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Zinc hydroxide is an amphoteric substance. Write equations that describe \(\mathrm{Zn}(\mathrm{OH})_{2}\) acting as a Br枚nsted-Lowry base toward \(\mathrm{H}^{+}\) and as a Lewis acid toward \(\mathrm{OH}^{-}\).

Short Answer

Expert verified
When Zn(OH)鈧 acts as a Br枚nsted-Lowry base, it accepts a proton (H鈦) and forms a complex ion, as shown in the equation: \(Zn(OH)_{2} + H^{+} \rightarrow Zn(OH)(H_{2}O)^{2+}\). When Zn(OH)鈧 acts as a Lewis acid, it accepts an electron pair from the OH鈦 ion, forming Zn(OH)鈧冣伝, with the equation: \(Zn(OH)_{2} + OH^{-} \rightarrow Zn(OH)_{3}^{-}\).

Step by step solution

01

Zn(OH)鈧 as a Br枚nsted-Lowry base

When Zn(OH)鈧 acts as a Br枚nsted-Lowry base, it will accept a proton (H鈦) from an acid. To write the equation for this reaction, we can start by writing the reactants on the left side and the products on the right side. In this case, the reactants are Zn(OH)鈧 and a source of H鈦, which we can represent as H鈦. Zn(OH)鈧 + H鈦 鈫 After accepting the proton, the hydroxide ion (OH鈦) in Zn(OH)鈧 is replaced by a water molecule (H鈧侽), resulting in the formation of Zn(OH)H鈧侽 or Zn(OH)(H鈧侽). This product is a complex ion, so we can write the equation as: Zn(OH)鈧 + H鈦 鈫 Zn(OH)(H鈧侽)虏鈦
02

Zn(OH)鈧 as a Lewis acid

When Zn(OH)鈧 acts as a Lewis acid, it will accept an electron pair, which in this case is donated by an OH鈦 ion. To write the equation for this reaction, we can again start by writing the reactants on the left side and the products on the right side. In this case, the reactants are Zn(OH)鈧 and OH鈦. Zn(OH)鈧 + OH鈦 鈫 After accepting the electron pair from the OH鈦 ion, the Zn(OH)鈧 will incorporate this extra OH鈦 ion, forming Zn(OH)鈧冣伝. We can now write the final equation as: Zn(OH)鈧 + OH鈦 鈫 Zn(OH)鈧冣伝

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Br枚nsted-Lowry Base
Br枚nsted-Lowry theory is one of the foundational concepts in acid-base chemistry. It defines acids as substances that can donate a proton (\( ext{H}^+ \)), while bases are substances that can accept a proton. In this case, zinc hydroxide, \( ext{Zn(OH)}_2 \), plays the role of a Br枚nsted-Lowry base. This means it acts as a proton acceptor.

When \( ext{Zn(OH)}_2 \) interacts with a proton donor (acid), it accepts a \( ext{H}^+ \) ion. Initially, the zinc hydroxide structure holds hydroxide ions (\( ext{OH}^- \)). When these ions accept a proton, a water molecule is formed in place of each. This transition leads to a new complex, \( ext{Zn(OH)}( ext{H}_2 ext{O})^2^+ \). This exchange demonstrates the versatility of zinc hydroxide to act as a base by intercepting protons during reactions.
Lewis Acid
The Lewis acid-base concept expands the definition beyond protons to include electron pairs. According to Lewis, an acid is any substance that can accept a pair of electrons. This ability reflects zinc hydroxide's role as a Lewis acid.

A notable feature of Lewis acids is their electron pair acceptance capability. In the context of \( ext{Zn(OH)}_2 \), it accepts an electron pair from \( ext{OH}^- \) ions. Unlike the Br枚nsted-Lowry model, this theory highlights the Lewis acid's role in forming bonds via electron pair acquisition. When \( ext{Zn(OH)}_2 \) absorbs this additional \( ext{OH}^- \), it forms \( ext{Zn(OH)}_3^- \), showcasing a clear path of interaction as a Lewis acid.
Acid-Base Reactions
Acid-base reactions are fundamental in chemistry, where acids donate protons and bases accept them. Alternatively, acids accepting electron pairs and bases donating them can define these reactions using the Lewis description.

Zinc hydroxide, as an amphoteric substance, illustrates the dual nature allowing it to play roles in these reactions. In the case of Br枚nsted-Lowry, it acts through proton acceptance, while in Lewis acid-base reactions, it displays acceptance of electron pairs.

Such reactions encourage a balance where substances like \( ext{Zn(OH)}_2 \) flexibly adjust roles based on the reacting partners. Observing these interactions enriches understanding of chemical behavior across different scenarios.
Complex Ions
Complex ions represent molecules formed by central atoms bonded to surrounding ligands via coordinate covalent bonds. In zinc hydroxide鈥檚 reactions, the resulting species are complex ions when it either accepts protons or electron pairs.

Acting as a Br枚nsted-Lowry base with acids, \( ext{Zn(OH)}_2 \) transforms to \( ext{Zn(OH)}( ext{H}_2 ext{O})^2^+ \), a complex ion. Conversely, as a Lewis acid with hydroxide ions, it turns into \( ext{Zn(OH)}_3^- \), which is also a complex ion.

Understanding complex ions includes recognizing the role of coordinate bonds in creating stable structures that allow versatile functions in different chemical environments.

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Most popular questions from this chapter

The pOH of a sample of baking soda dissolved in water is 5.74 at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH},\left[\mathrm{H}^{+}\right],\) and \(\left[\mathrm{OH}^{-}\right]\) for this sample. Is the solution acidic or basic?

What are the major species present in the following mixtures of bases? a. \(0.050 M \mathrm{NaOH}\) and \(0.050 \mathrm{M} \mathrm{LiOH}\) b. \(0.0010 M \mathrm{Ca}(\mathrm{OH})_{2}\) and \(0.020 \mathrm{M} \mathrm{RbOH}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

A solution is prepared by dissolving 0.56 g benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\) and the pH of this solution.

Which of the following represent conjugate acid-base pairs? For those pairs that are not conjugates, write the correct conjugate acid or base for each species in the pair. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) d. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

Calculate the \(\mathrm{pH}\) of a \(0.200-M\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is $$\begin{aligned} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \\ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q) &+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\end{aligned}$$

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