Q.If the solubility product of CuS is 6×10−16, calculate the maximum molarity of CuS in aqueous solution.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Solubility Rules
Solubility Rules – From Intuition to Precision
Imagine you drop a spoonful of sugar into a glass of water. Stir once, and it disappears. Now try the same with a spoonful of sand. It just sits at the bottom. Why? The sugar molecules are able to break apart and mingle with water molecules — we say sugar dissolves in water. Sand does not.
That "disappearing" act is solubility. But in chemistry, we don't just ask if something dissolves — we ask how much and under what conditions. For ionic compounds (salts), the answer is surprisingly predictable. That predictability is what we call the Solubility Rules.
The Core Idea: "Like Dissolves Like" — But Ionic Compounds Are Different
For molecular substances like sugar, the rule of thumb is "like dissolves like" — polar dissolves in polar, non-polar in non-polar. But ionic compounds are made of charged particles (cations and anions). When you drop an ionic solid into water, the water molecules (which are polar) try to pull the ions apart. Whether they succeed depends on a tug-of-war:
- The water molecules want to surround and separate the ions (hydration energy).
- The ions themselves are held together by electrostatic forces (lattice energy).
If the hydration energy wins, the salt dissolves. If the lattice energy wins, it stays solid.
You don't need to calculate these energies for exams. The Solubility Rules are a shortcut — a set of patterns discovered by observing thousands of salts.
The Precise Statement: The Solubility Rules
Here are the rules as you'll use them in exams. They are hierarchical — the first applicable rule overrides later ones.
Solubility Rules for Ionic Compounds in Water
| Rule | Statement | Examples |
|---|---|---|
| 1 | All nitrates (NO3−) are soluble. | NaNO3, AgNO3, Pb(NO3)2 |
| 2 | All acetates (CH3COO−) are soluble. | NaCH3COO, AgCH3COO |
| 3 | All chlorides (Cl−), bromides (Br−), and iodides (I−) are soluble, except with Ag+, Pb2+, and Hg22+. | Soluble: NaCl, KBr, CaI2 Insoluble: AgCl, PbI2, Hg2Cl2 |
| 4 | All sulfates (SO42−) are soluble, except with Ba2+, Pb2+, Sr2+, and Ca2+ (slightly soluble). | Soluble: Na2SO4, CuSO4 Insoluble: BaSO4, PbSO4 |
| 5 | All carbonates (CO32−), phosphates (PO43−), sulfides (S2−), and hydroxides (OH−) are insoluble, except with Group 1 metals (Li+, Na+, K+, etc.) and NH4+. | Insoluble: CaCO3, FePO4, CuS, Fe(OH)3 Soluble: Na2CO3, K3PO4, NaOH, NH4OH |
| 6 | All compounds of Group 1 metals (Li+, Na+, K+, Rb+, Cs+) and ammonium (NH4+) are soluble. | NaCl, KOH, NH4NO3 |
A common mistake: students remember "all chlorides are soluble" and forget the exceptions. AgCl is insoluble — that's why it's used in photography and qualitative analysis. Always check the exceptions first.
How to Use the Rules (Step-by-Step)
Suppose you need to predict whether PbSO4 dissolves in water.
- Identify the ions: Pb2+ and SO42−.
- Check the rules in order:
- Rule 1 (nitrates)? No. …
Why this formula?
Solubility Rules: Why They Work (Not Just What They Say)
Solubility rules are not arbitrary — they emerge from thermodynamics and electrostatic interactions between ions in water. Let's break down the why behind the key patterns.
1. The Core Idea: "Like Dissolves Like" at the Ionic Level
Water dissolves ionic compounds because it is polar. The δ+ hydrogen ends attract anions, and the δ− oxygen end attracts cations.
- Driving force: The lattice energy (energy holding the solid together) vs. the hydration energy (energy released when ions are surrounded by water).
- Net energy change:
ΔHsolution=Lattice Energy−Hydration Energy
If ΔHsolution is negative (exothermic) or small positive, the compound tends to dissolve.
2. Why Some Salts Are Always Soluble (Group 1 & NH₄⁺)
Rule: All salts of NaX+, KX+, NHX4X+ are soluble.
Why?
- These cations are large and have low charge density (charge/size ratio is small).
- Their lattice energies are relatively low because the ions are far apart in the crystal.
- Their hydration energies are high enough to overcome the lattice energy.
Key formula: For a cation like KX+, the hydration energy is roughly:
ΔHhyd∝rq2
where q is charge and r is ionic radius.
Since q=1 and r is large, ΔHhyd is moderate — but lattice energy is even smaller.
Result: The net ΔHsolution is negative → spontaneous dissolution.
