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

Short Answer

Expert verified
When zinc hydroxide (Zn(OH)鈧) acts as a Br酶nsted-Lowry base towards H鈦, it accepts a proton to form Zn(OH)鈧冣伜, resulting in the balanced equation: \(Zn(OH)_2 + H^+ \rightarrow Zn(OH)_3^+\). On the other hand, when Zn(OH)鈧 acts as a Lewis acid towards OH鈦, it accepts a hydroxide ion to form Zn(OH)鈧冣伝, with the balanced equation: \(Zn(OH)_2 + OH^- \rightarrow Zn(OH)_3^-\).

Step by step solution

01

Br酶nsted-Lowry Base Reaction with H鈦

To write the equation for the reaction of Zn(OH)鈧 with H鈦, we must first identify the species formed when Zn(OH)鈧 accepts a proton. The Zn(OH)鈧 molecule will take an H鈦 and become Zn(OH)鈧冣伜. The balanced equation for this process is: Zn(OH)鈧 + H鈦 鈫 Zn(OH)鈧冣伜
02

Lewis Acid Reaction with OH鈦

Next, we need to write the equation for the reaction of Zn(OH)鈧 with OH鈦. In this case, Zn(OH)鈧 acts as a Lewis acid and will accept a hydroxide ion. When Zn(OH)鈧 accepts an OH鈦, it becomes Zn(OH)鈧冣伝. The balanced equation for this process is: 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酶nsted-Lowry Base
In the quest to simplify complex chemical concepts, let's start by breaking down what a Br酶nsted-Lowry base is. According to the Br酶nsted-Lowry theory, a base is any substance that can accept protons, which are simply hydrogen ions (H^+). This approach gives us a broader perspective on acids and bases that goes beyond their behavior in water.
When studying chemical reactions, you might think of a 'base' as a kind of chemical sponge for H^+ ions. The given exercise illustrated zinc hydroxide (Zn(OH)_2) acting as a Br酶nsted-Lowry base by absorbing one of these H^+ ions, transforming into Zn(OH)_3^+. This reaction is pivotal to understanding amphoteric substances like zinc hydroxide that can behave as both a base and an acid depending on their chemical environment. Simply put, the ability to adapt its role makes zinc hydroxide a chemical chameleon of sorts.

Why is this important?

Understanding the flexible behavior of amphoteric substances is not just academic鈥攊t plays a crucial role in processes like industrial water treatment and the bioavailability of minerals in biology. For instance, within our bodies, amphoteric substances help regulate pH levels, facilitating vital enzymatic reactions.
Lewis Acid
Stepping into the arena of Lewis acids and bases, we dive into a more nuanced understanding of chemical reactions. A Lewis acid is defined as anything that can accept a pair of electrons to form a bond. Unlike the Br酶nsted-Lowry concept which focuses on protons, the Lewis definition broadens the scope to include a myriad of other reactions not involving H^+ ions.
In the case of zinc hydroxide acting as a Lewis acid, as seen in our exercise, it accepts an electron pair from an OH^- ion. This rendezvous results in the formation of Zn(OH)_3^-. This ability to snatch a lone pair of electrons from a donor makes it a versatile player in chemical interactions. It's like a molecular handshake where zinc hydroxide reaches out to form a new bond, expanding its connections within the chemical community.

Real-life Implications

Lewis acid-base chemistry is central to fields like catalysis and organic synthesis. In industrial settings, Lewis acids are used to catalyze reactions that produce everything from pharmaceuticals to plastics, showcasing their fundamental importance in science and industry.
Chemical Equations
Moving on to the language of chemistry: chemical equations. These are succinct notations that represent the reactants and products in a chemical reaction. They're a bit like recipes, telling chemists what ingredients to mix and what they'll end up with at the end of their experiment.
In the original exercise, two chemical equations were provided, one for the reaction of zinc hydroxide as a Br酶nsted-Lowry base, and the other for its function as a Lewis acid. Like sentences in a language, chemical equations have syntax and grammar rules, such as the law of conservation of mass which dictates that atoms must be balanced on both sides of the equation.

The Bigger Picture

Being able to interpret and balance chemical equations is a fundamental skill for scientists. It's crucial for predicting the outcomes of reactions and for scaling up a lab experiment to industrial production. It's also how chemists communicate their recipes internationally, transcending language barriers鈥攂ecause no matter where you are in the world, the elements and their reactions remain the same.

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Most popular questions from this chapter

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{~g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).

Calculate the \(\mathrm{pH}\) of an aqueous solution containing \(1.0 \mathrm{X}\) \(10^{-2} M \mathrm{HCl}, 1.0 \times 10^{-2} M \mathrm{H}_{2} \mathrm{SO}_{4}\), and \(1.0 \times 10^{-2} M \mathrm{HCN}\).

Calculate \(\left[\mathrm{CO}_{3}^{2-}\right]\) in a \(0.010-M\) solution of \(\mathrm{CO}_{2}\) in water (usually written as \(\mathrm{H}_{2} \mathrm{CO}_{3}\) ). If all the \(\mathrm{CO}_{3}{ }^{2-}\) in this solution comes from the reaction $$\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)$$ what percentage of the \(\mathrm{H}^{+}\) ions in the solution is a result of the dissociation of \(\mathrm{HCO}_{3}^{-} ?\) When acid is added to a solution of sodium hydrogen carbonate \(\left(\mathrm{NaHCO}_{3}\right)\), vigorous bubbling occurs. How is this reaction related to the existence of carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) molecules in aqueous 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. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}^{-}\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\) d. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

Calculate the concentrations of all species present in a \(0.25-M\) solution of ethylammonium chloride \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\right)\).

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