Q.Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.
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: …
The key idea is the Solubility Product Constant (Ksp): for a sparingly soluble salt AxBy, the product of ion concentrations (raised to stoichiometric coefficients) at saturation is constant at a given temperature.
General method (for a salt AxBy⇌xAy++yBx−):
Let molar solubility = s mol/L.
Then [Ay+]=xs, [Bx−]=ys.
Ksp=(xs)x(ys)y=xxyysx+y.
Solve for s, then compute individual ion molarities.
Using standard Ksp values at 298 K (from Table 6.9, NCERT):
The solubility of a sparingly soluble salt is found by relating its Ksp expression to the stoichiometric concentrations of its ions. For each salt, we set up the dissolution equilibrium, let s be the molar solubility, substitute into the Ksp formula, and solve for s. The individual ion molarities then follow from the stoichiometric coefficients. The results are tabulated below.
The key idea is that the solubility product constant Ksp is the equilibrium constant for the dissolution of a sparingly soluble salt. It is the product of the concentrations of the ions, each raised to the power of its stoichiometric coefficient in the balanced equation. For a salt AxBy that dissolves as:
AxBy(s)⇌xAy+(aq)+yBx−(aq)
the Ksp expression is:
Ksp=[Ay+]x[Bx−]y
If we let the molar solubility be s mol/L (the number of moles of salt that dissolve per litre of solution), then from the stoichiometry:
[Ay+]=xsand[Bx−]=ys
Substituting into the Ksp expression gives:
Ksp=(xs)x(ys)y=xxyysx+y
We then solve for s. The molarities of the individual ions are then xs and ys respectively.
Now, we apply this to each salt. The Ksp values at 298 K are taken from Table 6.9 (standard NCERT data). Let's work through each one.
1. Silver chromate, Ag2CrO4
Dissociation:Ag2CrO4(s)⇌2Ag+(aq)+CrO42−(aq)
Here, x=2, y=1. Let solubility = s mol/L.
Then [Ag+]=2s, [CrO42−]=s.
Ksp=[Ag+]2[CrO42−]=(2s)2(s)=4s3
From Table 6.9, Ksp(Ag2CrO4)=1.1×10−12.
4s3=1.1×10−12⟹s3=41.1×10−12=2.75×10−13
s=32.75×10−13=3275×10−15=3275×10−5
Since 3275≈6.5 (because 6.53=274.6), we get:
s≈6.5×10−5 mol/L
Ion molarities:[Ag+]=2s=1.3×10−4 M, [CrO42−]=s=6.5×10−5 M.
Watch out
A common mistake is to forget the coefficient 2 on Ag+ when squaring. The Ksp is (2s)2(s)=4s3, not s3. Always write the full expression from the balanced equation.
2. Barium chromate, BaCrO4
Dissociation:BaCrO4(s)⇌Ba2+(aq)+CrO42−(aq)
Here, x=1, y=1. Let solubility = s mol/L.
Then [Ba2+]=s, [CrO42−]=s.
Ksp=[Ba2+][CrO42−]=s⋅s=s2
From Table 6.9, Ksp(BaCrO4)=1.2×10−10.
s2=1.2×10−10⟹s=1.2×10−10=1.2×10−5
Since 1.2≈1.095, we get:
s≈1.1×10−5 mol/L
Ion molarities:[Ba2+]=s=1.1×10−5 M, [CrO42−]=s=1.1×10−5 M.
Tip
For a 1:1 salt like BaCrO4, the solubility is simply Ksp. This is the simplest case.
3. Ferric hydroxide, Fe(OH)3
Dissociation:Fe(OH)3(s)⇌Fe3+(aq)+3OH−(aq)
Here, x=1, y=3. Let solubility = s mol/L.
Then [Fe3+]=s, [OH−]=3s.
Ksp=[Fe3+][OH−]3=(s)(3s)3=s⋅27s3=27s4
From Table 6.9, Ksp(Fe(OH)3)=1.0×10−38.
27s4=1.0×10−38⟹s4=271.0×10−38≈3.70×10−40
s=43.70×10−40=43.70×10−10
Since 43.70≈1.39 (because 1.44=3.84, close enough), we get:
s≈1.39×10−10 mol/L
Ion molarities:[Fe3+]=s=1.39×10−10 M, [OH−]=3s=4.17×10−10 M.
Note
The exponent on s is x+y=1+3=4, so we take the fourth root. The very small Ksp reflects the extreme insolubility of Fe(OH)3.
4. Lead chloride, PbCl2
Dissociation:PbCl2(s)⇌Pb2+(aq)+2Cl−(aq)
Here, x=1, y=2. Let solubility = s mol/L.
Then [Pb2+]=s, [Cl−]=2s.
Ksp=[Pb2+][Cl−]2=(s)(2s)2=s⋅4s2=4s3
From Table 6.9, Ksp(PbCl2)=1.6×10−5.
4s3=1.6×10−5⟹s3=41.6×10−5=4.0×10−6
s=34.0×10−6=34.0×10−2
Since 34.0≈1.587, we get:
s≈1.59×10−2 mol/L
Ion molarities:[Pb2+]=s=1.59×10−2 M, [Cl−]=2s=3.18×10−2 M.
Watch out
PbCl2 has a relatively high Ksp compared to the others, so its solubility is in the 10−2 M range — it is not "insoluble" in the strict sense, but sparingly soluble. Always check the magnitude.