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Chapter 16: Problem 25

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow: (a) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q)\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Short Answer

Expert verified
(a) Products: \(\mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\); Equilibrium lies to the right. (b) Products: \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{2} \mathrm{S}(a q)\); Equilibrium lies to the left. (c) Products: \(\mathrm{HNO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\); Equilibrium lies to the right.

Step by step solution

01

Identify the acidic and basic components

In this reaction, \(\mathrm{O}^{2-}\) is the base as it can accept a proton, and \(\mathrm{H}_{2}\mathrm{O}\) is the acid, as it can donate a proton.
02

Predict the products of the reaction

The base, \(\mathrm{O}^{2-}\), will accept a proton from \(\mathrm{H}_{2}\mathrm{O}\), resulting in the formation of \(\mathrm{OH}^{-}\) and \(\mathrm{HOH}\) which is another \(\mathrm{H}_{2}\mathrm{O}\). The balanced reaction is as follows: \[ \mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \]
03

Predict the direction of equilibrium

To predict the direction of equilibrium, we need to compare the relative strengths of the acids and bases involved. The stronger the base, the larger the equilibrium constant, and the reaction will be directed toward products. \(\mathrm{O}^{2-}\) is a stronger base than \(\mathrm{OH}^{-}\). Therefore, the equilibrium will lie to the right of the reaction arrow. (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q)\)
04

Identify the acidic and basic components

In this reaction, \(\mathrm{CH}_{3} \mathrm{COOH}\) is the acid, and \(\mathrm{HS}^{-}\) is the base.
05

Predict the products of the reaction

The base, \(\mathrm{HS}^{-}\), will accept a proton from the acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), resulting in the formation of \(\mathrm{H}_{2}\mathrm{S}\) and \(\mathrm{CH}_{3} \mathrm{COO}^{-}\). The balanced reaction is: \[ \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{2} \mathrm{S}(a q) \]
06

Predict the direction of equilibrium

To predict the direction of the equilibrium, we compare the relative strengths of the acids and bases involved. Acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)) is a weaker acid than hydrosulfide (\(\mathrm{HS}^{-}\)), so the equilibrium will lie to the left of the reaction arrow. (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)
07

Identify the acidic and basic components

In this reaction, \(\mathrm{NO}_{2}^{-}\) is the base, and \(\mathrm{H}_{2}\mathrm{O}\) is the acid.
08

Predict the products of the reaction

The base, \(\mathrm{NO}_{2}^{-}\), accepts a proton from the acid, \(\mathrm{H}_{2}\mathrm{O}\), resulting in the formation of \(\mathrm{HNO}_{2}\) and \(\mathrm{HOH}\) which is another \(\mathrm{H}_{2}\mathrm{O}\). The balanced reaction is: \[ \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HNO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \]
09

Predict the direction of equilibrium

To predict the direction of equilibrium, we compare the relative strengths of the acids and bases involved. Nitrite (\(\mathrm{NO}_{2}^{-}\)) is a weaker base than hydroxide (\(\mathrm{OH}^{-}\)), and \(\mathrm{H}_{2}\mathrm{O}\) is a weaker acid than \(\mathrm{HNO}_{2}\). Therefore, the equilibrium will lie to the right of the reaction arrow.

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Key Concepts

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

Acid-Base Reaction Prediction
Understanding how to predict acid-base reactions is crucial for mastering chemistry. In these reactions, an acid donates a proton (a hydrogen ion, or H+) to a base, which accepts it. The key is to identify the acidic and basic components of the reactants.

For instance, in the reaction \(\mathrm{O}^{2-}(aq) + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons\), the oxide ion (\(\mathrm{O}^{2-}\)) acts as a base because it can accept a proton, while water (\(\mathrm{H}_2\mathrm{O}\)) is the acid, able to donate a proton. When these two react, the oxide ion accepts a hydrogen ion from water, forming hydroxide (\(\mathrm{OH}^{-}\)) and leaving behind another water molecule.

To successfully predict reaction products, always identify the prospective acid and base, and then determine the resulting products after they exchange the hydrogen ion.
Equilibrium Direction Prediction
Equilibrium direction in an acid-base reaction can be predicted by evaluating the relative strengths of the acids and bases involved. When a strong base reacts with a weaker acid, the equilibrium will favor the right side, forming more products.

For example, in the reaction involving \(\mathrm{CH}_{3}\mathrm{COOH}\) (acetic acid) and \(\mathrm{HS}^{-}\) (hydrosulfide), since acetic acid is a weaker acid than the conjugate acid of the hydrosulfide (\(\mathrm{H}_2\mathrm{S}\)), the equilibrium position is towards the reactants, or the left side.

Understanding how to predict the direction of an acid-base reaction's equilibrium is essential in determining the predominance of products or reactants after the reaction has reached its stable state.
Acid-Base Reaction Products
The products of an acid-base reaction are typically a conjugate base of the acid and a conjugate acid of the base. During the reaction, the acid loses a proton, becoming a conjugate base, and the base gains a proton, becoming a conjugate acid.

For instance, in the reaction between nitrite (\(\mathrm{NO}_2^{-}\)) and water, nitrite is the base and accepts a proton to become nitrous acid (\(\mathrm{HNO}_2\)), while water is the acid that donates a proton, essentially remaining as water. These reactions allow for the formation of new substances, and predicting the products involves understanding both the initial reactants and the concept of proton transfer in acid-base reactions.

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Most popular questions from this chapter

Calculate \(\left[\mathrm{OH}^{-}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: \((\mathbf{a})\left[\mathrm{H}^{+}\right]=0.0505 M (\mathbf{b})\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-10} M ;(\mathbf{c})\) a solution in which \(\left[\mathrm{H}^{+}\right]\) is 1000 times greater than \(\left[\mathrm{OH}^{-}\right] .\)

In many reactions, the addition of \(\mathrm{AlCl}_{3}\) produces the same effect as the addition of \(\mathrm{H}^{+} .\) (a) Draw a Lewis structure for \(\mathrm{AlCl}_{3}\) in which no atoms carry formal charges, and determine its structure using the VSEPR method. (b) What characteristic is notable about the structure in part (a) that helps us understand the acidic character of AlCl \(_{3} ?\) (c) Predict the result of the reaction between \(\mathrm{AlCl}_{3}\) and \(\mathrm{NH}_{3}\) in a solvent that does not participate as a reactant. (d) Which acid-base theory is most suitable for discussing the similarities between \(\mathrm{AlCl}_{3}\) and \(\mathrm{H}^{+}\) ?

(a) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\) , which is the stronger base? (b) Which is the stronger acid, the ammonium or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{3} \mathrm{NOH}^{+}.\)

The fluoride ion reacts with water to produce HF. (a) Write out the chemical equation for this reaction. (b) Will a concentrated solution of NaF in water be acidic, basic, or neutral? (c) Is fluoride acting as a Lewis acid or as a Lewis base when reacting with water?

Calculate the \(\mathrm{pH}\) of a solution made by adding 2.50 \(\mathrm{g}\) of lithium oxide \(\left(\mathrm{Li}_{2} \mathrm{O}\right)\) to enough water to make 1.500 \(\mathrm{L}\) of solution.
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