Ap Stat Review

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Sample Exercise 16.1 Identifying Conjugate Acids and Bases
(a) What is the conjugate base of each of the following acids: HClO4, H2S, PH4+, HCO3–? (b) What is the conjugate acid of each of the following bases: CN–, SO42–, H2O, HCO3– ? Solution Analyze: We are asked to give the conjugate base for each of a series of species and to give the conjugate acid for each of another series of species. Plan: The conjugate base of a substance is simply the parent substance minus one proton, and the conjugate acid of a substance is the parent substance plus one proton. Solve: (a) HClO4 less one proton (H+) is ClO4–. The other conjugate bases are HS–, PH3, and CO32–. (b) CN– plus one proton (H+) is HCN. The other conjugate acids are HSO4–, H3O+, and H2CO3. Notice that the hydrogen carbonate ion (HCO3–) is amphiprotic. It can act as either an acid or a Practice Exercise base. Write the formula for the conjugate acid of each of the following: HSO3–, F– , PO43–, CO. Answers: H2SO3, HF, HPO42–, HCO+

Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3–) is amphiprotic. (a) Write an equation for the reaction of HSO3– with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HSO3– with water, in which the ion acts as a base. In both cases identify the conjugate acid–base pairs. Solution Analyze and Plan: We are asked to write two equations representing reactions between HSO3– and water, one in which HSO3– should donate a proton to water, thereby acting as a Brønsted– Lowry acid, and one in which HSO3– should accept a proton from water, thereby acting as a base. We are also asked to identify the conjugate pairs in each equation. Solve:

The conjugate pairs in this equation are HSO3– (acid) and SO32– (conjugate base); and H2O (base) and H3O+ (conjugate acid). The conjugate pairs in this equation are H2O (acid) and OH– (conjugate base), and HSO3– (base) and H2SO3 (conjugate acid).

Practice Exercise
When lithium oxide (Li2O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O2–) with water. Write the reaction that occurs, and identify the conjugate acid–base pairs. Answer: O2–(aq) + H2O(l) → OH–(aq) + OH–(aq). OH– is the conjugate acid of the base O2–. OH– is also the conjugate base of the acid H2O.

Chemistry: The Central Science, Eleventh Edition By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy With contributions from Patrick Woodward

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Sample Exercise 16.3 Predicting the Position of a Proton-Transfer Equilibrium For the following proton-transfer reaction, use Figure 16.4 to predict whether the equilibrium lies predominantly to the left (that is, Kc < 1 ) or to the right (Kc > 1): Solution

Analyze: We are asked to predict whether the equilibrium shown lies to the right, favoring products, or to the left, favoring reactants. Plan: This is a proton-transfer reaction, and the position of the equilibrium will favor the proton going to the stronger of two bases. The two bases in the equation are CO32–, the base in the forward reaction as written, and SO42– the conjugate base of HSO4–. We can find the relative positions of these two bases in Figure 16.4 to determine which is the stronger base. Solve: CO32– appears lower in the right-hand column in Figure 16.4 and is therefore a stronger base than SO42–. CO32–, therefore, will get the proton preferentially to become HCO3–, while SO42– will remain mostly unprotonated. The resulting equilibrium will lie to the right, favoring products (that is, Kc > 1 ).

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