Ksp and Solubility — Converting Between Them Correctly

By Aniket Bhardwaj · 3 September 2026 · Calculator/Formula Guide

Two numbers describe how little a sparingly soluble salt dissolves: the solubility product K_sp, and the molar solubility s. Converting between them is a standard exam question, and it goes wrong for one reason — students assume K_sp = 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

For A_xB_y(s) ⇌ xA^y+(aq) + yB^x−(aq)

K_sp = [A^y+]^x × [B^x−]^y

The solid does not appear — its activity is 1. K_sp 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 A^y+ and y·s moles of B^x− appear. Substituting gives one formula per salt type. Learn the pattern, not the four results:

Type	Example	Ions produced	K_sp 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

K_sp(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 K_sp = 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

K_sp(PbCl₂) = 1.7 × 10⁻⁵. Find s.

PbCl₂ ⇌ Pb²⁺ + 2Cl⁻, so [Pb²⁺] = s and [Cl⁻] = 2s

K_sp = (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 K_sp does not rank solubility

K_sp(Ag₂CrO₄) = 1.1 × 10⁻¹². Find s and compare with AgCl.

Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻, so [Ag⁺] = 2s and [CrO₄²⁻] = s

K_sp = (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 K_sp is smaller by a factor of about 160.

The lesson: you may compare K_sp 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.

K_sp = [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 K_sp

The same expression, used in reverse. If CaF₂ has a molar solubility of 2.1 × 10⁻⁴ M:

K_sp = 4s³ = 4 × (2.1 × 10⁻⁴)³ = 4 × 9.261 × 10⁻¹² = 3.7 × 10⁻¹¹

Will a precipitate form? Compare Q with K_sp

Write the ionic product Q with the actual concentrations after mixing: Q > K_sp means precipitation, Q = K_sp means exactly saturated, and Q < K_sp 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 K_sp 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 K_sp calculation is only the starting point.

Where it appears in exams

Exam	Typical question
CBSE/ICSE Class 11–12	K_sp ↔ 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, K_sp from cell EMF

Skip the cube-root arithmetic. Choose the salt type, enter either K_sp 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 →