Answer:
Solubilities of [tex]Sr(OH)_{2},ZnS,Ca_{3}(PO_{4})_{2} and BaCO_{3}[/tex] increases with increasing acidity
Explanation:
[tex]Sr(OH)_{2}\rightleftharpoons Sr^{2+}+2 OH^{-}[/tex]
Concentration of [tex]OH^{-}[/tex] decreases with increasing acidity. Therefore solubility of [tex]Sr(OH)_{2}[/tex] increases to keep constant solubility product.
[tex]ZnS\rightleftharpoons Zn^{2+}+S^{2-}[/tex]
Concentration of [tex]S^{2-}[/tex] decreases with increasing acidity due to formation of [tex]H_{2}S[/tex]. therefore solubility of ZnS increases to keep constant solubility product
[tex]PbBr_{2}\rightleftharpoons Pb^{2+}+2Br^{-}[/tex]
Increasing acidity have no effect on this above equilibrium. Therefore solubility of [tex]PbBr_{2}[/tex] remains constant.
[tex]Ca_{3}(PO_{4})_{2}\rightleftharpoons 3Ca^{2+}+2PO_{4}^{3-}[/tex]
Concentration of [tex]PO_{4}^{3-}[/tex] decreases with increasing acidity due to formation of [tex]H_{3}PO_{4}[/tex]. Therefore solubility of [tex]Ca_{3}(PO_{4})_{2}[/tex] increases to keep constant solubility product.
[tex]NaI\rightleftharpoons Na^{+}+I^{-}[/tex]
Increasing acidity have no effect on this above equilibrium. Therefore solubility of NaI remains constant.
[tex]BaCO_{3}\rightleftharpoons Ba^{2+}+CO_{3}^{2-}[/tex]
Concentration of [tex]CO_{3}^{2-}[/tex] decreases with increasing acidity due to formation of [tex]H_{2}CO_{3}[/tex]. Therefore solubility of [tex]BaCO_{3}[/tex] increases to keep constant solubility product.