Q.Calculate the solubility of A 2X3 in pure water, assuming that neither kind of ion reacts with water. The solubility product of A 2X3, Ksp = 1.1 × 10⁻²³.
Concept understanding — Solubility Product Constant
Solubility Product Constant: From Intuition to Precision
Imagine you drop a pinch of salt into a glass of water. The salt crystals disappear — they dissolve. But what if you keep adding salt, spoonful after spoonful? At some point, the water can't hold any more; the extra salt just sits at the bottom, undissolved. That's a saturated solution — the maximum amount of solute has dissolved at that temperature.
Now, here's the key question: even in that saturated solution, is everything static? Not at all. At the microscopic level, salt ions are constantly leaving the solid crystal and entering the solution (dissolving), while other ions in solution bump into the crystal and stick back (precipitating). At saturation, these two processes happen at exactly the same rate. The system is in dynamic equilibrium.
A saturated solution is not "full" in a static sense — it's a busy, balanced dance between dissolving and precipitating.
The Intuition: A Crowded Dance Floor
Think of a dance hall with a capacity limit. The "dancers" are ions (like Na⁺ and Cl⁻ from table salt). The dance floor is the solution. When the floor is empty, dancers easily find space — dissolution is fast. As more dancers arrive, they start bumping into each other and some leave the floor (precipitate). At the maximum capacity, the number of dancers entering equals the number leaving. That equilibrium number of dancers is what we call solubility.
But here's the twist: for many salts, the "dancers" come in different types — say, positive ions and negative ions. The equilibrium isn't just about the total number; it's about the product of their concentrations. Why product? Because the chance of a positive and a negative ion meeting to form a solid depends on both their concentrations. If you double the concentration of positive ions, the chance of a collision doubles. If you double both, it quadruples.
That product — at equilibrium, for a saturated solution — is a constant. That's the solubility product constant, Ksp.
The Precise Statement
For a sparingly soluble salt that dissociates in water as:
AmBn(s)⇌mAn+(aq)+nBm−(aq)
the solubility product constant is defined as:
Ksp=[An+]m⋅[Bm−]n
where the square brackets denote molar concentrations (mol/L) at saturation (equilibrium with the solid).
The solid AmBn does not appear in the expression. Its concentration is constant (pure solid) and is absorbed into Ksp. Never write [AmBn] in the Ksp expression.
What Ksp Tells You
- Small Ksp (e.g., 10−30): The salt is very insoluble. Only a tiny amount dissolves.
- Large Ksp (e.g., 10−2): The salt is relatively soluble.
- Ksp is temperature-dependent — always quote the temperature (usually 25°C).
Ksp is an equilibrium constant. It only applies to saturated solutions in contact with undissolved solid. If no solid is present, the solution may be unsaturated (Q<Ksp) or supersaturated (Q>Ksp), but Ksp itself doesn't change.
A Concrete Example: Silver Chloride
Silver chloride, AgCl, is a classic sparingly soluble salt. Its dissolution:
AgCl(s)⇌Ag+(aq)+Cl−(aq)
The Ksp expression:
Ksp=[Ag+][Cl−]
At 25°C, Ksp=1.8×10−10. This tiny number means that in a saturated solution, the product of the two ion concentrations is only 1.8×10−10.
If you know the solubility of AgCl is s mol/L, then [Ag+]=s and [Cl−]=s, so:
s2=1.8×10−10⇒s=1.8×10−10≈1.34×10−5 M
That's about 0.0000134 moles per litre — barely any dissolves.
The Common Mistake: Forgetting the Stoichiometry
For a salt like calcium phosphate, Ca3(PO4)2:
Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO43−(aq)
The Ksp is:
Ksp=[Ca2+]3[PO43−]2
If the solubility is s mol/L, then [Ca2+]=3s and [PO43−]=2s, so:
Ksp=(3s)3(2s)2=108s5
Students often forget the coefficients as exponents and the stoichiometric factors in the concentrations. Always write the balanced dissociation equation first, then construct Ksp.
Why This Matters
Ksp is the foundation for:
- Predicting whether a precipitate will form when solutions are mixed (compare Q to Ksp)
- Understanding the common ion effect (adding one ion shifts equilibrium, reducing solubility)
- Designing qualitative analysis schemes in chemistry labs
- Controlling water hardness and scaling in pipes
Start with the dance floor analogy, remember the equilibrium nature, and always respect the stoichiometry. That's the solubility product constant.
