Question

The solubility of $BaS{O}_4$ in water is 2.42 $\times {10}^{-3}g{L}^{-1}$ at 298K. The value of solubility product will be:
(Given molar mass of $BaS{O}_4$ = 233 g $mo{l}^{-1}$)
a.) 1.08 $\times {10}^{-14}mo{l}^2{L}^{-2}$
b.) 1.08 $\times {10}^{-10}mo{l}^2{L}^{-2}$
c.) 1.08 $\times {10}^{-8}mo{l}^2{L}^{-2}$
d.) 1.08 $\times {10}^{-12}mo{l}^2{L}^{-2}$

Answer 1

Hint: The solubility product constant is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution.

Complete step by step answer:
The solubility product $$K_{sp}$$ for a solid substance dissolved in water is its equilibrium constant. It depicts the level of dissociation of the solute in the solution. The more soluble a substance is, the higher the value of $$K_{sp}$$ it has.
For any general reversible reaction
$$aA\to bB+cC$$

$K_{sp}$ = $[B{]}^b[C{]}^c$
Solids are not included in the equilibrium reactions as their active mass is 1.
Now, in the given question, $BaS{O}_4$ slightly decomposes to give $B{a}^{+2}$ and $S{O_4}^{-2}$ ions. The chemical reaction demonstrating the same is:
$BaS{O}_4\to B{a}^{+2}+S{O_4}^{-2}$
Solubility of $BaS{O}_4$ in gram per litres = $2.42\times {\text{10}}^{-3}g{L}^{-1}$

So, the solubility of $BaS{O}_4$ in moles per litres = s = ${\displaystyle \frac{2.42\times {10}^{-3}}{233}}mol{L}^{-1}$
s = 1.04 $\times {10}^{-5}mol{L}^{-1}$ = $[B{a}^{+2}]$ = $[S{O_4}^{-2}]$
Now,
The solubility product $K_{sp}$ = $[B{a}^{+2}][S{O_4}^{-2}]$
= $(1.04\times {10}^{-5}mol{L}^{-1}{)}^2$
= 1.08 $\times {10}^{-10}mo{l}^2{L}^{-2}$

Therefore, the solubility product equals 1.08 $\times {10}^{-10}mo{l}^2{L}^{-2}$.
Hence, the correct answer is (B) 1.08 $\times {10}^{-10}mo{l}^2{L}^{-2}$.

Note: Ensure that the solids do not appear in the equilibrium equation of solubility product. The active mass of solids is unity (1). Sometimes, a student can mistakenly take their concentration into consideration.