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Chapter 16: Problem 102

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

Step by step solution

01

Reaction (a) - Identifying the Lewis Acid and Base

For the reaction: \[\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(t)\] Here, the hydroxide ion (\(\mathrm{OH}^{-}\)) donates an electron pair to the \(\mathrm{HNO}_{2}\) molecule to form water (\(\mathrm{H}_{2} \mathrm{O}\)) and \(\mathrm{NO}_{2}^{-}\). Thus, the Lewis base is the hydroxide ion (\(\mathrm{OH}^{-}\)), and the Lewis acid is the \(\mathrm{HNO}_{2}\) molecule.
02

Reaction (b) - Identifying the Lewis Acid and Base

For the reaction: \[\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)\] Here, the Bromide ion (\(\mathrm{Br}^{-}\)) donates an electron pair to the \(\mathrm{FeBr}_{3}\) molecule to form \(\mathrm{FeBr}_{4}^{-}\). Thus, the Lewis base is the Bromide ion (\(\mathrm{Br}^{-}\)), and the Lewis acid is the \(\mathrm{FeBr}_{3}\) molecule.
03

Reaction (c) - Identifying the Lewis Acid and Base

For the reaction: \[\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\] Here, the ammonia molecules (\(\mathrm{NH}_{3}\)) donate electron pairs to the \(\mathrm{Zn}^{2+}\) ion to form the \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) complex. Thus, the Lewis base is the ammonia molecule (\(\mathrm{NH}_{3}\)), and the Lewis acid is the \(\mathrm{Zn}^{2+}\) ion.
04

Reaction (d) - Identifying the Lewis Acid and Base

For the reaction: \[\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\] Here, the water molecule (\(\mathrm{H}_{2} \mathrm{O}\)) donates an electron pair to the \(\mathrm{SO}_{2}\) molecule to form the \(\mathrm{H}_{2} \mathrm{SO}_{3}\) ion. Thus, the Lewis base is the water molecule (\(\mathrm{H}_{2} \mathrm{O}\)), and the Lewis acid is the \(\mathrm{SO}_{2}\) molecule.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Acid-Base Reactions
Understanding Lewis acid-base reactions is crucial for anyone studying chemistry. These reactions involve the transfer of electron pairs between two substances. A Lewis acid accepts an electron pair, while a Lewis base donates an electron pair.

Let's consider the reaction between hydroxide ions and nitrous acid as an example: \[\mathrm{HNO}_{2}(aq) + \mathrm{OH}^{-}(aq) \rightleftharpoons \mathrm{NO}_{2}^{-}(aq) + \mathrm{H}_{2} \mathrm{O}(l)\] Here, the \(\mathrm{OH}^{-}\) ion donates an electron pair to \(\mathrm{HNO}_{2}\), making \(\mathrm{OH}^{-}\) the Lewis base and \(\mathrm{HNO}_{2}\) the Lewis acid. Identifying which species is the acid or base in such reactions is key to predicting the products and understanding the reaction's direction.
Electron Pair Donation
The concept of electron pair donation plays a central role in the Lewis theory of acids and bases. In a Lewis acid-base reaction, the base donates a pair of electrons, which the acid accepts to form a new chemical bond.

For instance, in the reaction with iron(III) bromide and bromide ions: \[\mathrm{FeBr}_{3}(s) + \mathrm{Br}^{-}(aq) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(aq)\] the bromide ion \(\mathrm{Br}^{-}\) acts as a Lewis base by donating an electron pair to the \(\mathrm{FeBr}_{3}\), the Lewis acid. This interaction demonstrates how electron pair donation leads to the formation of a more complex ion. Understanding these interactions can elucidate the properties and behaviors of the resulting compounds in chemical reactions.
Chemical Equilibrium
In the context of Lewis acid-base chemistry, chemical equilibrium is the state in which the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time.

Let's analyze the reaction involving zinc ions and ammonia: \[\mathrm{Zn}^{2+}(aq) + 4 \mathrm{NH}_{3}(aq) \rightleftharpoons \mathrm{Zn}(\mathrm{NH}_{3})_{4}^{2+}(aq)\] The equilibrium is established as the zinc ion, the Lewis acid, binds with the Lewis base, ammonia, and this reversible reaction reaches a point where there are no further changes in concentration of the reactants and products. Understanding how equilibrium is established, and the conditions that affect it, such as temperature or concentration changes, is essential for predicting how a reaction will proceed and is a fundamental principle in chemical reaction dynamics.

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Most popular questions from this chapter

Hemoglobin plays a part in a series of equilibria involving protonation- deprotonation and oxygenation-deoxygenation. The overall reaction is approximately as follows $$ \mathrm{HbH}^{+}(a q)+\mathrm{O}_{2}(a q) \rightleftharpoons \mathrm{HbO}_{2}(a q)+\mathrm{H}^{+}(a q) $$ where Hb stands for hemoglobin, and \(\mathrm{HbO}_{2}\) for oxyhemoglobin. (a) The concentration of \(\mathrm{O}_{2}\) is higher in the lungs and lower in the tissues. What effect does high \(\left[\mathrm{O}_{2}\right]\) have on the position of this equilibrium? (b) The normal \(\mathrm{pH}\) of blood is \(7.4\). Is the blood acidic, basic, or neutral? (c) If the blood \(\mathrm{pH}\) is lowered by the presence of large amounts of acidic metabolism products, a condition known as acidosis results. What effect does lowering blood \(\mathrm{pH}\) have on the ability of hemoglobin to transport \(\mathrm{O}_{2}\) ?

Predict whether the equilibrium lies to the right or to the left in the following reactions: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{PO}_{4}{ }^{3-}(a q) \rightleftharpoons\) \(\mathrm{NH}_{3}(a q)+\mathrm{HPO}_{4}^{2-}(a q)\) (The ammonium ion is a stronger acid than the hydrogen phosphate ion.) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons\) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{HCN}(a q)\) (The cyanide ion is a stronger base than the acetate ion.)

Calculate [OH \(^{-}\) ] for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{H}^{+}\right]=0.0045 \mathrm{M}\); (b) \(\left[\mathrm{H}^{+}\right]=1.5 \times 10^{-9} \mathrm{M} ;\) (c) a solution in which \(\left[\mathrm{H}^{+}\right]\) is 10 times greater than \([\mathrm{OH}]\).

The amino acid glycine \(\left(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)\) can participate in the following equilibria in water: \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O}=\) \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \quad K_{a}=4.3 \times 10^{-3}\) \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) \({ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_{b}=6.0 \times 10^{-5}\) (a) Use the values of \(K_{a}\) and \(K_{b}\) to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}\) What assumptions did you need to make? (b) What is the pH of a \(0.050 \mathrm{M}\) aqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with pH 13? With pH 1?

If a substance is an Arrhenius base, is it necessarily a Bronsted-Lowry base? Is it necessarily a Lewis base? Explain.
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