Simultaneous Solubility of Sparingly Soluble Salts
In school chemistry and entrance examinations like JEE Main, NEET UG, and CUET UG, ionic equilibrium is a core scoring topic. While single-salt solubility is straightforward, problems with two sparingly soluble salts sharing a common ion require careful formulation. Because both salts supply the common ion, each suppresses the other through the common ion effect. Understanding simultaneous solubility equilibria ensures you solve these problems accurately.
The Common Ion Principle in Mixed Saturated Solutions
When a sparingly soluble salt dissolves in water, it reaches dynamic equilibrium between its solid lattice and solvated ions. If a second sparingly soluble salt sharing an ion is added to the container, the shared ion concentration is governed by both equilibria simultaneously.
Because the solution forms a single homogeneous phase, there is only one equilibrium concentration of the shared ion throughout the liquid. Both solubility product expressions must hold at that shared concentration. By Le Chatelier's principle, common ions supplied by the second salt drive the dissolution equilibrium of the first salt backward. Consequently, the equilibrium molar solubility of each salt in the mixture is lower than its solubility in pure water.
Mathematical Derivations and Governing Formulas
Consider two sparingly soluble 1:1 salts, MA and MB, sharing a common cation M+. Let the molar solubility of MA be x mol/L and that of MB be y mol/L.
The dissolution equilibria are:
MA(s) ⇔ M+(aq) + A−(aq), where [A−] = x mol/L
MB(s) ⇔ M+(aq) + B−(aq), where [B−] = y mol/L
Total common cation concentration is [M+] = (x + y) mol/L.
Ksp1 = [M+][A−] = (x + y) × x
Ksp2 = [M+][B−] = (x + y) × y
Summing expressions: Ksp1 + Ksp2 = (x + y)2
[M+]total = x + y = √(Ksp1 + Ksp2)
Solubility of MA: x = Ksp1 / [M+]total
Solubility of MB: y = Ksp2 / [M+]total
Solubility ratio: x / y = Ksp1 / Ksp2
Variables: x, y are molar solubilities in mol/L; Ksp is solubility product in mol2/L2.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 11, ISC Class 11, State Boards, JEE Main, NEET UG, CUET UG
- Topic: Ionic Equilibrium - Simultaneous Solubility of Sparingly Soluble Salts
- Difficulty: School / Foundation
Worked Numerical Examples
Question: Two sparingly soluble salts MA and MB are stirred in water at 298 K. Given Ksp(MA) = 3.60 × 10−11 and Ksp(MB) = 6.40 × 10−11, find equilibrium [M+] and individual solubilities of MA and MB.
Step 1: Define equilibrium concentrations.
Let x = solubility of MA and y = solubility of MB in mol/L.
[A−] = x, [B−] = y, and [M+] = x + y.
Step 2: Set up Ksp expressions.
Ksp(MA) = (x + y) × x = 3.60 × 10−11
Ksp(MB) = (x + y) × y = 6.40 × 10−11
Step 3: Calculate total cation concentration.
(x + y)2 = 3.60 × 10−11 + 6.40 × 10−11 = 1.00 × 10−10 mol2/L2
[M+] = x + y = √(1.00 × 10−10) = 1.00 × 10−5 mol/L.
Step 4: Compute individual solubilities.
x = (3.60 × 10−11) / (1.00 × 10−5) = 3.60 × 10−6 mol/L.
y = (6.40 × 10−11) / (1.00 × 10−5) = 6.40 × 10−6 mol/L.
Second Route Verification:
Sum check: x + y = 3.60 × 10−6 + 6.40 × 10−6 = 1.00 × 10−5 mol/L.
Ratio check: x / y = 3.60 / 6.40 = 0.5625; Ksp1 / Ksp2 = 0.5625. Verified.
Answer: [M+] = 1.00 × 10−5 mol/L, Solubility of MA = 3.60 × 10−6 mol/L, Solubility of MB = 6.40 × 10−6 mol/L
Question: A solution is saturated with AgCl and AgBr at 298 K. Given Ksp(AgCl) = 1.80 × 10−10 and Ksp(AgBr) = 5.40 × 10−13, calculate the simultaneous solubility of each salt.
Step 1: Set up equilibrium relations.
Let x = solubility of AgCl and y = solubility of AgBr in mol/L.
[Cl−] = x, [Br−] = y, and [Ag+] = x + y.
Step 2: Calculate total silver concentration.
[Ag+]2 = Ksp(AgCl) + Ksp(AgBr)
Sum = 1.80 × 10−10 + 0.0054 × 10−10 = 1.8054 × 10−10 mol2/L2
[Ag+] = √(1.8054 × 10−10) = 1.34365 × 10−5 mol/L.
Step 3: Calculate individual solubilities.
x = (1.80 × 10−10) / (1.34365 × 10−5) = 1.3396 × 10−5 mol/L ≈ 1.34 × 10−5 mol/L.
y = (5.40 × 10−13) / (1.34365 × 10−5) = 4.0189 × 10−8 mol/L ≈ 4.02 × 10−8 mol/L.
Second Route Verification:
Product check: [Ag+][Cl−] = (1.34365 × 10−5) × (1.3396 × 10−5) = 1.800 × 10−10.
Product check: [Ag+][Br−] = (1.34365 × 10−5) × (4.0189 × 10−8) = 5.400 × 10−13. Verified.
Answer: Solubility of AgCl = 1.34 × 10−5 mol/L, Solubility of AgBr = 4.02 × 10−8 mol/L
Specific Mistakes Students Make
- Calculating solubilities independently: Using s = √Ksp separately assumes pure water and ignores common ion suppression.
- Equating individual solubilities: Assuming x = y is incorrect unless both salts have identical Ksp values.
- Ignoring stoichiometry in unsymmetrical salts: For a 1:2 salt like Ag2CrO4, cation contribution is 2y rather than y.
- Omitting common ion summation: Forgetting that [M+] is the sum (x + y), not just x or y alone.
- Confusing simultaneous dissolution with selective precipitation: Simultaneous dissolution involves coexisting solids, whereas selective precipitation separates ions.
Exam Relevance and Question Patterns
| Examination | How the Topic Is Tested | Key Concepts Emphasised |
|---|---|---|
| CBSE Class 11 & State Boards | Descriptive questions on common ion effect and Ksp definitions. | Writing equilibrium expressions and calculating solubility from Ksp. |
| JEE Main | Numerical questions on simultaneous dissolution of halide pairs. | Finding total common ion concentration and individual solubilities. |
| NEET UG | Multiple-choice questions on solubility ratios and ion suppression. | Using x / y = Ksp1 / Ksp2 to find the more soluble salt. |
| CUET UG | Problems linking qualitative analysis to ionic product and precipitation. | Applying precipitation criteria and equilibrium concentrations. |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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