Let us now have a solid like barium sulphate in contact with its saturated aqueous solution. The equilibrium between the undisolved solid and the ions in a saturated solution can be represented by the equation:

BaSO_{4}(aq) ⇔ Ba^{2+}(aq) + SO_{4}^{2-}(aq)

The equilibrium constant is given by the equation:

K ={[Ba^{2+}][ SO_{4}^{2-}]}/[BaSO_{4}]

For a pure solid substance the concentration remains constant and we can write

K_{sp} = K[BaSO_{4}] = [Ba^{2+}][ SO_{4}^{2-}]

We call K_{sp} the solubility product constant or simply solubility product. The experimental value of K_{sp} in above equation at 298K is 1.1 × 10^{–10}.

This means that for solid barium sulphate in equilibrium with its saturated solution, the product of the concentrations of barium and sulphate ions is equal to its solubility product constant.

The concentrations of the two ions will be equal to the molar solubility of the barium sulphate. If molar solubility is S, then 1.1 × 10^{–10} = (S)(S) = S^{2}

S = 1.05 X 10^{-5}

Thus, molar solubility of barium sulphate will be equal to 1.05 × 10^{–5} mol L^{–1}.

A salt may give on dissociation two or more than two anions and cations carrying different charges. For example, consider a salt like zirconium phosphate of molecular formula (Zr^{4+})_{3}(PO_{4}^{3–})_{4}.

It dissociates into 3 zirconium cations of charge +4 and 4 phosphate anions of charge –3. If the molar solubility of zirconium phosphate is S, then it can be seen from the stoichiometry of the compound that

[Zr^{4+}]=3S and [PO_{4}^{3–}] = 4S

and K_{sp}= (3S)^{3}(4S)^{4}= 6912(S)^{7}

or S = { K_{sp} / (3S)^{3}(4S)^{4}}^{1/7} = { K_{sp} /6912}^{1/7}

A solid salt of the general formula M_{x}^{p+}X_{y}^{q-} with molar solubility S in equilibrium with its saturated solution may be represented by the equation:

M_{x}X_{y}(s) ⇔ xM^{p+}(aq) + yX^{q-}(aq) [where x p^{+} = y q^{–}]

And its solubility product constant is given by:

K_{sp} = [M^{p+}]^{x}[X^{q-}]^{y} = (xS)^{x}(yS)^{y} = x^{x} y^{y} S^{(x+y)}

S^{(x+y)} = K_{sp} / x^{x} y^{y}

S = (K_{sp} / x^{x} y^{y} )^{1/x+y}

The term K_{sp} in equation is given by Q_{sp} (section 7.6.2) when the concentration of one or more species is not the concentration under equilibrium. Obviously under equilibrium conditions K_{sp} = Q_{sp} but otherwise it gives the direction of the processes of precipitation or dissolution.