Question

Solubility product for salt $A{{B}_{2}}$ ​ is $4\times {{10}^{-12}}$ . Calculate its solubility.
A. \[1\times {{10}^{-3}}\] M
B. \[1\times {{10}^{-5}}\] M
C. \[1\times {{10}^{-4}}\] M
D. \[1\times {{10}^{-6}}\] M

Answer 1

Hint: Solubility product of a salt depends on the number of ions formed in a reaction or the concentration of the ions at equilibrium. A salt with chemical formula $A{{B}_{2}}$ forms one cation and two anions. Thus the solubility product depends on concentration of the cations and anions.

Complete Step by Step Answer:
Solubility product of a species is defined as the product of the concentration of the cation and anion formed from a species at equilibrium. At this stage the coefficients before the ions exist as a power to the concentration of the corresponding cation and anion.
$A{{B}_{2}}$dissociate at equilibrium as follows-
$A{{B}_{2}}\rightleftharpoons {{A}^{+2}}+2{{B}^{-}}$
Thus solubility product is given as follows-
$Ksp={{(2S)}^{2}}\times S$
Where $Ksp$ is defined as the solubility product and $S$ is defined as the solubility or concentration for the cation and anion.
The solubility product of the salt $A{{B}_{2}}$is $4\times {{10}^{-12}}$. Putting the value of solubility product in the above equation we get-
$4\times {{10}^{-12}}=4\times {{(S)}^{3}}$
$S=1\times {{10}^{-4}}$ M
Thus the value of solubility product of the salt $A{{B}_{2}}$ is $1\times {{10}^{-4}}$moles per liter or M.
Thus the correct option is C.

Note: The unit of solubility product depends on the unit of concentration of the ions. As the concentration is expressed in moles per liter or in M unit thus the solubility product also has the same unit.