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Chapter 16: 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
The balanced dissolution reactions and solubility product expressions for the given solids are: a. \[ \mathrm{Ag}_{2}\mathrm{CO}_{3} (s) \rightarrow 2 \mathrm{Ag}^{+} (aq) + \mathrm{CO}_{3}^{2-} (aq) \] Solubility Product Expression: \[ K_{sp} = [\mathrm{Ag}^{+}]^{2} [\mathrm{CO}_{3}^{2-}] \] b. \[ \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} (s) \rightarrow \mathrm{Ce}^{3+} (aq) + 3 \mathrm{IO}_{3}^{-} (aq) \] Solubility Product Expression: \[ K_{sp} = [\mathrm{Ce}^{3+}] [\mathrm{IO}_{3}^{-}]^{3} \] c. \[ \mathrm{BaF}_{2} (s) \rightarrow \mathrm{Ba}^{2+} (aq) + 2 F^{-} (aq) \] Solubility Product Expression: \[ K_{sp} = [\mathrm{Ba}^{2+}] [F^{-}]^{2} \]

Step by step solution

01

Dissolution Reaction for \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\)

First, we have to write the dissolution reaction for \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\), which is a solid. This compound will dissociate into its respective ions when dissolving in water. The balanced equation for the dissolution reaction is: \[ \mathrm{Ag}_{2}\mathrm{CO}_{3} (s) \rightarrow 2 \mathrm{Ag}^{+} (aq) + \mathrm{CO}_{3}^{2-} (aq) \] ##Step 2: Find the Solubility Product Expression for \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\)##
02

Solubility Product Expression for \(\mathrm{Ag}_{2}\mathrm{CO}_{3}\)

Now that we have the balanced equation for the dissolution reaction, we can write the solubility product expression. The solubility product, \(K_{sp}\), is the product of the equilibrium concentrations of the ions raised to the power of their stoichiometric coefficients in the balanced equation: \[ K_{sp} = [\mathrm{Ag}^{+}]^{2} [\mathrm{CO}_{3}^{2-}] \] ##Step 3: Write the Dissolution Reaction for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\)##
03

Dissolution Reaction for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\)

Next, we have to write the dissolution reaction for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\), which is also a solid. This compound will dissociate into its respective ions when dissolving in water. The balanced equation for the dissolution reaction is: \[ \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3} (s) \rightarrow \mathrm{Ce}^{3+} (aq) + 3 \mathrm{IO}_{3}^{-} (aq) \] ##Step 4: Find the Solubility Product Expression for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\)##
04

Solubility Product Expression for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\)

Using the balanced equation for the dissolution reaction of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\), we can write the solubility product expression: \[ K_{sp} = [\mathrm{Ce}^{3+}] [\mathrm{IO}_{3}^{-}]^{3} \] ##Step 5: Write the Dissolution Reaction for \(\mathrm{BaF}_{2}\)##
05

Dissolution Reaction for \(\mathrm{BaF}_{2}\)

Lastly, we have to write the dissolution reaction for \(\mathrm{BaF}_{2}\), which is a solid. This compound will dissociate into its respective ions when dissolving in water. The balanced equation for the dissolution reaction is: \[ \mathrm{BaF}_{2} (s) \rightarrow \mathrm{Ba}^{2+} (aq) + 2 F^{-} (aq) \] ##Step 6: Find the Solubility Product Expression for \(\mathrm{BaF}_{2}\)##
06

Solubility Product Expression for \(\mathrm{BaF}_{2}\)

Using the balanced equation for the dissolution reaction of \(\mathrm{BaF}_{2}\), we can write the solubility product expression: \[ K_{sp} = [\mathrm{Ba}^{2+}] [F^{-}]^{2} \]

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Most popular questions from this chapter

The \(K_{\mathrm{sp}}\) of hydroxyapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\), is \(6.8 \times 10^{-37}\). Calculate the solubility of hydroxyapatite in pure water in moles per liter. How is the solubility of hydroxyapatite affected by adding acid? When hydroxyapatite is treated with fluoride, the mineral fluorapatite, \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\), forms. The \(K_{\mathrm{sp}}\) of this substance is \(1 \times\) \(10^{-60}\). Calculate the solubility of fluorapatite in water. How do these calculations provide a rationale for the fluoridation of drinking water?

A \(50.0-\mathrm{mL}\) sample of \(0.00200 \mathrm{M} \mathrm{AgNO}_{3}\) is added to \(50.0 \mathrm{~mL}\) of \(0.0100 M \mathrm{NaIO}_{3-}\) What is the equilibrium concentration of \(\mathrm{Ag}^{+}\) in solution? \(\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{AgIO}_{3}\) is \(3.0 \times 10^{-8}\).)

A solution contains \(1.0 \times 10^{-5} \mathrm{M} \mathrm{Ag}^{+}\) and \(2.0 \times 10^{-6} \mathrm{M} \mathrm{CN}^{-}\). Will \(\mathrm{AgCN}(s)\) precipitate? \(\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{AgCN}(s)\) is \(2.2 \times 10^{-12}\).)

\(\mathrm{Mg}(\mathrm{OH})_{2}\) is the main ingredient in the antacid TUMS and has a \(K_{\text {sp }}\) value of \(8.9 \times 10^{-12}\). If a \(10.0-\mathrm{g}\) sample of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is placed in \(500.0 \mathrm{~mL}\) of solution, calculate the moles of \(\mathrm{OH}^{-}\) ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is small, not a lot of solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acids.

A solution contains \(1.0 \times 10^{-5} \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\). What is the minimum concentration of \(\mathrm{AgNO}_{3}\) that would cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)
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