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Chapter 13: Problem 140

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

Short Answer

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

Step by step solution

01

Zn(OH)₂ as a Brönsted-Lowry base

When Zn(OH)₂ acts as a Brönsted-Lowry base, it will accept a proton (H⁺) 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)₂ and a source of H⁺, which we can represent as H⁺. Zn(OH)₂ + H⁺ → After accepting the proton, the hydroxide ion (OH⁻) in Zn(OH)₂ is replaced by a water molecule (H₂O), resulting in the formation of Zn(OH)H₂O or Zn(OH)(H₂O). This product is a complex ion, so we can write the equation as: Zn(OH)₂ + H⁺ → Zn(OH)(H₂O)²⁺
02

Zn(OH)₂ as a Lewis acid

When Zn(OH)₂ acts as a Lewis acid, it will accept an electron pair, which in this case is donated by an OH⁻ 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)₂ and OH⁻. Zn(OH)₂ + OH⁻ → After accepting the electron pair from the OH⁻ ion, the Zn(OH)₂ will incorporate this extra OH⁻ ion, forming Zn(OH)₃⁻. We can now write the final equation as: Zn(OH)₂ + OH⁻ → Zn(OH)₃⁻

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Most popular questions from this chapter

Calculate the percent dissociation for a \(0.22-M\) solution of chlorous acid (HClO_, \(K_{\mathrm{a}}=1.2 \times 10^{-2}\) ).

The \(\mathrm{pH}\) of a \(0.016-M\) aqueous solution of \(p\) -toluidine \(\left(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NH}_{2}\right)\) is \(8.60 .\) Calculate \(K_{\mathrm{b}}\).

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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