Q. The solubility product of barium sulphate is $1.5 \times 10^{- 9}\ at\ 18℃$. Its solubility in water at $18℃$ is
A) $1.5 \times 10^{- 9}$
B) $1.5 \times 10^{- 5}$
C) $3.9 \times 10^{- 9}$
D) $3.9 \times 10^{- 5}$
Explanation
(d)
Let solubility of $BaSO_{4} = mol\ L^{- 1}$
$BaSO_{4} \rightleftharpoons Ba^{2 +} + SO_{4}^{2 -}$
$x$ $x$ $x$
$\therefore$ Ions at equilibrium
$K_{sp} = \left\lbrack Ba^{2 +} \right\rbrack\left\lbrack SO_{4}^{2 -} \right\rbrack$
$= x \times x$
$x^{2}$
Given, $K_{sp} = 1.5 \times 10^{- 9}$
$\left( 1.5 \times 10^{- 9} \right) = x^{2}$
or $\sqrt{1.5 \times 10^{- 9}} = x$
$x = 3.9 \times 10^{- 5}$ mol/L
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.