3. Why Nitrates, Acetates, and Chlorates Are Always Soluble
Rule: All nitrates (NOX3X−), acetates (CHX3COOX−), and chlorates (ClOX3X−) are soluble.
Why?
- These anions are large and polyatomic — their charge is spread over many atoms.
- This delocalization of charge means they have low charge density.
- They form weak ionic bonds with cations (low lattice energy).
- Water can easily hydrate them because the negative charge is not concentrated.
Key insight: The lattice energy for NaNOX3 is much smaller than for NaCl because NOX3X− is larger and more polarizable.
4. The "Exceptions" — Why Some Salts Are Insoluble
4.1. Carbonates, Phosphates, Sulfides (Except with Group 1 & NH₄⁺)
Rule: Most carbonates (COX3X2−), phosphates (POX4X3−), and sulfides (SX2−) are insoluble.
Why?
- These anions have high charge (−2 or −3) and are small (especially SX2−).
- This gives them very high charge density.
- They form extremely strong ionic bonds with cations (very high lattice energy).
- The hydration energy, though large, is not enough to overcome the lattice energy.
Example: For CaCOX3:
Lattice energy≈−2800 kJ/mol
Hydration energy≈−2400 kJ/mol
Net ΔHsolution≈+400 kJ/mol → insoluble
4.2. Silver, Lead, Mercury Halides
Rule: AgCl, PbClX2, HgX2ClX2 are insoluble (most other chlorides are soluble).
Why?
- AgX+, PbX2+, HgX2X2+ are soft (polarizable) cations.
- They form covalent character in their bonds with halides (especially ClX−, BrX−, IX−).
- This covalent contribution increases the effective lattice energy beyond what simple ionic models predict.
- Water cannot break these partially covalent bonds.
Key formula: The polarizing power of a cation is:
ϕ=rq …
The key idea is that for a sparingly soluble salt like CuS, the molar solubility is directly related to its solubility product Ksp.
-
CuS dissociates in water as:
CuS(s)⇌Cu2+(aq)+S2−(aq)
-
If the molar solubility of CuS is s mol/L, then [Cu2+]=s and [S2−]=s.
-
The solubility product expression is:
Ksp=[Cu2+][S2−]=s×s=s2 …
The maximum molarity of CuS in water is simply the square root of its Ksp, because the salt dissociates in a 1:1 ratio. The answer is 2.45×10−8M.
The key to this problem is understanding what "maximum molarity" means in the context of a sparingly soluble salt. When you drop CuS into water, only a tiny amount dissolves. The dissolved CuS breaks apart completely into ions:
CuS(s)⇌Cu2+(aq)+S2−(aq)
The solution becomes saturated when no more solid can dissolve. At that point, the product of the ion concentrations hits a fixed ceiling — the solubility product constant, Ksp. That ceiling is what we use to find the molar solubility.
For a salt AxBy, Ksp=[Ay+]x[Bx−]y. For a 1:1 salt like CuS, Ksp=s2, where s is the molar solubility.
Now, let's walk through the calculation.
- Write the dissociation equilibrium. Every mole of CuS that dissolves gives one mole of Cu2+ and one mole of S2−. If the molar solubility is s mol/L, then:
[Cu2+]=s,[S2−]=s
- Write the Ksp expression. From the equilibrium:
Ksp=[Cu2+][S2−]=s⋅s=s2
- Plug in the given value. We know Ksp=6×10−16. So:
s2=6×10−16
- Solve for s. Take the square root of both sides:
s=6×10−16=6×10−8
Since 6≈2.449, we get: …
Method: Solubility Product (Ksp) to Molar Solubility
Concept (Why this works)
For a sparingly soluble salt, the solubility product Ksp is the equilibrium constant for its dissolution. At saturation, the product of ion concentrations (each raised to the power of its stoichiometric coefficient) equals Ksp. The maximum molarity of the salt that can stay dissolved is its molar solubility, s.
Steps
Step 1: Write the dissociation equation
CuS(s)⇌Cu2+(aq)+S2−(aq)
CuS is a 1:1 (AB-type) salt -- one mole of CuS gives one mole of Cu2+ and one mole of S2−.
Step 2: Express ion concentrations in terms of molar solubility s
If s mol/L of CuS dissolves:
[Cu2+]=s,[S2−]=s
Step 3: Write the Ksp expression
Ksp=[Cu2+][S2−]=s×s=s2
Step 4: Substitute and solve …
Here are the common mistakes students make when solving this type of solubility product problem, along with how to avoid each.
