Question

The solubility product of barium sulphate is $1.5 \times 10^{-9}$ at 18$^\circ$C. Its solubility in water at 18$^\circ$C is

• A. $1.5 \times 10^{-9} \, mol L^{-1}$
• B. $1.5 \times 10^{-5} mol \,L^{-1}$
• C. $3.9 \times 10^{-9} mol \,L^{-1}$
• D. $3.9 \times 10^{-5} mol \, L^{-1}$.

Answer 1

Answer: (D)

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