Question

The solubility of $BaS{{O}_{4}}$ in water, is $2.33\times {{10}^{-3}}g{{L}^{-1}}$. Its solubility product will be: (molecular weight of $BaS{{O}_{4}}$=233).
(A) $1\times {{10}^{-5}}$
(B) $1\times {{10}^{-10}}$
(C) $1\times {{10}^{-15}}$
(D) $1\times {{10}^{-20}}$

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${{K}_{sp}}$. As the given solubility of $BaS{{O}_{4}}$ is in grams per litre, so, we will first convert it into moles per litre by dividing it with the molecular mass of $BaS{{O}_{4}}$ then, calculating the solubility product.

Formula Used: For a compound$BaS{{O}_{4}}$, the solubility product is given by
${{K}_{sp}}=\left[ B{{a}^{2+}} \right]\left[ S{{O}_{4}}^{2-} \right]$

Complete Step by Step Solution:
We will first convert the solubility of $BaS{{O}_{4}}$from grams per litre to moles per litre.
Solubility of $BaS{{O}_{4}}$= $2.33\times {{10}^{-3}}g{{L}^{-1}}$
Molecular weight of $BaS{{O}_{4}}$= 233
Hence, the solubility of $BaS{{O}_{4}}$, $s=\frac{2.33\times {{10}^{-3}}}{233}$
$s=1\times {{10}^{-5}}mol{{L}^{-1}}$
The compound forms ion pairs as
 

Now, the solubility product, ${{K}_{sp}}=\left[ B{{a}^{2+}} \right]\left[ S{{O}_{4}}^{2-} \right]$
Let solubility be s, then the solubility product is ${{K}_{sp}}=s\times s$
${{K}_{sp}}={{s}^{2}}$
${{K}_{sp}}={{(1\times {{10}^{-5}})}^{2}}$
${{K}_{sp}}=1\times {{10}^{-10}}mo{{l}^{2}}{{L}^{-2}}$
Correct Option: (B) $1\times {{10}^{-10}}$.

Note: The units of each of the terms should be carefully noted. The units of solubility product depend on each of the compounds depending upon the molar concentration of ions formed in the reaction. It measures the solubility of a compound. The greater the value of the solubility product, the greater the solubility of the compound and vice-versa. The solubility product depends upon temperature, common-ion effect, diverse-ion effect, and the presence of ion pairs.