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Chapter 14: Problem 140

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

Short Answer

Expert verified
When Zinc hydroxide (Zn(OH)₂) acts as a Bronsted-Lowry base toward H⁺ ions, the equation is: \( Zn(OH)_{2} + H^{+} \rightarrow Zn(OH)_3^- \). And, when Zn(OH)₂ acts as a Lewis acid toward OH⁻ ions, the equation is: \( Zn(OH)_{2} + 2OH^{-} \rightarrow Zn(OH)_{4}^{2-} \).

Step by step solution

01

Recalling the Definitions of Bronsted-Lowry Base and Lewis Acid

In order to write the equations, it is necessary to understand the meaning of Bronsted-Lowry base and Lewis acid. - A Bronsted-Lowry base is a substance that can accept a proton (H+ ion) in a chemical reaction. - A Lewis acid is a substance that can accept a lone pair of electrons, forming a coordinate covalent bond.
02

Writing the Equation for Zn(OH)2 as a Bronsted-Lowry Base

When Zn(OH)2 acts as a Bronsted-Lowry base, it accepts a proton (H+ ion) and subsequently forms Zinc hydroxide ion (Zn(OH)₃⁻). The equation can be written as: \[ Zn(OH)_{2} + H^{+} \rightarrow Zn(OH)_3^- \]
03

Writing the Equation for Zn(OH)2 as a Lewis Acid

When Zn(OH)2 acts as a Lewis acid, it accepts a lone pair of electrons from the hydroxide ion (OH⁻). As a result, a new bond with the hydroxide ion is formed, producing the Zinc hydroxide complex ion (Zn(OH)₄²⁻). The equation for this reaction can be written as: \[ Zn(OH)_{2} + 2OH^{-} \rightarrow Zn(OH)_{4}^{2-} \] In conclusion, the two equations that describe Zn(OH)2 acting as a Bronsted-Lowry base toward H+ ions and as a Lewis acid toward OH- ions are: 1. Bronsted-Lowry base: \( Zn(OH)_{2} + H^{+} \rightarrow Zn(OH)_3^- \) 2. Lewis acid: \( Zn(OH)_{2} + 2OH^{-} \rightarrow Zn(OH)_{4}^{2-} \)

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Most popular questions from this chapter

a. The principal equilibrium in a solution of \(\mathrm{NaHCO}_{3}\) is $$ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ Calculate the value of the equilibrium constant for this reaction. b. At equilibrium, what is the relationship between $\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\( and \)\left[\mathrm{CO}_{3}^{2-}\right] ?$ c. Using the equilibrium $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ derive an expression for the pH of the solution in terms of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) using the result from part b. d. What is the pH of a solution of \(\mathrm{NaHCO}_{3} ?\)

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. $\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}$

A codeine-containing cough syrup lists codeine sulfate as a major ingredient instead of codeine. The Merck Index gives $\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{N}_{2} \mathrm{O}_{10} \mathrm{S}$ as the formula for codeine sulfate. Describe the composition of codeine sulfate. (See Exercise \(104 . )\) Why is codeine sulfate used instead of codeine?

Making use of the assumptions we ordinarily make in calculating the \(\mathrm{pH}\) of an aqueous solution of a weak acid, calculate the pH of a \(1.0 \times 10^{-6}-\mathrm{M}\) solution of hypobromous acid \(\left(\mathrm{HBrO}, K_{\mathrm{a}}=2 \times 10^{-9}\right) .\) What is wrong with your answer? Why is it wrong? Without trying to solve the problem, explain what has to be included to solve the problem correctly.

Identify the Lewis acid and the Lewis base in each of the following reactions. a. $\mathrm{Fe}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)$ b. $\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q) \Longrightarrow \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q)$ c. $\mathrm{HgI}_{2}(s)+2 \mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{Hgl}_{4}^{2-}(a q)$
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