Equilibrium Ques 68
68. 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 $ $mol^{-1}$ )
[2018]
(a) $1.08 \times 10^{-10} $ $mol^{2} $ $L^{-2}$
(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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Answer:
Correct Answer: 68.(a)
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} $ $L^{-2}$
BETA
