Question

Solid $BaC{{O}_{3}}$ is gradually dissolved in a $1\text{ x 1}{{\text{0}}^{-4}}\text{ M}$ $N{{a}_{2}}C{{O}_{3}}$ solution. At what concentration of $B{{a}^{2+}}$ will a precipitate form? (${{K}_{sp}}$ for $BaC{{O}_{3}}$ = $5.1\text{ x 1}{{\text{0}}^{-9}}$)
(A) 4.1 x $\text{1}{{\text{0}}^{-5}}$ M
(B) 5.1 x $\text{1}{{\text{0}}^{-5}}$ M
(C) 8.1 x $\text{1}{{\text{0}}^{-8}}$ M
(D) 8.1 x $\text{1}{{\text{0}}^{-7}}$ M

Answer 1

Hint: It is important to know that there are two dissociation reactions taking place. It is clear that sodium carbonate is a strong electrolyte and thus undergoes complete dissociation. Consider the effect of dissociation of sodium carbonate on the dissociation of barium carbonate. This is because they have an ion common to both compounds.

Complete step by step solution:
Solubility is defined as the property of a substance (solute) to get dissolved in a solvent in order to form a solution.
The solubility product constant is the equilibrium constant for the dissolution of a solid solute in a solvent to form a solution. It is denoted by the symbol ${{K}_{sp}}$. We can form an idea about the dissolution of an ionic compound by the value of its solubility product.
We will write the dissociation of $N{{a}_{2}}C{{O}_{3}}$and$BaC{{O}_{3}}$.
$N{{a}_{2}}C{{O}_{3}}\ \to \text{ }\ 2N{{a}^{2+}}+C{{O}_{3}}^{2-}$
As it is a strong electrolyte, complete dissociation takes place and the concentration of carbonate ion becomes,
$[C{{O}_{3}}^{2-}]=1\text{ x 1}{{\text{0}}^{-4}}$
The dissociation of the weak electrolyte $BaC{{O}_{3}}$ is given below:
$BaC{{O}_{3}}\text{ }\to \text{ }B{{a}^{2+}}+C{{O}_{3}}^{2-}$
${{K}_{sp}}=[B{{a}^{2+}}][C{{O}_{3}}^{2-}]$
The concentration of carbonate ion will be from the dissociation of sodium carbonate as the dissociation of $BaC{{O}_{3}}$ is very less and thus concentration of carbonate ion from $BaC{{O}_{3}}$ is insignificant and thus ignored.
Substituting the values in the equation, we get
$\text{5}\text{.1 x 1}{{\text{0}}^{-9}}=[B{{a}^{2+}}][1\text{ x 1}{{\text{0}}^{-4}}]$
Thus, the value of $[B{{a}^{2+}}]$ becomes,
$[B{{a}^{2+}}]=5.1\text{ x 1}{{\text{0}}^{-5}}\text{ M}$

Therefore, the correct answer is option (B).

Note: The common ion effect describes the effect of adding a common ion on the ​equilibrium of the new solution. The common ion effect generally decreases ​solubility of a solute. The equilibrium shifts to the left to relieve off the excess product formed. When a strong electrolyte is added to a solution of weak electrolyte, the solubility of weak electrolyte decreases leading to the formation of precipitate at the bottom of the testing apparatus.