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Chapter 16: Problem 26

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons\) (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\)

Short Answer

Expert verified
(a) \(\mathrm{NH}_{4}^{+}(a q) + \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NH}_3 (a q) + \mathrm{H}_2 \mathrm{O}(l)\), equilibrium lies to the right. (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(a q) + \mathrm{H}_2 \mathrm{O}(l)\), equilibrium lies to the left. (c) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3 (a q) + \mathrm{HF}(a q)\), equilibrium lies to the right.

Step by step solution

01

a) Reaction equation and identify acid/base reactants

First, we need to complete the reaction equation for the reactants given: \(\mathrm{NH}_{4}^{+}(a q) + \mathrm{OH}^{-}(a q) \rightleftharpoons\) Next, we need to identify the acid and base reactants. The \(\mathrm{NH}_{4}^{+}\) ion acts as an acid, as it donates a proton (H\(^+\)), and the \(\mathrm{OH}^{-}\) ion acts as a base, as it accepts a proton.
02

a) Reaction products

Now that we've identified the acid and base reactants, we can predict the reaction products. In this case, the \(\mathrm{NH}_{4}^{+}\) ion donates a proton to \(\mathrm{OH}^{-}\), forming NH3 and H2O: \(\mathrm{NH}_{4}^{+}(a q) + \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NH}_3 (a q) + \mathrm{H}_2 \mathrm{O}(l)\)
03

a) Equilibrium position

To predict if the equilibrium lies to the left or the right, we need to consider the strength of the acid and the base reacting and compare their conjugate acid-base pairs. The reaction between ammonium ions (\(\mathrm{NH}_{4}^{+}\)) and hydroxide ions (\(\mathrm{OH}^{-}\)) gives ammonia (\(\mathrm{NH}_3\)) and water (\(\mathrm{H}_2\mathrm{O}\)). Since the hydroxide ion is a stronger base than ammonia and the ammonium ion is a stronger acid than water, the equilibrium lies to the right due to the formation of the weaker acid and base.
04

b) Reaction equation and identify acid/base reactants

First, we need to complete the reaction equation for the reactants given: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons\) Next, we need to identify the acid and base reactants. The \(\mathrm{CH}_{3} \mathrm{COO}^{-}\) ion (acetate ion) acts as a base, as it accepts a proton, and the \(\mathrm{H}_{3} \mathrm{O}^{+}\) ion (hydronium ion) acts as an acid, as it donates a proton.
05

b) Reaction products

We can predict the reaction products by having the acetate ion accept a proton from the hydronium ion, forming acetic acid (CH3COOH) and water (H2O): \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}(a q) + \mathrm{H}_2 \mathrm{O}(l)\)
06

b) Equilibrium position

When considering the strength of the acid and the base reacting and comparing their conjugate acid-base pairs, the equilibrium position is to the left because hydronium ion (\(\mathrm{H}_{3}\mathrm{O}^{+}\)) is a stronger acid than acetic acid (\(\mathrm{CH}_3 \mathrm{COOH}\)) and the acetate ion (\(\mathrm{CH}_{3} \mathrm{COO}^{-}\)) is a stronger base than water (\(\mathrm{H}_2\mathrm{O}\)).
07

c) Reaction equation and identify acid/base reactants

First, we need to complete the reaction equation for the reactants given: \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) Next, we need to identify the acid and base reactants. The \(\mathrm{HCO}_{3}^{-}\) ion (bicarbonate ion) acts as an acid, as it donates a proton, and the \(\mathrm{F}^{-}\) ion (fluoride ion) acts as a base, as it accepts a proton.
08

c) Reaction products

Now that we've identified the acid and base reactants, we can predict the reaction products. The bicarbonate ion donates a proton to the fluoride ion, forming carbonic acid (H2CO3) and hydrogen fluoride (HF): \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{H}_2 \mathrm{CO}_3 (a q) + \mathrm{HF}(a q)\)
09

c) Equilibrium position

To determine the equilibrium position, we analyze the strength of the acid and the base reactants and compare their conjugate acid-base pairs. Bicarbonate ion (\(\mathrm{HCO}_{3}^{-}\)) is a weaker acid than hydrogen fluoride (\(\mathrm{HF}\)), and fluoride ion (\(\mathrm{F}^{-}\)) is a stronger base than the resulting carbonic acid (\(\mathrm{H}_2 \mathrm{CO}_3\)). Therefore, the equilibrium lies to the right due to the formation of the weaker acid and base.

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Most popular questions from this chapter

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{p} K_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: \(\mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q)\) What is the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of this substance?

(a) What is the difference between the Arrhenius and the Bronsted-Lowry definitions of an acid? (b) \(\mathrm{NH}_{3}(g)\) and \(\mathrm{HCl}(g)\) react to form the ionic solid \(\mathrm{NH}_{4} \mathrm{Cl}(s) .\) Which substance is the Bronsted-Lowry acid in this reaction? Which is the Bronsted-Lowry base?

Predict whether aqueous solutions of the following substances are acidic, basic, or neutral: (a) \(\mathrm{AlCl}_{3}\), (b) \(\mathrm{NaBr}\), (c) \(\mathrm{NaClO},(\mathrm{d})\) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}\right] \mathrm{NO}_{3},(\mathrm{e}) \mathrm{Na}_{2} \mathrm{SO}_{3}\)

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) In general, the acidity of binary acids increases from left to right in a given row of the periodic table. (b) In a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom. (c) Hydrotelluric acid \(\left(\mathrm{H}_{2} \mathrm{Te}\right)\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{~S}\) because Te is more electronegative than \(\mathrm{S}\).

(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.
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