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.
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.
Note
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).
Watch out
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).
Important
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
Watch out
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
✓Final answer
The solubility of A2X3 in pure water is 1.0×10−5mol/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.
Watch out
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
Tip
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
Note
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.
✓Final answer
The solubility of A2X3 in pure water is approximately 1.0×10−5 mol/L.