This topic is commonly searched as "Solubility Product Constant 11 chemistry important questions" or "Solubility Product Constant formula and examples", and it maps cleanly onto the Class 11 Chemistry portion of the NCERT/CBSE syllabus. Because solubility product constant shows up repeatedly in JEE Main, NEET and state CET Chemistry papers, mastering the underlying idea (not just the formula) is genuinely worth the extra time.
The key idea is the solubility product constant (Ksp), which relates the equilibrium concentrations of the ions in a saturated solution.
Step 1: Write the dissolution equilibrium.
For A2X3:
A2X3(s)⇌2A3+(aq)+3X2−(aq)
Step 2: Relate solubility to ion concentrations.
Let the molar solubility of A2X3 be s mol/L. Then:
[A3+]=2s,[X2−]=3s
Step 3: Write and solve the Ksp expression.
Ksp=[A3+]2[X2−]3=(2s)2(3s)3=4s2⋅27s3=108s5
Given Ksp=1.1×10−23:
108s5=1.1×10−23
s5=1081.1×10−23≈1.0185×10−25
s=(1.0185×10−25)1/5
Step 4: Compute the fifth root.
Since 10−25=10−5×5, the fifth root of 10−25 is 10−5.
1.01851/5≈1.0037 (very close to 1).
Thus:
s≈1.0×10−5 mol/L
The solubility of A2X3 in pure water is 1.0×10−5 mol/L.
The solubility of A2X3 in pure water is found by relating its dissociation stoichiometry to the Ksp expression. For A2X3(s)⇌2A3++3X2−, if solubility is s mol/L, then [A3+]=2s, [X2−]=3s, and Ksp=(2s)2(3s)3=108s5. Solving 108s5=1.1×10−23 gives s≈1.0×10−5 M.
Why the solubility product approach works
When a sparingly soluble salt like A2X3 dissolves in water, it establishes an equilibrium between the solid and its ions in solution. The solubility product constant Ksp is the equilibrium constant for this dissolution. The key insight: Ksp is not the solubility itself — it’s the product of ion concentrations at saturation, each raised to the power of its stoichiometric coefficient. To find solubility, we must connect the ion concentrations to the amount of salt that dissolved.
For A2X3, each formula unit releases 2 cations (A3+) and 3 anions (X2−). So if s moles of A2X3 dissolve per litre, the ion concentrations are directly proportional to s — but not equal to s. This stoichiometric link is the heart of the calculation.
A common mistake is to set [A3+]=s or [X2−]=s. Always check the subscripts: the ion concentrations are multiples of s, not s itself.
Step-by-step solution
1. Write the dissolution equilibrium
A2X3(s)⇌2A3+(aq)+3X2−(aq)
The solid does not appear in the Ksp expression (its activity is 1).
2. Define the variable
Let s = solubility of A2X3 in mol/L. This means s moles of the salt dissolve per litre of water.
3. Express ion concentrations in terms of s
From the stoichiometry:
- Each mole of A2X3 gives 2 moles of A3+, so [A3+]=2s
- Each mole of A2X3 gives 3 moles of X2−, so [X2−]=3s
Think of it as: the concentration of each ion equals (coefficient) × (solubility). The coefficients come from the balanced equation.
4. Write the Ksp expression
Ksp=[A3+]2[X2−]3
Substitute the expressions from step 3:
Ksp=(2s)2(3s)3
5. Simplify the algebra
(2s)2=4s2
(3s)3=27s3
Ksp=4s2×27s3=108s5
Ksp=108s5
6. Insert the given Ksp value and solve for s
108s5=1.1×10−23
s5=1081.1×10−23
Compute the division:
1081.1≈0.010185
So s5≈1.0185×10−25
Now take the fifth root. Since 10−25=(10−5)5, we expect s to be around 10−5.
s=(1.0185×10−25)1/5
s=(1.0185)1/5×10−5
Now (1.0185)1/5 is very close to 1 (since 15=1 and 1.0185 is only 1.85% above 1). A quick check: 1.00375≈1.0186, so the factor is about 1.0037.
Thus:
s≈1.0×10−5 mol/L
The fifth root of 10−25 is exactly 10−5, and the small numerical factor (1.0037) rounds to 1.0 given the single significant figure in Ksp=1.1×10−23.