1. Forgetting the Stoichiometry of Dissociation
The Mistake:
Students often write Ksp=[Cu2+][S2−] and then set [Cu2+]=[S2−]=s, but then incorrectly write Ksp=s2 without checking the dissociation equation.
Why it happens:
They memorise the formula Ksp=s2 for all 1:1 salts, but CuS dissociates as:
CuS(s)⇌Cu2+(aq)+S2−(aq)
Here, one mole of CuS gives one mole of each ion, so s=[Cu2+]=[S2−] and Ksp=s2 is actually correct for this case. The mistake is assuming this holds for every salt (e.g., for Ag2CrO4, Ksp=4s3).
How to avoid:
Always write the balanced dissociation equation first. Then express each ion concentration in terms of s (molar solubility). Only then substitute into the Ksp expression.
2. Confusing Molar Solubility with Ksp
The Mistake:
Students think Ksp is the solubility, so they answer 6×10−16 M directly.
Why it happens:
They see “solubility product” and “maximum molarity” and assume they are the same number.
How to avoid:
Remember:
- Ksp = equilibrium constant (product of ion concentrations at saturation).
- Molar solubility (s) = concentration of the salt that dissolves (in mol/L).
For CuS:
Ksp=s2⇒s=Ksp
So the correct calculation is:
s=6×10−16=6×10−8≈2.45×10−8 M
Key result: The maximum molarity is 2.45×10−8 M, not 6×10−16 M.
3. Incorrect Square Root Calculation
The Mistake:
Taking 6×10−16 and writing 3×10−8 (forgetting to take square root of 6) or 6×10−8 (taking square root of exponent only).
Why it happens:
Rushing the arithmetic or not knowing that a×10b=a×10b/2.
How to avoid:
Break it down step by step:
6×10−16=6×10−16=6×10−8
Then approximate 6≈2.45.
Always check: the exponent should be halved (from −16 to −8).
4. Ignoring Units or Significant Figures
The Mistake:
Writing the answer as 2.45×10−8 without units, or using too many/few decimal places.
Why it happens:
Carelessness in final presentation.
How to avoid:
- Always include units: M (mol/L) or mol L−1. …
- AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQQ.Which of the following is insoluble in water? (A) CH3(CH2)10CH2OSO3Na (B) [CH3(CH2)15-N⊕(CH3)3]Br⊖ (C) C17H35COONa (D) (C17H35COO)2Ca
›Reveal solutionSolution
This tests the classic hard-water problem: sodium soaps are water-soluble, but their calcium/magnesium salts are insoluble scum.
Concept and Intuition
Soaps and detergents are amphiphilic (a long hydrophobic hydrocarbon tail plus a polar/ionic head). With a monovalent counter-ion like Na⁺, the ionic head keeps the whole molecule soluble/dispersible in water via micelle formation. But when the counter-ion is a divalent metal ion like Ca²⁺ or Mg²⁺ (as found in hard water), the resulting salt is much less soluble and precipitates out — this precipitate is the familiar bathtub/clothes 'scum' and is the reason ordinary soap doesn't lather well in hard water.
Step-by-Step Solution
- (A) CH3(CH2)10CH2OSO3Na is sodium lauryl sulfate, an anionic synthetic detergent — water-soluble.
- (B) is a cationic (quaternary ammonium) detergent with Br⁻ counter-ion — water-soluble.
- (C) C17H35COONa is sodium stearate, ordinary soap — water-soluble, forms micelles. …
- AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQQ.Assertion (A):- MgSO4 is readily soluble in water. Reason (R):- The greater hydration enthalpy of Mg2+ ions overcome its lattice enthalpy. (A) A and R both are correct and R is the correct explanation of A. (B) A and R both are correct but R is not the correct explanation of A. (C) A is correct but R is not correct (D) A is incorrect but R is correct
›Reveal solutionSolution
Both statements are true, and the reason correctly explains the assertion: Mg2+'s large hydration enthalpy beats its lattice enthalpy, driving dissolution.
Concept and Intuition
Whether an ionic solid dissolves in water is a balance of two competing energies: the lattice enthalpy (energy needed to separate the ions from the crystal) and the hydration enthalpy (energy released when the separated ions are surrounded by water molecules). Using a Born-Haber-type cycle,
ΔHsolution≈ΔHlattice+ΔHhydration
where ΔHlattice is endothermic (positive, energy required to break the lattice) and ΔHhydration is exothermic (negative, energy released on hydration). A salt is soluble when the hydration enthalpy released is large enough (in magnitude) to compensate for, or exceed, the lattice enthalpy that must be supplied.
Step-by-Step Solution
- Mg2+ has a small ionic radius and a high charge (+2), so its charge density is very high.
