[1.08 × 10^{–10} mol^2L^{...] The solubility of BaSO_4 in water is 2.42 × 10^{–3} gL^{

For Competitive Exams

The solubility of $BaSO_4$ in water is 2.42 × $10^{–3} gL^{–1}$ at 298 K. The value of its solubility product $(K_{sp})$ will be
(Given molar mass of $BaSO_4 = 233 g mol–1)$

$1.08 × 10^{–14} mol^2L^{–2}$

$1.08 × 10^{–12} mol^2L^{–2}$

$1.08 × 10^{–10} mol^2L^{–2}$

$1.08 × 10^{–8} mol^2L^{–2}$

Explanation:

Solubility of $BaSO_4, s =\dfrac{2.42 \times 10^{-3}}{233} (mol L^{-1})$
=$1.04 \times 10^{-5}(mol L^{-1})$

$K_{sp}=\left[Ba^{2+}\right]\left[SO_4^{2-}\right]= s^2$
$(1.04 \times 10^{-5})^2$
$1.08 \times 10^{-10} mol^2 L^{-2}$

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