The solubility of A2X3 in pure water is approximately 1.0×10−5 mol/L.
- COMEDK 2025Set 2025-M1 markMCQQ.Solubility product of the sparingly soluble salt AgBrO3 in aqueous medium is 9.3×10−10 Calculate the mass in gram of AgBrO3 present in 100 ml of its saturated solution. (Molar mass of AgBrO3 is 236 g/mol ) (A) 3.0495×10−4 (B) 4.962×10−4 (C) 6.248×10−5 (D) 7.198×10−4
›Reveal solutionSolution
s=Ksp=3.05×10−5 mol L−1; in 100 mL this is 7.20×10−4 g of AgBrO3 — option (D).
Dissolution equilibrium
AgBrO3(s)⇌Ag+(aq)+BrO3−(aq),Ksp=s2.
Molar solubility
s=9.3×10−10=3.05×10−5 mol L−1.
Mass in 100 mL (0.1 L)
n=s×0.1=3.05×10−6 mol,
mass=n×M=3.05×10−6×236=7.20×10−4 g.
(The value 3.05×10−4 in option A is the trap — that is the solubility per litre, not the mass in 100 mL.)
✓Final answermass=7.198×10−4 g — option (D).
- KCET 2024Set B-21 markMCQQ.Solubility product of CaC2O4 at a given temperature in pure water is 4×10−9 (mol L−1)2. Solubility of CaC2O4 at the same temperature is : (A) 6.3×10−5 mol L−1 (B) 2×10−5 mol L−1 (C) 2×10−4 mol L−1 (D) 6.3×10−4 mol L−1
›Reveal solutionSolution
For a sparingly soluble salt like CaC2O4 that dissociates into two ions, solubility s is the square root of Ksp. Here s=4×10−9=2×10−4.5=6.3×10−5 mol L−1, so the answer is (A).
The key idea is that solubility and solubility product are linked by the stoichiometry of dissociation. For a salt like CaC2O4, which breaks into one Ca2+ and one C2O42− ion, the relationship is especially simple: if s is the solubility in mol L−1, then at saturation [Ca2+]=s and [C2O42−]=s, so Ksp=s⋅s=s2.
This means you don't need to set up an ICE table or worry about common ions — it's a direct square root. The trap many students fall into is forgetting that Ksp is given in units of (mol L−1)2, which already tells you it's the product of two concentrations, each equal to s.
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Write the dissociation equilibrium:
CaC2O4(s)⇌Ca2+(aq)+C2O42−(aq)
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Let the solubility be s mol L−1. Then:
[Ca2+]=s, [C2O42−]=s
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The solubility product expression is:
Ksp=[Ca2+][C2O42−]=s⋅s=s2
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Substitute the given value:
s2=4×10−9
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Take the square root:
s=4×10−9=4×10−9=2×10−4.5
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Now 10−4.5=10−5×100.5=10−5×10≈10−5×3.16=3.16×10−5
So s=2×3.16×10−5=6.32×10−5 mol L−1
Watch outA common mistake is to take s=Ksp but then forget to handle the 10−4.5 correctly — some students write 2×10−4 by treating 10−4.5 as 10−4, which gives option (C). Always remember that 10−4.5=10−5×10, not 10−4.
TipIf you memorise that 10≈3.16, then 4×10−9=2×10−4.5=2×10−5×10≈6.3×10−5. This matches option (A) exactly.
✓Final answerThe solubility is 6.3×10−5 mol L−1, which corresponds to option (A).
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- COMEDK 2024Set 2024-M1 markMCQQ.A3 B4 is a sparingly soluble salt with a solubility of sg/L. If the Molar mass of A3 B4 is Mg/mol, what is the expression for its Ksp ? (A) 6912( s/M)7 (B) 8413( sM)8 (C) 5184( s/M)7 (D) 5185( sM)6
›Reveal solutionSolution
The key is to convert the given solubility in g/L to molar solubility (mol/L), then write the Ksp expression for the salt A₃B₄ in terms of that molar solubility. The result is Ksp=6912(s/M)7, which corresponds to option (A).
Concept & Intuition
For a sparingly soluble salt like A₃B₄, the dissolution equilibrium is:
A3B4(s)⇌3An+(aq)+4Bm−(aq)
The solubility product Ksp is the product of the ion concentrations at equilibrium, each raised to the power of its stoichiometric coefficient. If we know the molar solubility (mol/L), we can directly find the ion concentrations. Here, solubility is given in g/L, so we first convert to molar solubility using the molar mass M.
