Ksp and Solubility — Converting Between Them Correctly
Two numbers describe how little a sparingly soluble salt dissolves: the solubility product Ksp, and the molar solubility s. Converting between them is a standard exam question, and it goes wrong for one reason — students assume Ksp = s² for every salt. That is true only for salts of the AB type. Get the stoichiometry into the expression and the rest is arithmetic.
The definition
Ksp = [Ay+]x × [Bx−]y
The solid does not appear — its activity is 1. Ksp is just an equilibrium constant for a dissolution that has reached saturation.
Building the s-expression
If s moles per litre dissolve, then x·s moles of Ay+ and y·s moles of Bx− appear. Substituting gives one formula per salt type. Learn the pattern, not the four results:
| Type | Example | Ions produced | Ksp in terms of s |
|---|---|---|---|
| AB | AgCl, BaSO₄ | s, s | s² |
| AB₂ or A₂B | PbCl₂, Ag₂CrO₄, Ca(OH)₂ | s, 2s | (s)(2s)² = 4s³ |
| AB₃ or A₃B | Fe(OH)₃ | s, 3s | (s)(3s)³ = 27s⁴ |
| A₂B₃ | Ca₃(PO₄)₂ type | 2s, 3s | (2s)²(3s)³ = 108s⁵ |
Two things happen when you substitute: the coefficient becomes a multiplier (2s, 3s) and it also becomes the power. Doing only one of the two is the most common error in this topic.
Worked example 1 — AgCl, an AB salt
Ksp(AgCl) = 1.8 × 10⁻¹⁰ at 298 K. Find the molar solubility and the solubility in g/L.
AgCl ⇌ Ag⁺ + Cl⁻, so [Ag⁺] = [Cl⁻] = s and Ksp = s²
s = √(1.8 × 10⁻¹⁰) = √1.8 × 10⁻⁵ = 1.342 × 10⁻⁵ M
M(AgCl) = 107.868 + 35.45 = 143.32 g mol⁻¹
Solubility = 1.342 × 10⁻⁵ × 143.32 = 1.92 × 10⁻³ g/L
s = 1.34 × 10⁻⁵ M, i.e. about 1.9 mg per litre.
Note the square-root trick: split 1.8 × 10⁻¹⁰ into 1.8 × (10⁻⁵)² so the exponent is even and comes out cleanly.
Worked example 2 — PbCl₂, an AB₂ salt
Ksp(PbCl₂) = 1.7 × 10⁻⁵. Find s.
PbCl₂ ⇌ Pb²⁺ + 2Cl⁻, so [Pb²⁺] = s and [Cl⁻] = 2s
Ksp = (s)(2s)² = 4s³
s³ = 1.7 × 10⁻⁵ ÷ 4 = 4.25 × 10⁻⁶
s = (4.25 × 10⁻⁶)1/3 = (4.25)1/3 × 10⁻² = 1.62 × 10⁻² M
s = 1.62 × 10⁻² M (check: 4 × (1.62 × 10⁻²)³ = 4 × 4.25 × 10⁻⁶ = 1.7 × 10⁻⁵ ✓)
Always write 10⁻⁶ as (10⁻²)³ before taking the cube root, so the exponent divides by 3 exactly.
Worked example 3 — Ag₂CrO₄, and why Ksp does not rank solubility
Ksp(Ag₂CrO₄) = 1.1 × 10⁻¹². Find s and compare with AgCl.
Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻, so [Ag⁺] = 2s and [CrO₄²⁻] = s
Ksp = (2s)²(s) = 4s³
s³ = 1.1 × 10⁻¹² ÷ 4 = 2.75 × 10⁻¹³ = 275 × 10⁻¹⁵
s = (275)1/3 × 10⁻⁵ = 6.50 × 10⁻⁵ M
s(Ag₂CrO₄) = 6.50 × 10⁻⁵ M, which is about 4.8 times larger than s(AgCl) = 1.34 × 10⁻⁵ M — even though its Ksp is smaller by a factor of about 160.
The lesson: you may compare Ksp values directly only for salts of the same type. Across types, convert to s first.
Worked example 4 — the common ion effect
What is the solubility of AgCl in 0.010 M NaCl?
NaCl is fully dissociated, so [Cl⁻] ≈ 0.010 M from the salt, plus a negligible s from the AgCl itself. [Ag⁺] = s.
Ksp = [Ag⁺][Cl⁻] = s × 0.010 = 1.8 × 10⁻¹⁰
s = 1.8 × 10⁻¹⁰ ÷ 0.010 = 1.8 × 10⁻⁸ M
Compared with 1.34 × 10⁻⁵ M in pure water, the solubility fell by a factor of about 745. The approximation "0.010 + s ≈ 0.010" is safe here because s came out four orders of magnitude smaller — always check that afterwards.
Going the other way — s to Ksp
The same expression, used in reverse. If CaF₂ has a molar solubility of 2.1 × 10⁻⁴ M:
Will a precipitate form? Compare Q with Ksp
Write the ionic product Q with the actual concentrations after mixing: Q > Ksp means precipitation, Q = Ksp means exactly saturated, and Q < Ksp means the solution is unsaturated and nothing forms.
Mixing 100 mL of 0.010 M AgNO₃ with 100 mL of 0.010 M NaCl doubles the total volume, so each ion is halved to 0.0050 M. Q = 0.0050 × 0.0050 = 2.5 × 10⁻⁵, which is vastly larger than 1.8 × 10⁻¹⁰, so AgCl precipitates. Forgetting to halve the concentrations on mixing is the classic trap in this question.
Common mistakes
- Using s² for everything. Only AB salts are s². PbCl₂, Ca(OH)₂ and Ag₂CrO₄ are all 4s³.
- Applying the coefficient once instead of twice. [Cl⁻] = 2s must then be squared, giving 4s², not 2s².
- Comparing Ksp across different salt types. Convert to molar solubility first.
- Not halving concentrations on mixing. Equal volumes mixed means every concentration is halved before you compute Q.
- Quoting solubility in the wrong unit. Molar solubility is mol/L; multiply by the molar mass for g/L.
- Ignoring pH for hydroxides and weak-acid salts. The solubility of Ca(OH)₂ or CaCO₃ depends strongly on pH, so a plain Ksp calculation is only the starting point.
Where it appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 11–12 | Ksp ↔ s conversions; simple common ion effect |
| JEE / NEET | Will a precipitate form, selective precipitation, solubility vs pH |
| IIT-JAM / CUET-PG | Simultaneous equilibria, complex-ion effect on solubility |
| GATE / CSIR-NET | Qualitative analysis group separations, Ksp from cell EMF |
Skip the cube-root arithmetic. Choose the salt type, enter either Ksp or the solubility, and the calculator returns the other along with the expression it used — so you can confirm whether the salt was really 4s³ and not s².
Open the Ksp & Solubility Calculator →Ionic equilibrium carries heavy weight in Class 12 and in every entrance paper that follows. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in Delhi-NCR — details at abcchemistry.in.