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Chapter 15: Problem 20

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. \(\mathrm{BaF}_{2}\)

Short Answer

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

Step by step solution

01

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: \(\mathrm{K}_{sp} = [\mathrm{Ag}^{+}]^2[\mathrm{CO}_{3}^{2-}]\)
02

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: \(\mathrm{K}_{sp} = [\mathrm{Ce}^{3+}][\mathrm{IO}_{3}^{-}]^3\)
03

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: \(\mathrm{K}_{sp} = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2\)

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Most popular questions from this chapter

Describe how you could separate the ions in each of the following groups by selective precipitation. a. \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\) c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\)

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(6.2 \times 10^{-12} \mathrm{mol} / \mathrm{L}\). b. The solubility of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) is \(7.4 \times 10^{-2} \mathrm{mol} / \mathrm{L}\).

\(K_{\mathrm{f}}\) for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7} . K_{\mathrm{sp}}\) for \(\mathrm{AgCl}\) is \(1.6 \times 10^{-10} .\) Calculate the molar solubility of \(\mathrm{AgCl}\) in \(1.0 M \mathrm{NH}_{3}\).

The copper(I) ion forms a chloride salt that has \(K_{\mathrm{sp}}=1.2 \times\) \(10^{-6} .\) Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}:\) $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 M\) NaCl.

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}, \mathrm{Na}_{2} \mathrm{S},\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are \(K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)\) \(=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},\) and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=\) \(1 \times 10^{-54}\).
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