Equilibrium - Result Question 67
####70. The solubility of $BaSO_4$ in water is $2.42 \times 10^{-3} gL^{-1}$ at $298 K$. The value of its solubility product $(K _{s p})$ will be
(Given molar mass of $BaSO_4=233 g mol^{-1}$ )
(a) $1.08 \times 10^{-10} mol^{2} L^{-2}$
[2018]
(b) $1.08 \times 10^{-12} mol^{2} L^{-2}$
(c) $1.08 \times 10^{-8} mol^{2} L^{-2}$
(d) $1.08 \times 10^{-14} mol^{2} L^{-2}$
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Solution:
- (a) Solubility of $BaSO_4=2.42 \times 10^{-3} gL^{-1}$
$\therefore \quad s=\frac{2.42 \times 10^{-3}}{233}=1.038 \times 10^{-5} mol L^{-1}$
$K _{s p}=s^{2}=(1.038 \times 10^{-5})^{2}=1.08 \times 10^{-10} mol^{2} L^{-2}$
