/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Predict the products of the foll... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 27

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) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Short Answer

Expert verified
(a) \(O^{2-}(aq) + H_2O(l) \rightleftharpoons OH^-(aq) + H_3O^+(aq)\); equilibrium to the right (b) \(CH_3COOH(aq) + HS^-(aq) \rightleftharpoons CH_3COO^-(aq) + H_2S(aq)\); equilibrium (c) \(NO_2^-(aq) + H_2O(l) \rightleftharpoons HNO_2(aq) + OH^-(aq)\); equilibrium

Step by step solution

01

Identify the acid and base in the reaction

In this reaction, oxide ion (O虏鈦�) is the base, and water (H鈧侽) is the acid.
02

Predict the products of the reaction

The products of the acid-base reaction will be the conjugate acid of the base (O虏鈦�) and the conjugate base of the acid (H鈧侽). Since oxide ion accepts a proton, its conjugate acid will be hydroxide ion (OH鈦�). And since water donates a proton, its conjugate base will be hydronium ion (H鈧僌鈦�).
03

Write the balanced equation

Using the reactants and products, we can write the balanced equation: \[ O^{2-}(aq) + H_2O(l) \rightleftharpoons OH^-(aq) + H_3O^+(aq) \]
04

Determine the direction of equilibrium

Since water is amphoteric, it can act as an acid or a base. It varies in proton donation or acceptance depending on the context. As the oxide ion is a strong base and water is a weak acid, the equilibrium will lie to the right, favoring the formation of the conjugate acid (OH鈦�) and conjugate base (H鈧僌鈦�). (b)
05

Identify the acid and base in the reaction

In this reaction, acetic acid (CH鈧僀OOH) is the acid, and the hydrogen sulfide ion (HS鈦�) is the base.
06

Predict the products of the reaction

The products of the acid-base reaction will be the conjugate base of the acid (CH鈧僀OOH) and the conjugate acid of the base (HS鈦�). The conjugate base of CH鈧僀OOH is CH鈧僀OO鈦� and the conjugate acid of HS鈦� is H鈧係.
07

Write the balanced equation

Using the reactants and products, we can write the balanced equation: \[ CH_3COOH(aq) + HS^-(aq) \rightleftharpoons CH_3COO^-(aq) + H_2S(aq) \]
08

Determine the direction of equilibrium

The equilibrium will lie in the direction of the weaker acid and base. Acetic acid is a weak acid, and the hydrogen sulfide ion is a relatively weak base. Since their strengths are comparable, the equilibrium lies neither to the left nor the right, and the reaction is considered to be in equilibrium. (c)
09

Identify the acid and base in the reaction

In this reaction, nitrite ion (NO鈧傗伝) is the base, and water (H鈧侽) is the acid.
10

Predict the products of the reaction

The products of the acid-base reaction will be the conjugate acid of the base (NO鈧傗伝) and the conjugate base of the acid (H鈧侽). The conjugate acid of NO鈧傗伝 is HNO鈧�, and the conjugate base of H鈧侽 is OH鈦�.
11

Write the balanced equation

Using the reactants and products, we can write the balanced equation: \[ NO_2^-(aq) + H_2O(l) \rightleftharpoons HNO_2(aq) + OH^-(aq) \]
12

Determine the direction of equilibrium

Nitrite ion (NO鈧傗伝) is a weak base, and water (H鈧侽) is a weak acid. The equilibrium will lie in the direction of the weaker acid and base. Since the nitrite ion and water have comparable strengths, the equilibrium lies neither to the left nor the right, and the reaction is considered to be in equilibrium.

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Most popular questions from this chapter

How does the acid strength of an oxyacid depend on (a) the electronegativity of the central atom; (b) the number of nonprotonated oxygen atoms in the molecule?

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\). Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.082 \mathrm{M} \mathrm{KOH}\), (b) \(1.065 \mathrm{~g}\) of \(\mathrm{KOH}\) in \(500.0 \mathrm{~mL}\) of solution, (c) \(10.0 \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{~mL}\), (d) a solution formed by mixing \(7.5 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\)

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 \(\mathrm{S}\).

The iodate ion is reduced by sulfite according to the following reaction: $$ \mathrm{IO}_{3}^{-}(a q)+3 \mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+3 \mathrm{SO}_{4}^{2-}(a q) $$ The rate of this reaction is found to be first order in \(\mathrm{IO}_{3}^{-}\), first order in \(\mathrm{SO}_{3}^{2-}\), and first order in \(\mathrm{H}^{+}\). (a) Write the rate law for the reaction. (b) By what factor will the rate of the reaction change if the \(\mathrm{pH}\) is lowered from \(5.00\) to 3.50? Does the reaction proceed faster or slower at the lower pH? (c) By using the concepts discussed in Section 14.6, explain how the reaction can be pH- dependent even though \(\mathrm{H}^{+}\) does not appear in the overall reaction.
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