Answer
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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 = $\dfrac{{2.42 \times {{10}^{ - 3}}}}{{233}}mol{L^{ - 1}}$
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.
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 = $\dfrac{{2.42 \times {{10}^{ - 3}}}}{{233}}mol{L^{ - 1}}$
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.
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