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Chemistry Question on Solutions

The solubility product of barium sulphate is 1.5×10−91.5 \times 10^{-9} at 18∘^\circC. Its solubility in water at 18∘^\circC is

A

1.5×10−9 molL−11.5 \times 10^{-9} \, mol L^{-1}

B

1.5×10−5mol L−11.5 \times 10^{-5} mol \,L^{-1}

C

3.9×10−9mol L−13.9 \times 10^{-9} mol \,L^{-1}

D

3.9×10−5mol L−13.9 \times 10^{-5} mol \, L^{-1}.

Answer

3.9×10−5mol L−13.9 \times 10^{-5} mol \, L^{-1}.

Explanation

Solution

BaSO4→Ba+2+SO42−BaSO_{4} \to Ba^{+2}+SO^{2-}_{4} ∴\therefore Solubility =(Ksp)1/2=(K_{sp})^{1/2} =(1.5×10−9)1/2=3.87×10−5 mol L−1=(1.5 \times 10^{-9})^{1/2}=3.87\times10^{-5}\,mol\,L^{-1}