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Chapter 14: Problem 136

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

Short Answer

Expert verified
Zinc hydroxide, \(\mathrm{Zn}(\mathrm{OH})_{2}\), can act as a Brønsted-Lowry base and a Lewis acid. In the Brønsted-Lowry base reaction, it accepts a proton (\(\mathrm{H}^{+}\)): \[\mathrm{Zn}(\mathrm{OH})_{2} + \mathrm{H}^{+} \rightarrow \mathrm{Zn}(\mathrm{OH})_{3}^{+}\] In the Lewis acid reaction, it accepts an electron pair from a hydroxide ion (\(\mathrm{OH}^{-}\)): \[\mathrm{Zn}(\mathrm{OH})_{2} + \mathrm{OH}^{-} \rightarrow \mathrm{Zn}(\mathrm{OH})_{3}^{-}\]

Step by step solution

01

Acting as a Brønsted-Lowry base

Zinc hydroxide, \(\mathrm{Zn}(\mathrm{OH})_{2}\), can accept a proton (\(\mathrm{H}^{+}\)) to form a new compound. In this case, the reaction would be: \[\mathrm{Zn}(\mathrm{OH})_{2} + \mathrm{H}^{+} \rightarrow \mathrm{Zn}(\mathrm{OH})_{3}^{+}\]
02

Acting as a Lewis acid

Zinc hydroxide, \(\mathrm{Zn}(\mathrm{OH})_{2}\), can accept an electron pair from a hydroxide ion (\(\mathrm{OH}^{-}\)) to form a new compound. In this case, the reaction would be: \[\mathrm{Zn}(\mathrm{OH})_{2} + \mathrm{OH}^{-} \rightarrow \mathrm{Zn}(\mathrm{OH})_{3}^{-}\] These equations show how zinc hydroxide can act as a Brønsted-Lowry base by accepting a proton when it reacts with a hydrogen ion, and as a Lewis acid by accepting an electron pair when it reacts with a hydroxide ion.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of the following solutions. a. \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(1.0 \times 10^{-10} \mathrm{M} \mathrm{NaOH}\) c. \(2.0 \mathrm{M} \mathrm{NaOH}\)

What mass of \(\mathrm{NaOH}(s)\) must be added to \(1.0 \mathrm{~L}\) of \(0.050 \mathrm{M} \mathrm{NH}_{3}\) to ensure that the percent ionization of \(\mathrm{NH}_{3}\) is no greater than \(0.0010 \%\) ? Assume no volume change on addition of \(\mathrm{NaOH}\).

Consider \(50.0 \mathrm{~mL}\) of a solution of weak acid \(\mathrm{HA}=K_{\mathrm{a}}(1.00 \times\) \(10^{-6}\) ), which has a pH of \(4.000\). What volume of water must be added to make the \(\mathrm{pH}=5.000 ?\)

Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{S}\), \(363 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{~kJ} / \mathrm{mol})\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}{ }^{+}\) d. \(\mathrm{NH}_{4}{ }^{+}, \mathrm{PH}_{4}{ }^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H}, 322\) \(\mathrm{kJ} / \mathrm{mol}\) ) Give reasons for the orders you chose.

Calculate the \(\mathrm{pH}\) of a solution that contains \(1.0 \mathrm{M} \mathrm{HF}\) and \(1.0 \mathrm{M}\) \(\mathrm{HOC}_{6} \mathrm{H}_{5} .\) Also calculate the concentration of \(\mathrm{OC}_{6} \mathrm{H}_{5}^{-}\) in this solution at equilibrium.
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