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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 stepwise formation constants for a complex ion are all generally values much greater than 1 . What is the significance of this?

Use the following data to calculate the \(K_{\mathrm{sp}}\) value for each solid. a. The solubility of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) is \(4.8 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{BiI}_{3}\) is \(1.32 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\).

\(\mathrm{Ag}_{2} \mathrm{~S}(s)\) has a larger molar solubility than \(\mathrm{CuS}\) even though \(\mathrm{Ag}_{2} \mathrm{~S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.

Calculate the solubility (in moles per liter) of \(\mathrm{Fe}(\mathrm{OH})_{3}\left(K_{\mathrm{sp}}=4 \times\right.\) \(10^{-33}\) ) in each of the following. a. water b. a solution buffered at \(\mathrm{pH}=5.0\) c. a solution buffered at \(\mathrm{pH}=11.0\)

Consider a solution made by mixing \(500.0 \mathrm{~mL}\) of \(4.0 \mathrm{M} \mathrm{NH}_{3}\) and \(500.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{AgNO}_{3} \cdot \mathrm{Ag}^{+}\) reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{AgNH}_{3}^{+}\) and \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}:\) \(\mathrm{Ag}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{AgNH}_{3}^{+}(a q) \quad K_{1}=2.1 \times 10^{3}\) \(\mathrm{AgNH}_{3}^{+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K_{2}=8.2 \times 10^{3}\)
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