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Question: The solubility of \(BaS{O_4}\) in water is 2.42 \(\times {10^{ - 3}}g{L^{ - 1}}\) at 298K. The value...

The solubility of BaSO4BaS{O_4} in water is 2.42 ×10−3gL−1\times {10^{ - 3}}g{L^{ - 1}} at 298K. The value of solubility product will be:
(Given molar mass of BaSO4BaS{O_4} = 233 g mol−1mo{l^{ - 1}})
a.) 1.08 ×10−14mol2L−2\times {10^{ - 14}}mo{l^2}{L^{ - 2}}
b.) 1.08 ×10−10mol2L−2 \times {10^{ - 10}}mo{l^2}{L^{ - 2}}
c.) 1.08 ×10−8mol2L−2 \times {10^{ - 8}}mo{l^2}{L^{ - 2}}
d.) 1.08 ×10−12mol2L−2\times {10^{ - 12}}mo{l^2}{L^{ - 2}}

Explanation

Solution

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 Ksp{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 Ksp{K_{sp}} it has.
For any general reversible reaction
aA→bB+cCaA \to bB + cC

Ksp{K_{sp}} = [B]b[C]c{[B]^b}{[C]^c}
Solids are not included in the equilibrium reactions as their active mass is 1.
Now, in the given question, BaSO4BaS{O_4} slightly decomposes to give Ba+2B{a^{ + 2}} and SO4−2S{O_4}^{ - 2} ions. The chemical reaction demonstrating the same is:
BaSO4→Ba+2+SO4−2BaS{O_4} \to B{a^{ + 2}} + S{O_4}^{ - 2}
Solubility of BaSO4BaS{O_4} in gram per litres = 2.42×10−3gL−12.42\times{{\text{10}}^{-3}}g{{L}^{-1}}

So, the solubility of BaSO4BaS{O_4} in moles per litres = s = 2.42×10−3233molL−1\dfrac{{2.42 \times {{10}^{ - 3}}}}{{233}}mol{L^{ - 1}}
s = 1.04 ×10−5molL−1 \times {10^{ - 5}}mol{L^{ - 1}} = [Ba+2][B{a^{ + 2}}] = [SO4−2][S{O_4}^{ - 2}]
Now,
The solubility product Ksp{K_{sp}} = [Ba+2][SO4−2][B{a^{ + 2}}][S{O_4}^{ - 2}]
= (1.04×10−5molL−1)2{(1.04 \times {10^{ - 5}}mol{L^{ - 1}})^2}
= 1.08 ×10−10mol2L−2 \times {10^{ - 10}}mo{l^2}{L^{ - 2}}

Therefore, the solubility product equals 1.08 ×10−10mol2L−2\times {10^{ - 10}}mo{l^2}{L^{ - 2}}.
Hence, the correct answer is (B) 1.08 ×10−10mol2L−2 \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.