Solubility and Solubility Product MCQ - Practice Questions & Answers

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Solubility and Solubility Product

Solubility

The maximum amount of a particular solute in grams, which can dissolve in 100 grams of solvent at a given temperature is called solubility. It is denoted by 's' and is expressed in g. The number of moles of solute in 1 L of saturated solution is known as molar solubility.

Solubility decreases with the increase in the concentration of common ion. It increases with temperature and increases in case the ions formed from the sparingly soluble salt undergo some sort of reaction like complexation.

Example, The solubility of AgCl in water in presence of AgNO₃

Solubility increases due to complex ion formation. For example, AgCl has more solubility in ammonia due to complex formation
$\mathrm{AgCl + 2NH_3 \rightleftharpoons Ag(NH_3)_2Cl}$

Solubility Product: It is the product of the molar concentrations of ions of an electrolyte in a saturated solution at a particular temperature. It is denoted by Ksp or S.

General Representation
$\mathrm{A_xB_y} \rightleftharpoons \mathrm{xA}^{+y}+\mathrm{yB}^{-\mathrm{x}}$

$\mathrm{Ksp =[ A^{+y}]^x \times[ B^{-x}]^y}$

Relation between Solubility(s) and Solubility Product (Ksp)
$\mathrm{A_xB_y} \rightleftharpoons \mathrm{xA}^{+y}+\mathrm{yB}^{-\mathrm{x}}$
s 0 0
- xs ys
$\text { Thus, } \mathrm{Ksp = x ^{x} y ^{y}( s )^{ x + y }}$

Solubility Product and Precipitation

If K_sp ≈ Ionic product
The solution is saturated and for precipitation, more solute is to be added.
If Ionic product > K_sp
The solution is supersaturated so easily precipitated.
If Ionic product < K_sp
The solution is unsaturated so no precipitation takes place.