Problem 20 - Chemistry An Atoms First Approach

Question

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) b. \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) c. \(\mathrm{BaF}_{2}\)

Step-by-Step Solution

Answer

\(a. \mathrm{Ag}_{2}\mathrm{CO}_{3}\left(s\right) \longleftrightarrow 2\mathrm{Ag}^{+}\left(aq\right) + \mathrm{CO}_{3}^{2-}\left(aq\right)\), \(\mathrm{K}_{sp} = [\mathrm{Ag}^{+}]^2[\mathrm{CO}_{3}^{2-}]\) \(b. \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_3\left(s\right) \longleftrightarrow \mathrm{Ce}^{3+}\left(aq\right) + 3\mathrm{IO}_3^-\left(aq\right)\), \(\mathrm{K}_{sp} = [\mathrm{Ce}^{3+}][\mathrm{IO}_{3}^{-}]^3\) \(c. \mathrm{BaF}_{2}\left(s\right) \longleftrightarrow \mathrm{Ba}^{2+}\left(aq\right) + 2\mathrm{F}^{-}\left(aq\right)\), \(\mathrm{K}_{sp} = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2\)

Step 1: a. Solve for Silver Carbonate (\(\mathrm{Ag}_{2} \mathrm{CO}_{3}\))

Step 1: Write the dissociation/dissolution reaction of the given solid. Silver Carbonate will dissociate into Silver ions and Carbonate ions: \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\left(s\right) \longleftrightarrow 2\mathrm{Ag}^{+}\left(aq\right) + \mathrm{CO}_{3}^{2-}\left(aq\right)\) Step 2: Write the solubility product expression. The solubility product expression for Silver Carbonate is: \(\mathrm{K}_{sp} = [\mathrm{Ag}^{+}]^2[\mathrm{CO}_{3}^{2-}]\)

Step 2: b. Solve for Cerium(III) Iodate (\(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\))

Step 1: Write the dissociation/dissolution reaction of the given solid. Cerium(III) Iodate dissociates into Cerium ions and Iodate ions: \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_3\left(s\right) \longleftrightarrow \mathrm{Ce}^{3+}\left(aq\right) + 3\mathrm{IO}_3^-\left(aq\right)\) Step 2: Write the solubility product expression. The solubility product expression for Cerium(III) Iodate is: \(\mathrm{K}_{sp} = [\mathrm{Ce}^{3+}][\mathrm{IO}_{3}^{-}]^3\)

Step 3: c. Solve for Barium Fluoride (\(\mathrm{BaF}_{2}\))

Step 1: Write the dissociation/dissolution reaction of the given solid. Barium Fluoride dissociates into Barium ions and Fluoride ions: \(\mathrm{BaF}_{2}\left(s\right) \longleftrightarrow \mathrm{Ba}^{2+}\left(aq\right) + 2\mathrm{F}^{-}\left(aq\right)\) Step 2: Write the solubility product expression. The solubility product expression for Barium Fluoride is: \(\mathrm{K}_{sp} = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2\)