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Chapter 16: Problem 42

Write a net ionic equation for the reaction with \(\mathrm{OH}^{-}\) by which (a) \(\mathrm{Ni}^{2+}\) forms a precipitate. (b) \(\mathrm{Sn}^{4+}\) forms a complex ion. (c) \(\mathrm{Al}(\mathrm{OH})_{3}\) dissolves.

Short Answer

Expert verified
Question: Write the net ionic equation for the dissolution of aluminum hydroxide with the hydroxide ion. Answer: \(\mathrm{Al}(\mathrm{OH})_{3}(s) + \mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Al}(\mathrm{OH})_{4}]^{-}(aq)\)

Step by step solution

01

(Net ionic equation for Nickel precipitation reaction)

For the precipitation of nickel ion (\(\mathrm{Ni}^{2+}\)) with hydroxide ion, the balanced molecular equation is: \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)\) Now, split the reaction into its respective ions: \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(s)\) The net ionic equation cancels out the spectator ions: \(\mathrm{Ni}^{2+}(aq) + 2\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)\)
02

(Net ionic equation for Tin complex-ion formation reaction)

When tin ion forms a complex ion with hydroxide ions, we can represent the balanced molecular equation as: \(\mathrm{Sn}^{4+}(aq) + 4\mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Sn}(\mathrm{OH})_{4}]^{2-}(aq)\) Split the reaction into its respective ions: \(\mathrm{Sn}^{4+}(aq) + 4\mathrm{OH}^{-}(aq) \rightarrow \mathrm{Sn}^{4+}(aq) + 4\mathrm{OH}^{-}(aq)\) The net ionic equation cancels out the spectator ions: \(\mathrm{Sn}^{4+}(aq) + 4\mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Sn}(\mathrm{OH})_{4}]^{2-}(aq)\)
03

(Net ionic equation for Aluminum hydroxide dissolution reaction)

To represent the dissolution of aluminum hydroxide (\(\mathrm{Al}(\mathrm{OH})_{3}\)) with hydroxide ion, the balanced molecular equation is: \(\mathrm{Al}(\mathrm{OH})_{3}(s) + \mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Al}(\mathrm{OH})_{4}]^{-}(aq)\) Split the reaction into its respective ions: \(\mathrm{Al}^{3+}(aq) + 3\mathrm{OH}^{-}(s) + \mathrm{OH}^{-}(aq) \rightarrow \mathrm{Al}^{3+}(aq) + 4\mathrm{OH}^{-}(aq)\) The net ionic equation cancels out the spectator ions: \(\mathrm{Al}(\mathrm{OH})_{3}(s) + \mathrm{OH}^{-}(aq) \rightarrow [\mathrm{Al}(\mathrm{OH})_{4}]^{-}(aq)\)

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Most popular questions from this chapter

Fill in the blanks in the following table. (a) \(\mathrm{CoCO}_{3}\) (b) \(\mathrm{LaF}_{3}\) (c) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Given \(K_{s p}\) and the equilibrium concentration of one ion, calculate the equilibrium concentration of the other ion. (a) cadmium(II) hydroxide: \(K_{\text {sp }}=2.5 \times 10^{-14} ;\left[\mathrm{Cd}^{2+}\right]=1.5 \times 10^{-6} M\) (b) copper(II) arsenate \(\left(\mathrm{Cu}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\right): K_{\mathrm{sp}}=7.6 \times 10^{-36} ;\left[\mathrm{AsO}_{4}^{3-}\right]=\) \(2.4 \times 10^{-4} M\) (c) zinc oxalate: \(K_{s p}=2.7 \times 10^{-8} ;\left[\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right]=8.8 \times 10^{-3} M\)

\(\mathrm{K}_{\mathrm{sp}}\) for silver acetate \(\left(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) at \(80^{\circ} \mathrm{C}\) is estimated to be \(2 \times 10^{-2}\). Ten grams of silver acetate are added to \(1.0 \mathrm{~L}\) of water at \(25^{\circ} \mathrm{C}\). (a) Will all the silver acetate dissolve at \(25^{\circ} \mathrm{C} ?\) (b) If the solution (assume the volume to be \(1.0 \mathrm{~L}\) ) is heated to \(80^{\circ} \mathrm{C}\), will all the silver acetate dissolve?

Write the equilibrium equations on which the following \(K_{s p}\) expressions are based. (a) \(\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{CO}_{3}{ }^{2-}\right]\) (b) \(\left[\mathrm{Co}^{3+}\right]\left[\mathrm{OH}^{-}\right]^{3}\) (c) \(\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{~S}^{2-}\right]\) (d) \(\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}\)

Shown below is a representation of the ionic solid \(\mathrm{MX}\), where \(\mathrm{M}\) cations are represented by squares and \(X\) anions are represented by circles. Fill in the box after the arrow to represent what happens to the solid after it has been completely dissolved in water. For simplicity, do not represent the water molecules.
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