Question

The solubility product of $AgCl$ is $1.44\times {{10}^{-4}}$ at ${{100}^{o}}C$ . The solubility of silver chloride in boiling water is:
(A) $0.72\times {{10}^{-4}}M$
(B) $1.20\times {{10}^{-2}}M$
(C) $0.72\times {{10}^{-2}}M$
(D) $1.20\times {{10}^{-4}}M$

Answer 1

Hint: When a salt is stirred in water and only a small amount of it gets dissolved but a large amount of it remains undissolved, then the salt is known as sparingly soluble salt. 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 given by${{K}_{sp}}$.

Formula Used: The formula for the solubility product depends on the number of ions formed after dissociation of the compound. The solubility product is given by ${{K}_{sp}}={{\left[ {{A}^{+}} \right]}^{a}}{{\left[ {{B}^{-}} \right]}^{b}}$ where ${{A}^{+}}$ are the cations, ${{B}^{-}}$are the anions, and a and b are the numbers of cations and anions formed in the reaction.

Complete Step by Step Solution:
$AgCl$gets dissociated as

Let solubility be s
${{K}_{sp}}=\left[ A{{g}^{+}} \right]\left[ C{{l}^{-}} \right]$
${{K}_{sp}}=s\times s$
${{K}_{sp}}={{s}^{2}}$

The given solubility product is ${{K}_{sp}}=1.44\times {{10}^{-4}}$ .
${{s}^{2}}=1.44\times {{10}^{-4}}$
$s=\sqrt{1.44\times {{10}^{-4}}}$
$s=1.2\times {{10}^{-2}}$
Hence, the solubility of $AgCl$ is $1.2\times {{10}^{-2}}$ moles/litre.
Correct Option: (B) $1.20\times {{10}^{-2}}M$.

Note: The solubility product depends upon the temperature. It increases with an increase in temperature. This is because, with an increase in temperature, solubility increases. Solubility means the tendency of a solute to get dissolved in a solvent to form a solution. It also depends on the common-ion effect; that is, if a common ion is present in the solution, then the solubility product gets lowered.