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Introduction to Solubility Equilibria
The dissociation of ionic compounds is a reversible reaction. For example, here is the dissociation equation for MgCl₂:
MgCl₂ (s) ⇌ Mg²⁺ (aq) + 2Cl⁻ (aq)
The reversibility of these reactions differ in terms of solubility:
More soluble → equilibrium shifted rightwards → more ions in solution.
Less soluble → equilibrium shifted leftwards → less ions in solution.
All solutes have a solubility-product constant, abbreviated as Ksp. These constants tell us how soluble a compound is at a specific temperature by showing how much it has dissolved at equilibrium. The smaller the constant, the less soluble a compound is.
The formula for Ksp is the same as the formula for Kc. However, there is no denominator, as solids (the salt) and liquids don't count towards the equilibrium constant.
Common Ion Effect
Common Ion: An ion that exists in both a solution and the solute.
For example, if I were to dissolve NaNO3 in a solution containing aqueous NaCl, the common ion would be Na⁺.
The presence of a common ion in a solution decreases the solubility of a compound, compared to dissolving the same compound in pure water.
Recall Le Chatelier's Principle, if you were to add more of that common ion, the reaction would shift to the left, causing a decrease in solubility and potentially even creating a precipitate if oversaturated.
Example Questions
Example #1: Solid Ag₂CrO₄ is dissolved in a 0.100M solution of AgNO₃. Given that the Ksp of Ag₂CrO₄ is 9.0×10⁻¹², find the concentration of the ions in both pure water and the solution of AgNO₃.
Concentration of ions in pure water:
Write the dissociation equation for the salt: Ag₂CrO₄ (s) ⇌ 2Ag⁺ (aq) + CrO₄²⁻ (aq)
Write the equilibrium expression: K = [Ag⁺]²[CrO₄²⁻]
Find the concentration of ions:
[Ag⁺] becomes 2x, [CrO₄²⁻] becomes x.
9.0×10⁻¹² = (2x)²(x)
9.0×10⁻¹² = 4x³
x³ = 2.25×10⁻¹²
x = 1.31×10⁻⁴
s = 1.31×10⁻⁴ M
[Ag⁺] = 2x = 2.62×10⁻⁴ M
[CrO₄²⁻] = x = 1.31×10⁻⁴ M
Concentration of ions in solution of AgNO₃:
Write the equilibrium expression: K = [Ag⁺]²[CrO₄²⁻]
Find the concentration of ions:
[Ag⁺] = 0.100 M, [CrO₄²⁻] becomes x.
9.0×10⁻¹² = (0.100)²(x)
x = 9.0×10⁻¹⁰
s = 9.0×10⁻¹⁰ M
[CrO₄²⁻] = x = 9.0×10⁻¹⁰ M
[Ag⁺] = 0.100 M
Example #2: Solid CaF₂ is dissolved in a 0.025M solution of NaF. Given that the Ksp of CaF₂ is 4.0×10⁻¹¹, calculate the solubility of the salt in both pure water and the solution of NaF.
Concentration of ions in pure water:
Write the dissociation equation for the salt: CaF₂ (s) ⇌ Ca²⁺ (aq) + 2F⁻ (aq)
Write the equilibrium expression: K = [Ca²⁺][F⁻]²
Find the concentration of ions:
[Ca²⁺] becomes x, [F⁻] becomes 2x.
4.0×10⁻¹¹ = (x)(2x)²
4.0×10⁻¹¹ = 4x³
x³ = 1.0×10⁻¹¹
x = 4.64×10⁻⁴
s = 4.64×10⁻⁴ M
[Ca²⁺] = x = 4.64×10⁻⁴ M
[F⁻] = 2x = 9.28×10⁻⁴ M
Concentration of ions in solution of NaF:
Write the equilibrium expression: K = [Ca²⁺][F⁻]²
Find the concentration of ions:
[Ca²⁺] becomes x, [F⁻] = 0.025 M.
4.0×10⁻¹¹ = (x)(0.025)²
x = 4.0×10⁻¹¹/(0.025)²
x = 6.4×10⁻⁷
s = 6.4×10⁻⁷ M
[Ca²⁺] = 6.4×10⁻⁷ M
[F⁻] = 0.025 M
pH and Solubility
Example: NaOH (s) ⇌ Na⁺ (aq) + OH⁻ (aq)
a. What would happen to the solubility of NaOH when the pH increases?
The solubility would decrease, as more OH⁻ would be present, shifting the reaction in reverse.
b. What would happen to the solubility of NaOH when the pH decreases?
The solubility would increase, as H⁺ ions would get donated to the OH⁻ to form water, shifting the reaction forward.
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