Question

The solubility of $Ba{F_2}$in a solution of $Ba{(N{O_3})_2}$ will be represented by the concentration term:
A.${[Ba]^{2 + }}$
B.$[{F^ - }]$
C.$\dfrac{1}{2}[{F^ - }]$
D.$2{[N{O_3}]^ - }$

Answer 1

The solubility product ${K_{sp}}$ is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in a solution. The more soluble the substance is, the higher is the ${K_{sp}}$ value.
Formula used:
${K_{sp}} =$[Cation] [Anions]
i.e. it is equal to the concentration of cations $\times$ concentration of anions

Complete step by step answer:
The solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Further, the solubility product constant (${K_{sp}}$) describes the equilibrium between a solid and its constituent ions in a solution. These are sparingly soluble salts i.e. they are very less soluble in the given solvent.
Now, in the given question, $Ba{F_2}$ dissociates into $B{a^{2 + }}$ and $2{F^ - }$ ions.
i.e.
Now, in case of $Ba{(N{O_3})_2}$, it will dissociate as:
$Ba{(N{O_3})_2}$ $B{a^{ + 2}} + 2N{O_3}^ -$
Now, the concentration of $B{a^{ + 2}}$ in $Ba{(N{O_3})_2}$ is ‘c’ and concentration of $B{a^{2 + }}$ in $Ba{F_2}$ is ‘s’ but ‘s ‘ is very less than ‘c’ so we will finally take the concentration of $B{a^{2 + }}$ as ‘c’ in $Ba{F_2}$
Now, we will determine ${K_{sp}}$
Therefore, ${K_{sp}}$ $= [B{a^{ + 2]}}]{[{F^ - }]^2}$
Now, $[{F^ - }] = 2s$
So, $s = \dfrac{{[{F^ - }]}}{2}$

Hence, option C is correct.

Note:
Some important factors that have an impact on the solubility product constant are, common ion effect i.e. the presence of a common ion lowers the value of ${K_{sp}}$, the diverse ion effect which means that if the ions of the solutes are uncommon, then the value of ${K_{sp}}$ will be high , and the presence of ion-pairs.