Step-by-step solution
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Convert solubility from g/L to mol/L
Solubility in g/L = s. Molar mass = M g/mol.
Molar solubility (mol/L) = Ms.
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Write the dissolution equilibrium
A3B4(s)⇌3An++4Bm−
If the molar solubility is x=s/M, then:
[An+]=3x,[Bm−]=4x
- Write the Ksp expression
Ksp=[An+]3[Bm−]4
Substitute the concentrations:
Ksp=(3x)3⋅(4x)4
- Simplify the expression
(3x)3=27x3,(4x)4=256x4
Multiply:
Ksp=27×256×x3+4=6912x7
- Substitute back x=s/M
Ksp=6912(Ms)7
TipA common mistake is to forget that the coefficients 3 and 4 become exponents and multipliers. The factor 6912 comes from 33×44=27×256. Always check: for a salt AₐBₓ, the numerical factor is aa⋅bb.
Watch outDo not confuse the given solubility in g/L with molar solubility. If you mistakenly treat s as molar solubility, you would get 6912s7, which is not among the options — the correct form must have s/M.
✓Final answerThe correct option is (A).
ANSWER: A
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- KCET 2023Set D-21 markMCQQ.If 'a' stands for the edge length of the cubic systems - The ratio of radii in simple cubic, body centered cubic and face centered cubic unit cells is (A) 1a:3a:2a (B) 21a:43a:221a (C) 21a:23a:22a (D) 21a:3a:21a
›Reveal solutionSolution
In each cubic lattice, find the direction along which the spheres actually touch, count how many radii lie along it, and equate to the length of that line in terms of a.
1. Simple cubic (SC)
Atoms sit only at the corners and touch along the cube edge. The edge of length a contains two half-atoms ⇒ 2 radii:
2r=a⟹rSC=2a
2. Body-centred cubic (BCC)
The corner atoms do not touch each other; the contact is corner–body-centre–corner, i.e. along the body diagonal, whose length is 3a. That diagonal contains 4 radii (r+2r+r):
4r=3a⟹rBCC=43a
3. Face-centred cubic (FCC)
Here the contact is corner–face-centre–corner, i.e. along the face diagonal, of length 2a, which contains 4 radii:
4r=2a⟹rFCC=42a=22a
4. The required ratio
rSC:rBCC:rFCC=21a:43a:221a
Numerically 0.500a:0.433a:0.354a — the radius shrinks as the packing gets denser (52% → 68% → 74%), which is the physical sanity check: a tighter packing needs smaller spheres in the same cube.
✓Final answerThe correct option is (B) 21a:43a:221a.
ANSWER: B
- KCET 2023Set D-21 markMCQQ.When FeCl3 is added to excess of hot water gives a sol ‘X’. When FeCl1 is added to NaOH(aq) solution, gives sol ‘Y’. X and Y formed in the above processes respectively are (A) Fe2O3⋅xH2O / OH− and Fe2O3⋅xH2O/Fe3+ (B) Fe2O3⋅xH2O / H+ and Fe2O3⋅xH2O/Na+ (C) Fe2O3⋅xH2O / Cl− and Fe2O3⋅xH2O/OH− (D) Fe2O3⋅xH2O / Fe3+ and Fe2O3⋅xH2O/OH−
›Reveal solutionSolution
Both routes give the same hydrated ferric oxide sol; what differs is the peptising ion adsorbed — Fe3+ from hot-water hydrolysis (positive sol) versus OH− from the alkaline medium (negative sol).
1. The dispersed phase in both cases
Both reactions produce the same colloidal particle — hydrated ferric oxide:
FeCl3+3H2OhotFe(OH)3/Fe2O3⋅xH2O+3HCl
So every option has the same dispersed phase; the discriminator is the charge-conferring (adsorbed) ion, written after the slash.
2. The rule that decides the charge — preferential adsorption
A colloidal particle preferentially adsorbs the ion common to its own lattice that is present in the medium (Hardy–Schulze / preferential-adsorption idea).
Sol X — FeCl3 added to excess HOT WATER:
The medium is rich in Fe3+ (from the ferric chloride itself). The Fe2O3⋅xH2O particles adsorb Fe3+ ions, which are common to their lattice.
