Solubility ProductJEE Main

Ksp → molar solubility — interactive Chemistry simulation for IIT-JEE.

Concept

For a sparingly-soluble salt, the solubility product $K_{sp}$ is the equilibrium constant for its dissolution. From $K_{sp}$ you can predict the molar solubility $s$ — how many moles dissolve per litre before the solution is saturated.

Key formula

A_xB_y \rightleftharpoons x\,A^{y+} + y\,B^{x-}, \qquad K_{sp} = x^x\,y^y\,s^{\,x+y}

Derivation

At saturation, dissolving $s$ mol/L gives $[A^{y+}] = x s$ and $[B^{x-}] = y s$ .

Substituting into $K_{sp} = [A^{y+}]^x[B^{x-}]^y$ gives $K_{sp} = (xs)^x(ys)^y = x^x y^y\,s^{x+y}$ , which you invert to solve for $s$ .

Scenarios to explore

Solubility Product — Ksp to molar solubility for 1:1, 1:2 and 1:3 salts.

Real-world applications

Predicting whether a precipitate forms (compare Qsp with Ksp).
Selective precipitation in qualitative analysis.
The common-ion effect lowering solubility.

JEE exam tips

AB: s = √Ksp. AB₂ or A₂B: s = ∛(Ksp/4). AB₃: s = ⁴√(Ksp/27).
Precipitate forms when the ionic product Qsp > Ksp.
Higher xˣyʸ ⇒ a more soluble salt for the same Ksp.

Common mistakes

Forgetting the xˣyʸ factor for non-1:1 salts (e.g. Ksp = 4s³ for AB₂).
Equating Ksp directly with solubility.
Ignoring that a common ion suppresses solubility.

Exam traps to avoid

Comparing solubilities across different salt types by Ksp alone is wrong — convert to s first.
Common-ion problems: the added ion's concentration dominates, not s.