Simultaneous Solubility: Complete Guide & Solved Problems | ChemCa

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Simultaneous Solubility

A comprehensive guide to solving equilibrium problems involving multiple sparingly soluble salts. Master the Common Ion Effect and complex stoichiometry.

Ionic Equilibrium Advanced Chemistry

What is Simultaneous Solubility?

Simultaneous solubility refers to the condition where two or more electrolytes (usually sparingly soluble salts) are present in the same solution and dissolve until equilibrium is established.

The defining characteristic of these problems is usually the presence of a Common Ion. According to Le Chatelier's Principle, the presence of this common ion shifts the equilibrium backward, decreasing the solubility of both salts compared to their solubility in pure water.

"The solubility of a salt is suppressed in the presence of another salt containing a common ion."

Governing Principles

1 Constraint: All Solubility Product ($K_{sp}$) expressions must be satisfied simultaneously.
2 Summation: The concentration of the common ion is the sum of contributions from all sources.
3 Electroneutrality: Total positive charge must equal total negative charge in the solution.

Case Study 1: Simple Stoichiometry (1:1)

Problem Statement

Find the simultaneous solubility of Silver Chloride ($AgCl$) and Silver Bromide ($AgBr$) in water.
Given: $K_{sp}(AgCl) = 10^{-10}$, $K_{sp}(AgBr) = 5 \times 10^{-13}$.

Step 1: Set up variables

Let solubility of $AgCl$ be $x$ and $AgBr$ be $y$. The common ion is $Ag^+$.

$$ \begin{align} [Ag^+]_{total} &= [Ag^+]_{from AgCl} + [Ag^+]_{from AgBr} = x + y \\ [Cl^-] &= x \\ [Br^-] &= y \end{align} $$

Step 2: Formulate Equations

Apply the $K_{sp}$ equation for both salts using the total silver concentration.

$$ \begin{align} (1) \quad K_{sp}(AgCl) &= [Ag^+_{total}][Cl^-] = (x+y)(x) = 10^{-10} \\ (2) \quad K_{sp}(AgBr) &= [Ag^+_{total}][Br^-] = (x+y)(y) = 5 \times 10^{-13} \end{align} $$

Step 3: Solve

Divide equation (1) by equation (2) to find the relationship between $x$ and $y$. This eliminates the common term $(x+y)$.

$$ \frac{(x+y)x}{(x+y)y} = \frac{10^{-10}}{5 \times 10^{-13}} \implies \frac{x}{y} = \frac{1000}{5} = 200 \implies x = 200y $$

Substitute $x = 200y$ back into Equation (1):

$$ \begin{align} (200y + y)(200y) &= 10^{-10} \\ (201y)(200y) &= 10^{-10} \\ 40200y^2 &= 10^{-10} \\ y^2 &\approx 2.48 \times 10^{-15} \\ y &= \sqrt{24.8 \times 10^{-16}} \approx 5 \times 10^{-8} \text{ M} \end{align} $$

Final Answer

Solubility of AgBr (y) = 5 × 10⁻⁸ M
Solubility of AgCl (x) = 1 × 10⁻⁵ M

Case Study 2: Complex Stoichiometry

Problem Statement

Calculate simultaneous solubility of Silver Chromate ($Ag_2CrO_4$) and Silver Chloride ($AgCl$).
Given: $K_{sp}(Ag_2CrO_4) = 1.1 \times 10^{-12}$, $K_{sp}(AgCl) = 1.8 \times 10^{-10}$.

Analysis

Salt 1: Ag₂CrO₄ (Solubility x)

$$ [Ag^+] = 2x, \quad [CrO_4^{2-}] = x $$

Salt 2: AgCl (Solubility y)

$$ [Ag^+] = y, \quad [Cl^-] = y $$

Warning: The total silver concentration is the sum: $[Ag^+]_{total} = 2x + y$

Equations

$$ \begin{align} (1) \quad [Ag^+]^2[CrO_4^{2-}] &= (2x+y)^2(x) = 1.1 \times 10^{-12} \\ (2) \quad [Ag^+][Cl^-] &= (2x+y)(y) = 1.8 \times 10^{-10} \end{align} $$

Strategic Solution

Direct division is difficult here because of the square term. A better strategy is to check if one concentration is negligible. Let's calculate individual solubilities ($S$) in pure water first.

$$ S_{AgCl} = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} $$ $$ S_{Ag_2CrO_4} = \sqrt[3]{\frac{1.1 \times 10^{-12}}{4}} \approx 6.5 \times 10^{-5} $$

Since $6.5 \times 10^{-5} > 1.34 \times 10^{-5}$, the chromate contributes significantly more silver ions. However, they are close enough that we should try to solve exactly or use the division method carefully.

The Division Trick: Divide Eq(1) by the Square of Eq(2).

$$ \frac{K_{sp1}}{(K_{sp2})^2} = \frac{(2x+y)^2(x)}{[(2x+y)(y)]^2} = \frac{(2x+y)^2 x}{(2x+y)^2 y^2} = \frac{x}{y^2} $$ $$ \frac{x}{y^2} = \frac{1.1 \times 10^{-12}}{(1.8 \times 10^{-10})^2} = \frac{1.1 \times 10^{-12}}{3.24 \times 10^{-20}} \approx 3.39 \times 10^7 $$ $$ x = 3.39 \times 10^7 y^2 $$

Now substitute $x$ back into the simpler equation (Eq 2) or solve iteratively.

Summary of Formulas

Case Type	Salt Types	Key Relation (Approx)
Type AB + Type AC	AgCl + AgBr	x/y = Ksp1 / Ksp2
Type A₂B + Type AC	Ag₂CrO₄ + AgCl	x/y² = Ksp1 / (Ksp2)²
Strong Electrolyte	AgCl + AgNO₃ (C)	s = Ksp / C

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