Question

The solubility product of $BaS{{O}_{4}}$ is $1.3\times {{10}^{-9}}$ . The solubility of this salt in pure water will be
A. $1.69\times {{10}^{-5}}M$
B. $1.69\times {{10}^{-2}}M$
C. $3.6\times {{10}^{-6}}M$
D. $3.6\times {{10}^{-5}}M$

Answer 1

Hint: Solubility product depends on the number of ions formed in a reaction at equilibrium. Barium sulphate dissociates into one barium cation and one sulphate anion.

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.

Barium sulphate dissociation at equilibrium as follows:
$BaS{{O}_{4}}\rightleftharpoons B{{a}^{+2}}+S{{O}_{4}}^{2-}$
Thus Barium sulphate dissociates into one barium cation and one sulphate anion at equilibrium.
Thus solubility product is given as follows:
$K_{sp}=(S)\times S$; Where $Ksp$ is defined as the solubility product and $S$ is defined as the solubility for the cation and anion.

The value of the solubility product of Barium sulphate is $1.3\times {{10}^{-9}}$. Putting the value of solubility product in the above equation we get-
$1.3\times {{10}^{-9}}={{S}^{2}}$
$S=3.6\times {{10}^{-5}}M$
Thus the value of solubility of barium sulphate is $3.6\times {{10}^{-5}}M$.
Thus the correct option is D.

Note: The chemical name of $BaS{{O}_{4}}$ is barium sulphate. It is an ionic compound. It dissociates into the corresponding ions in solution state. It is a white colored solid and it is insoluble in water.