Answer to Question #172524 in General Chemistry for Mayuka

Question

I. ESSAY: Answer the following questions as briefly and concisely as possible.

Use the following acids to answer the questions in this examination.

CH3CH2CH2COOH Ka = 4.4 X 10-7

HOBr Ka = 2.50 x 10-9

HNO2 Ka = 7.2 x 10-4

1. There are three different theories of acids and bases.

(a) Define each theory and give a specific example to support this theory.

(b) Also identify the conjugates of your examples when possible.

(c) From the acids given above, arrange them in order of increasing acid strength.

2. From the acids given, choose an acid. For this acid:

(a) show how the acid dissociates in water. Write the equation.

(b) Give the equilibrium expression for this acid.

(c) determine the equilibrium concentrations of a 1.0 M acid.

(d) What is the pH of the solution in (c) if it contains 0.50M of the salt of its conjugate?

Answer 1

a) Arrhenius theory, states that acids are substances that dissociate in water to yield electrically charged atoms or molecules, called ions, one of which is a hydrogen ion (H⁺), and that bases ionize in water to yield hydroxide ions (OH⁻).

H⁺Cl^- (acid)+ Na⁺OH^- (base)----> NaCl + H₂O

Brønsted–Lowry theory, also called proton theory of acids and bases stating that any compound that can transfer a proton to any other compound is an acid, and the compound that accepts the proton is a base. A proton is a nuclear particle with a unit positive electrical charge; it is represented by the symbol H⁺ because it constitutes the nucleus of a hydrogen atom.

CH₃COOH + H₂O -----> CH₃COO^- (congugate base)+ H₃O⁺

Lewis theory of acid-base reactions, bases donate pairs of electrons and acids accept pairs of electrons. A Lewis acid is therefore any substance, such as the H⁺ ion, that can accept a pair of nonbonding electrons. In other words, a Lewis acid is an electron-pair acceptor. A Lewis base is any substance, such as the OH^- ion, that can donate a pair of nonbonding electrons. A Lewis base is therefore an electron-pair donor.

Na* + *Cl^**_*** ---->Na**cl_****^**

CH₃CH₂CH₂COOH < HOBr < HNO₂

(b) CH₃CH₂CH₂COOH +H₂O -----> CH₃CH₂CH₂COO^- + H₃O⁺

K= [CH₃CH₂CH₂COO^-][H₃O⁺] /[CH₃CH₂CH₂COOH ]

pH = - log [H⁺]