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Chapter 16: Problem 98

Identify the Lewis acid and Lewis base in each of the following reactions: (a) \(\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)\) (c) \(\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\) (d) \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\)

Short Answer

Expert verified
(a) Lewis acid: $\mathrm{HNO}_{2}$, Lewis base: $\mathrm{OH}^{-}$; (b) Lewis acid: $\mathrm{FeBr}_{3}$, Lewis base: $\mathrm{Br}^{-}$; (c) Lewis acid: $\mathrm{Zn}^{2+}$, Lewis base: $\mathrm{NH}_{3}$; (d) Lewis acid: $\mathrm{SO}_{2}$, Lewis base: $\mathrm{H}_{2} \mathrm{O}$.

Step by step solution

01

Identification of Lewis Acid and Lewis Base

In this reaction, the hydroxide ion \(\mathrm{OH}^{-}\) donates its electron pair to \(\mathrm{HNO}_{2}\). This makes \(\mathrm{OH}^{-}\) the Lewis base, while \(\mathrm{HNO}_{2}\) acts as the Lewis acid. L2) Reaction (b): \(\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)\)
02

Identification of Lewis Acid and Lewis Base

In this reaction, the bromide ion \(\mathrm{Br}^{-}\) donates its electron pair to \(\mathrm{FeBr}_{3}\). This makes \(\mathrm{Br}^{-}\) the Lewis base, while \(\mathrm{FeBr}_{3}\) acts as the Lewis acid. 3) Reaction (c): \(\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\)
03

Identification of Lewis Acid and Lewis Base

In this reaction, the ammonia molecules \(\mathrm{NH}_{3}\) donate their electron pairs to the \(\mathrm{Zn}^{2+}\) ion. This makes \(\mathrm{NH}_{3}\) the Lewis base, while \(\mathrm{Zn}^{2+}\) acts as the Lewis acid. 4) Reaction (d): \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\)
04

Identification of Lewis Acid and Lewis Base

In this reaction, the water molecule \(\mathrm{H}_{2} \mathrm{O}\) donates its electron pair to \(\mathrm{SO}_{2}\). This makes \(\mathrm{H}_{2} \mathrm{O}\) the Lewis base, while \(\mathrm{SO}_{2}\) acts as the Lewis acid.

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Most popular questions from this chapter

(a) Which of the following is the stronger Bronsted-Lowry acid, \(\mathrm{HClO}_{3}\) or \(\mathrm{HClO}_{2} ?\) (b) Which is the stronger Bronsted-Lowry base, \(\mathrm{HS}^{-}\) or \(\mathrm{HSO}_{4}^{-}\) ? Briefly explain your choices.

The average \(\mathrm{pH}\) of normal arterial blood is \(7.40 .\) At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14} \cdot\) Calculate \(\left[\mathrm{H}^{+}\right]\) \(\left[\mathrm{OH}^{-}\right],\) and \(\mathrm{pOH}\) for blood at this temperature. $$ \begin{array}{llll} \hline \text { pH } & \text { pOH } & \text { [H }^{+} \text {] } & \text { [OH }^{-} \text {] } & \text { Acidic or basic? } \\ \hline 5.25 & & & \\ & 2.02 & & \\ & & 4.4 \times 10^{-10} M & & \\ & & & 8.5 \times 10^{-2} M & \\ \hline \end{array} $$

Calculate the molar concentration of \(\mathrm{OH}^{-}\) ions in a \(0.724 \mathrm{M}\) solution of hypobromite ion \(\left(\mathrm{BrO}^{-} ; K_{b}=4.0 \times 10^{-6}\right) .\) What is the \(\mathrm{pH}\) of this solution?

Write the chemical equation and the \(K_{a}\) expression for the ionization of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{HBrO}_{2}\), (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\).

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base of each on the right side: $$ \text { (a) } \mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q) $$ (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) $$ \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q) $$
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