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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

Write net ionic equations for the reactions of each of the following with strong acid. (a) \(\mathrm{CaCO}_{3}\) (b) NiS (c) \(\mathrm{Al}(\mathrm{OH})_{3}\) (d) \(\mathrm{Sb}(\mathrm{OH})_{4}^{-}\) (e) \(\mathrm{AgCl}\)

Consider the reaction \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) (a) Calculate \(K\) given that for \(\mathrm{Cu}(\mathrm{OH})_{2} K_{s p}=2 \times 10^{-19}\) and for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} K_{\mathrm{f}}=2 \times 10^{12}\) (b) Determine the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) (in \(\mathrm{mol} / \mathrm{L}\) ) in \(4.5 \mathrm{M} \mathrm{NH}_{2}\).

A solution is made up by adding \(0.925 \mathrm{~g}\) of silver(I) nitrate and \(6.25 \mathrm{~g}\) of magnesium nitrate to enough water to make \(375 \mathrm{~mL}\) of solution. Solid sodium carbonate is added without changing the volume of the solution. (a) Which salt, \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) or \(\mathrm{MgCO}_{3}\), will precipitate first? (b) What is the concentration of the carbonate ion when the first salt starts to precipitate?

The box below represents one liter of a saturated solution of the species \(\square\), where squares represent the cation and circles represent the anion. Water molecules, though present, are not shown. Complete the next three figures below by filling one-liter boxes to the right of the arrow, showing the state of the ions after water is added to form saturated solutions. The species represented to the left of the arrow is the solid form of the ions represented above. Do not show the water molecules.

One gram of \(\mathrm{PbCl}_{2}\) is dissolved in \(1.0 \mathrm{~L}\) of hot water. When the solution is cooled to \(25^{\circ} \mathrm{C}\), will some of the \(\mathrm{PbCl}_{2}\) crystallize out? If so, how much?
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