Question

why is the solubility of barium sulfate increased by the presence of acids? (2.5marks)​

Answer 1

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WE address the equilibrium…

WE address the equilibrium…[math]BaSO_{4}(s) \stackrel{H_{2}O} ⇌ Ba^{2+} + SO_{4}^{-2}[/math]

WE address the equilibrium…[math]BaSO_{4}(s) \stackrel{H_{2}O} ⇌ Ba^{2+} + SO_{4}^{-2}[/math]And a value of [math]K_{sp}=[Ba^{2+}][SO_{4}^{2-}][/math] will be posted SOMEWHERE, and its magnitude will be (i) LOW, and (ii) depend on temperature. But if we add an acid [math]HX[/math] to the give system, it will protonate the sulfate dianion to give bisulfate….

WE address the equilibrium…[math]BaSO_{4}(s) \stackrel{H_{2}O} ⇌ Ba^{2+} + SO_{4}^{-2}[/math]And a value of [math]K_{sp}=[Ba^{2+}][SO_{4}^{2-}][/math] will be posted SOMEWHERE, and its magnitude will be (i) LOW, and (ii) depend on temperature. But if we add an acid [math]HX[/math] to the give system, it will protonate the sulfate dianion to give bisulfate….[math]SO_{4}^{2-} + H_{3}O^{+} → HSO_{4}^{-} + H_{2}O[/math]

WE address the equilibrium…[math]BaSO_{4}(s) \stackrel{H_{2}O} ⇌ Ba^{2+} + SO_{4}^{-2}[/math]And a value of [math]K_{sp}=[Ba^{2+}][SO_{4}^{2-}][/math] will be posted SOMEWHERE, and its magnitude will be (i) LOW, and (ii) depend on temperature. But if we add an acid [math]HX[/math] to the give system, it will protonate the sulfate dianion to give bisulfate….[math]SO_{4}^{2-} + H_{3}O^{+} → HSO_{4}^{-} + H_{2}O[/math]And thus the concentration of sulfate dianion WILL REDUCE, and the first solubility equilibrium will be driven to the RIGHT as we face the page, and the barium ion will be more soluble in the acidified solvent. Capisce? And as a result of this equilibrium, while many sulfates are insoluble, for instance, calcium sulfate, and barium sulfate, MANY BISULFATES, i.e. salts of [math]HSO_{4}^{-}[/math], are soluble in aqueous solution….

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