mole of each compound produces 3 moles of dissolved ions, the respective Ksp values can be used directly to compare their relative solubilities.
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Q1: What is solubility equilibrium and when does it form?
Solubility equilibrium forms when the rate of dissolution equals the rate of precipitation. For example, when lead chloride is added to water, some solid dissolves while dissolved ions recombine to form precipitate. When these opposing processes occur at equal rates, the system reaches equilibrium, with undissolved solid coexisting with dissolved ions in solution.
Q2: What is the solubility product constant and how is it calculated?
The solubility product constant, or Ksp, is the equilibrium constant for dissolution reactions of sparingly soluble salts. For lead chloride, Ksp equals the molar concentration of lead ions multiplied by the square of chloride ion concentration. The solid's concentration is excluded because it remains constant. At 25°C, lead chloride's Ksp is 1.17 × 10⁻⁵, representing the extent to which the compound dissolves.
Q3: How do you calculate molar solubility from Ksp?
Molar solubility is calculated using an ICE table and the Ksp expression. For lead chloride, if x represents molar solubility, then lead ion concentration is x and chloride concentration is 2x. Substituting into Ksp = x(2x)² = 4x³ = 1.17 × 10⁻⁵ gives x = 1.43 × 10⁻² molar, the compound's solubility in moles per liter.
Q4: How does the reaction quotient predict whether precipitation will occur?
When mixing solutions containing Ca²⁺ and CO₃²⁻ ions, compare the reaction quotient Q to Ksp. If Q < Ksp, no precipitation occurs because the solution is unsaturated. If Q > Ksp, the solution is supersaturated and precipitation will occur, lowering ion concentrations until Q equals Ksp and equilibrium is established.
Q5: Why is the concentration of solid excluded from the Ksp expression?
Solids have constant concentration and do not participate in equilibrium expressions. Only dissolved ions and aqueous species are included in Ksp calculations. For lead chloride, the solid's concentration remains constant regardless of how much dissolves, so only the molar concentrations of lead and chloride ions appear in the Ksp equation.
Q6: What factors can affect the solubility of a compound?
Compound solubility varies depending on solution pH and the presence of other ions. At a given temperature, Ksp remains constant, but the molar solubility can change based on these factors. Understanding how pH and ion concentration influence solubility is essential for managing natural and technological processes like water purification and tooth decay prevention.
Q7: How can you compare the solubility of compounds with the same stoichiometry?
For compounds with identical dissociation stoichiometry, such as lead chloride and calcium fluoride, where one mole produces three moles of dissolved ions, their Ksp values can be directly compared to determine relative solubilities. The compound with the larger Ksp value is more soluble under the same conditions.