⇒X=Fe2O3⋅xH2O/Fe3+— a positively charged sol
Sol Y — FeCl3 added to NaOH(aq):
Now the medium is alkaline, full of OH−, which is the ion common to Fe(OH)3. The particles adsorb OH−.
⇒Y=Fe2O3⋅xH2O/OH−— a negatively charged sol
3. Why this matters
This is the textbook illustration that the same substance can form either a positive or a negative sol depending on the medium in which it is prepared — the charge comes from what is adsorbed, not from the particle itself. (The adsorbed layer also provides the electrostatic repulsion that keeps the sol from coagulating.)
4. Matching the options
We need X with Fe3+ and Y with OH−, in that order — which is exactly option (D). Option (C) reverses the roles (Cl− is not the lattice ion here), and (A)/(B) attach ions that are not preferentially adsorbed.
✓Final answerThe correct option is (D) Fe2O3⋅xH2O/Fe3+ and Fe2O3⋅xH2O/OH−.
ANSWER: D
- COMEDK 2023Set 2023-E1 markMCQQ.What would be the volume of water required to dissolve 0.2 g of PbCl2 of molar mass 278 g/mol to prepare a saturated solution of the salt? (KSP of PbCl2=3.2×10−8) (A) 1000 ml (B) 359.7 ml (C) 278.8 ml (D) 360.4 ml
›Reveal solutionSolution
Volume of a saturated solution that holds this many moles: V = n / s = 7.194 x 10^-4 / 2 x 10^-3 = 0.3597 L = 359.7 mL
Concept: solubility from Ksp of an AB2-type salt, then volume = moles / solubility.
PbCl2 -> Pb2+ + 2Cl-
Ksp = (s)(2s)^2 = 4 s^3
4 s^3 = 3.2 x 10^-8
s^3 = 8 x 10^-9
s = 2 x 10^-3 mol/L
Moles of PbCl2 to be dissolved:
n = 0.2 / 278 = 7.194 x 10^-4 mol
Volume of a saturated solution that holds this many moles:
V = n / s = 7.194 x 10^-4 / 2 x 10^-3 = 0.3597 L = 359.7 mL
✓Final answerThe correct option is (B) — 359.7 ml
ANSWER: B
- KCET 2019Set A-11 markMCQQ.Critical Micelle concentration for a soap solution is 1.5×10−4 mol L−1. Micelle formation is possible only when the concentration of soap solution in mol L−1 is (A) 2.0×10−3 (B) 7.5×10−5 (C) 4.6×10−5 (D) 1.1×10−4
›Reveal solutionSolution
Micelles form only above the critical micelle concentration (CMC). Since the CMC is 1.5×10−4 mol L−1, the only concentration above it is 2.0×10−3 mol L−1. The correct option is (A).
The key idea is simple: micelle formation is not spontaneous at any concentration. Soap molecules (surfactants) exist as individual ions or molecules in dilute solution. As concentration increases, they eventually reach a threshold called the critical micelle concentration (CMC). Above this value, the molecules cluster into micelles — spherical aggregates with hydrophobic tails inward and hydrophilic heads outward. Below the CMC, no micelles form.
So the question reduces to: which of the given concentrations is greater than the CMC of 1.5×10−4 mol L−1?
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Compare each option to the CMC:
- (A) 2.0×10−3 — this is 20×10−4, clearly larger than 1.5×10−4.
- (B) 7.5×10−5 — this is 0.75×10−4, smaller than the CMC.
- (C) 4.6×10−5 — this is 0.46×10−4, also smaller.
- (D) 1.1×10−4 — this is 1.1×10−4, still less than 1.5×10−4.
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Only option (A) exceeds the CMC. Therefore, micelle formation is possible only at that concentration.
Watch outA common mistake is to pick the concentration closest to the CMC, like 1.1×10−4, thinking it is "near enough". But micelle formation is a threshold phenomenon — it does not begin until the concentration strictly exceeds the CMC. Below it, no micelles exist.
TipIn exam problems, the CMC is often given in scientific notation. To compare quickly, write all numbers with the same exponent. Here, convert everything to 10−4: CMC = 1.5×10−4, (A) = 20×10−4, (B) = 0.75×10−4, (C) = 0.46×10−4, (D) = 1.1×10−4. Only (A) is larger.
✓Final answerThe correct option is (A) 2.0×10−3 mol L−1.
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