7. write solubility product constant expressions for each of the following slightly soluble substances.\n a…

Explanation:

Step1: Write dissociation equation for iron(III) hydroxide

The dissociation of iron(III) hydroxide ($Fe(OH)3$) in water is $Fe(OH)3(s)\rightleftharpoons Fe^{3 +}(aq)+3OH^-(aq)$. The solubility - product constant ($K{sp}$) is the product of the molar concentrations of the ions in the saturated solution, each raised to the power equal to its coefficient in the balanced chemical equation. So, $K{sp}=[Fe^{3 +}][OH^-]^3$.

Step2: Write dissociation equation for magnesium carbonate

The dissociation of magnesium carbonate ($MgCO_3$) in water is $MgCO_3(s)\rightleftharpoons Mg^{2+}(aq)+CO_3^{2 -}(aq)$. Then $K_{sp}=[Mg^{2+}][CO_3^{2 -}]$.

Step3: Write dissociation equation for aluminum phosphate

The dissociation of aluminum phosphate ($AlPO_4$) in water is $AlPO_4(s)\rightleftharpoons Al^{3+}(aq)+PO_4^{3 -}(aq)$. So, $K_{sp}=[Al^{3+}][PO_4^{3 -}]$.

Answer:

a. $K_{sp}=[Fe^{3 +}][OH^-]^3$ b. $K_{sp}=[Mg^{2+}][CO_3^{2 -}]$ c. $K_{sp}=[Al^{3+}][PO_4^{3 -}]$