Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

The solubility of BaSO4BaSO _{4} in water is 2.42×10−3 gL−12.42 \times 10^{-3}\, gL ^{-1} at 298 K298 \,K. The value of its solubility product (Ksp)\left( K _{ sp }\right) will be (Given molar mass of BaSO4=233 g mol−1BaSO _{4}=233 \,g\, mol ^{-1} )

A

1.08×10−8 mol2 L−21.08 \times 10^{-8} \, mol^{2} \, L^{-2}

B

1.08×10−10 mol2 L−21.08 \times 10^{-10} \, mol^{2} \, L^{-2}

C

1.08×10−14 mol2 L−21.08 \times 10^{-14} \, mol^{2} \, L^{-2}

D

1.08×10−12 mol2 L−21.08 \times 10^{-12} \, mol^{2} \, L^{-2}

Answer

1.08×10−10 mol2 L−21.08 \times 10^{-10} \, mol^{2} \, L^{-2}

Explanation

Solution

The correct answer is B:1.08×10−10 mol2 L−21.08 \times 10^{-10} \, mol^{2} \, L^{-2}
Sol. Solubility of BaSO4,s=2.42×10−3233(mol L−1)BaSO _{4}, s =\frac{2.42 \times 10^{-3}}{233}\left( mol\, L ^{-1}\right)
=1.04×10−5(mol L−1)=1.04 \times 10^{-5}\left( mol \,L ^{-1}\right)
BaSO4(s)⇌Ba2+s(aq)+SO42−s(aq)BaSO _{4}( s ) \rightleftharpoons \underset{s}{{ Ba }^{2+}}( aq )+ \underset{s}{SO _{ 4 }^{2-}}( aq )
Ksp=[Ba2+][SO42−]=s2K _{ sp }=\left[ Ba ^{2+}\right]\left[ SO _{4}^{2-}\right]= s ^{2}
=(1.04×10−5)2=\left(1.04 \times 10^{-5}\right)^{2}
=1.08×10−10 mol2 L−2=1.08 \times 10^{-10} \,mol ^{2}\,L ^{-2}