1. 1.08×10⁻¹⁰ mol ² L ⁻² 2. 1.08×10⁻¹⁴ mol ² L ⁻² 3. 1.08×10⁻¹² mol ² L ⁻² 4. 1.08×10⁻⁸ mol ² L ⁻²

Answer: 1. 1.08×10⁻¹⁰ mol ² L ⁻² Convert solubility to mol/L, then use K_sp=s² for a 1:1 salt. s=dfrac2.42×10⁻³233=1.04×10⁻⁵\ mathrmmol L⁻¹ K_sp=[mathrmBa²⁺][mathrmSO₄²⁻]=s²=(1.04×10⁻⁵)²≈1.08×10⁻¹⁰\ mathrmmol²L⁻² .

Topic: Solubility product· Equilibrium

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. \(1.08\times10^{-10}\) mol\(^2\) L\(^{-2}\)

2. \(1.08\times10^{-14}\) mol\(^2\) L\(^{-2}\)

3. \(1.08\times10^{-12}\) mol\(^2\) L\(^{-2}\)

4. \(1.08\times10^{-8}\) mol\(^2\) L\(^{-2}\)

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Answer: 1. \(1.08\times10^{-10}\) mol\(^2\) L\(^{-2}\)

Convert solubility to mol/L, then use \(K_{sp}=s^2\) for a 1:1 salt.

\(s=\dfrac{2.42\times10^{-3}}{233}=1.04\times10^{-5}\ \mathrm{mol\,L^{-1}}\)

\(K_{sp}=[\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}]=s^2=(1.04\times10^{-5})^2\approx1.08\times10^{-10}\ \mathrm{mol^2L^{-2}}\).

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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)

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