Question

If the solubility of a sparingly soluble salt of the type $B{{A}_{2}}$ ​ (giving three ions on dissociation of a molecule) is x moles per litre, then its solubility product is given by:
(A) x²
(B) x³
(C) 4x²
(D) 4x³

Answer 1

Hint: The solubility product of a sparingly soluble salt forming a saturated solution in water is calculated as the product of the concentrations of the ions, raised to a power equal to the number of the ions occurring in the equation representing the dissociation of the electrolyte. The solubility product is denoted by ${{K}_{sp}}$.

Formula Used: For a compound$B{{A}_{2}}$, the solubility product is given by
${{K}_{sp}}=\left[ {{B}^{+}} \right]{{\left[ {{A}^{-}} \right]}^{2}}$ where ${{B}^{+}}$are the cations and ${{A}^{-}}$ are the anions.

Complete Step by Step Solution:
A sparingly soluble salt of the type $B{{A}_{2}}$ dissociates as

$B{{A}_{2}}$, on dissociation gives 1 ${{B}^{+}}$ion and 2 ${{A}^{-}}$ions.

Let solubility be x. The solubility product is, thus given as ${{K}_{sp}}=\left[ {{B}^{+}} \right]{{\left[ {{A}^{-}} \right]}^{2}}$
${{K}_{sp}}=x\times {{\left( 2x \right)}^{2}}$
${{K}_{sp}}=4{{x}^{3}}$
Hence, the solubility product of $B{{A}_{2}}$ is given by 4x3.
Correct Option: (D) 4x³.

Additional Information: The solubility product depends upon temperature. With an increase in temperature, the solubility of a substance is increased. Hence, the solubility product increases. The solubility product also depends upon the common-ion effect, the diverse-ion effect, and the presence of ion pairs.

Note: The units of the solubility product depend upon the number of ions formed after dissociation of the electrolyte. It is generally given as ${{(mol{{L}^{-1}})}^{n}}$where n is the number of ions formed in the reaction. The greater the solubility product, the greater the solubility of the compound and vice-versa.