Explanation:
(6.1). The reaction equation will be as follows.
[tex]BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO^{2-}_{4}(aq)[/tex]
Assuming the value of [tex]K_{sp}[/tex] as [tex]1.1 \times 10^{-10}[/tex] and let the solubility of each specie involved in this reaction is "s". The expression for [tex]K_{sp}[/tex] will be as follows.
[tex]K_{sp} = [Ba^{2+}][SO^{-}_{2}][/tex] (Solids are nor considered)
= [tex]s \times s[/tex]
s = [tex]\sqrt{K_{sp}}[/tex]
= [tex]\sqrt{1.1 \times 10^{-10}}[/tex]
= [tex]1.05 \times 10^{-5}[/tex]
Therefore, solubility of barium sulfate in water is [tex]1.05 \times 10^{-5}[/tex].
(6.2). As the molar mass of [tex]BaSO_{4}[/tex] is 233.38 g/mol
Therefore, the solubility is g/L will be calculated as follows.
[tex]233.38 g/mol \times 1.05 \times 10^{-5}[/tex]
= [tex]2.45 \times 10^{-3} g/L[/tex]
Therefore, solubility of barium sulfate in grams per liter is [tex]2.45 \times 10^{-3} g/L[/tex].