- Charge density directly determines the strength of ion-dipole interactions with water, so Mg2+ has an unusually large (very negative) hydration enthalpy compared to larger group-2 cations like Ca2+, Sr2+, Ba2+.
- Even though MgSO4 has a substantial lattice enthalpy, the magnitude of Mg2+'s hydration enthalpy exceeds it, so the net enthalpy of solution is favourable (exothermic or only mildly endothermic, with entropy assisting). …
- AP EAPCET 2021Set ap-2021-10-05-FN1 markMCQQ.Which of the following is arranged in increasing order of solubilities? (A) CaCO3<KHCO3<NaHCO3 (B) NaHCO3<KHCO3<CaCO3 (C) KHCO3<NaHCO3<CaCO3 (D) CaCO3<NaHCO3<KHCO3
›Reveal solutionSolution
Alkaline-earth carbonates are far less soluble than alkali-metal bicarbonates, and within alkali bicarbonates, potassium salts are generally more soluble than the corresponding sodium salt: CaCO3<NaHCO3<KHCO3.
Concept and Intuition
Solubility of s-block carbonates/bicarbonates in water is governed largely by lattice enthalpy versus hydration enthalpy. Ca2+'s high charge density gives CaCO3 a very high lattice enthalpy that hydration cannot overcome, so it is nearly insoluble. Among the singly-charged alkali cations, the larger K+ has lower lattice enthalpy with the bicarbonate ion than the smaller Na+, so KHCO3 dissolves more readily.
Step-by-Step Solution
- CaCO3: a 2+/2- ionic solid with very high lattice energy — essentially insoluble in water.
- NaHCO3: moderately soluble (~9–10 g/100 mL near room temperature). …
- AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQQ.When some ether is added to an aqueous solution of a mixture of LiCl, NaCl and AlCl3, which of these will be extracted into ether? (A) LiCl,NaCl,AlCl3 (B) LiCl,AlCl3 only (C) LiCl,NaCl only (D) NaCl,AlCl3 only
›Reveal solutionSolution
Ether extracts covalent chlorides; among LiCl, NaCl, AlCl3, only LiCl and AlCl3 show enough covalent character to be pulled into the ether layer, while NaCl stays purely ionic in water.
Concept and Intuition
Ether is a non-polar organic solvent, so it can only dissolve (extract) species with sufficient covalent character — pure ionic salts have no affinity for it and stay in the aqueous (polar) phase. Whether a chloride is "ionic" or has covalent character is governed by Fajan's rules: small, highly-charged cations polarize the anion's electron cloud strongly, giving the bond partial covalent character.
Step-by-Step Solution
- Al3+ is small and carries a high +3 charge, giving it very strong polarizing power (classic Fajan's-rule example) — AlCl3 is substantially covalent, forms a dimer Al2Cl6, and dissolves readily in organic solvents like ether/benzene.
- Li+, though only +1, is exceptionally small (smallest alkali metal cation), giving it unusually high charge density for its group — this is the basis of lithium's well-known "diagonal relationship" with magnesium. LiCl shows measurable covalent character (e.g. its solubility in organic solvents like ethanol/acetone), unlike the other alkali chlorides. …
- AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQQ.BeSO4 is soluble in water, whereas BaSO4 is insoluble. This is because ________ (A) The lattice energy of BeSO4 is more than its hydration enthalpy. (B) The hydration enthalpy of BeSO4 is more than its lattice enthalpy. (C) The ionization enthalpy of Be is comparatively higher than Ba. (D) In BaSO4, the lattice enthalpy and hydration enthalpy are equal.
›Reveal solutionSolution
BeSO4 dissolves because Be2+'s very high hydration enthalpy outweighs the lattice enthalpy; BaSO4 stays insoluble because Ba2+'s much smaller hydration enthalpy can't overcome the (still large, anion-dominated) lattice enthalpy.
Concept and Intuition
Solubility of an ionic salt in water is essentially a tug-of-war between two competing energies: the lattice enthalpy (energy that must be supplied to break apart the ionic solid) and the hydration enthalpy (energy released as water molecules solvate the separated ions). If hydration enthalpy exceeds lattice enthalpy, dissolution is energetically favourable (net exothermic/favourable process); if lattice enthalpy wins, the salt stays undissolved.
Step-by-Step Solution
- For sulfates down Group 2 (Be, Mg, Ca, Sr, Ba), the anion SO42− is large, so lattice enthalpy changes relatively slowly down the group (dominated by the big anion's size, not much affected by the smaller cation size differences).
- Hydration enthalpy, however, depends strongly on the cation's size: Be2+ is the smallest, most highly charged-density cation in the group, so it has an exceptionally large (very exothermic) hydration enthalpy. …
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