Q. If the solubility product of $BaSO_{4}$ is $1.5 \times 10^{- 10}$ in water. Its solubility, in moles per litre, is
A) $1.5 \times 10^{- 9}$
B) $3.9 \times 10^{- 5}$
C) $7.5 \times 10^{- 5}$
D) $1.5 \times 10^{- 5}$
Explanation
(b)
$BaSO_{4} \rightleftharpoons Ba^{2 +} + SO_{4}^{2 -}$
Solubility product $= s \times s$
$15 \times 10^{- 10} = s^{2}:s = \sqrt{15 \times 10^{- 10}}:s = 3.9 \times 10^{- 5}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.