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Ch.17 - Additional Aspects of Aqueous Equilibria
Chapter 17, Problem 17a

a. Calculate the percent ionization of 0.007 M butanoic acid (πΎπ‘Ž=1.5Γ—10βˆ’5).

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<strong>Step 1:</strong> Write the ionization equation for butanoic acid (C<sub>3</sub>H<sub>7</sub>COOH) in water: C<sub>3</sub>H<sub>7</sub>COOH (aq) β‡Œ C<sub>3</sub>H<sub>7</sub>COO<sup>-</sup> (aq) + H<sup>+</sup> (aq).
<strong>Step 2:</strong> Set up the expression for the acid dissociation constant (K<sub>a</sub>): K<sub>a</sub> = [C<sub>3</sub>H<sub>7</sub>COO<sup>-</sup>][H<sup>+</sup>] / [C<sub>3</sub>H<sub>7</sub>COOH].
<strong>Step 3:</strong> Assume that the initial concentration of butanoic acid is 0.007 M and that the change in concentration due to ionization is x. Therefore, at equilibrium, [C<sub>3</sub>H<sub>7</sub>COO<sup>-</sup>] = x, [H<sup>+</sup>] = x, and [C<sub>3</sub>H<sub>7</sub>COOH] = 0.007 - x.
<strong>Step 4:</strong> Substitute these equilibrium concentrations into the K<sub>a</sub> expression: 1.5 \times 10^{-5} = (x)(x) / (0.007 - x).
<strong>Step 5:</strong> Solve the equation for x, which represents the concentration of ionized butanoic acid. Then, calculate the percent ionization using the formula: \text{Percent Ionization} = \left( \frac{x}{0.007} \right) \times 100\%.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ionization of Weak Acids

Weak acids, like butanoic acid, do not completely dissociate in solution. Instead, they establish an equilibrium between the undissociated acid and its ions. The degree of ionization is crucial for calculating the concentration of ions in solution, which directly affects the acid's strength and its pH.
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Equilibrium Constant (Ka)

The acid dissociation constant (Ka) quantifies the strength of an acid in solution. It is defined as the ratio of the concentration of the products (ions) to the concentration of the reactants (undissociated acid) at equilibrium. A smaller Ka value indicates a weaker acid, which is essential for determining the extent of ionization in weak acids like butanoic acid.
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Percent Ionization

Percent ionization is a measure of the extent to which an acid ionizes in solution, expressed as a percentage. It is calculated by dividing the concentration of ionized acid by the initial concentration of the acid, then multiplying by 100. This concept helps in understanding the effectiveness of an acid in solution and is particularly useful for comparing the strengths of different acids.
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Related Practice
Textbook Question

Three cations, Ni2+ , Cu2+ , and Ag+, are separated using two different precipitating agents. Based on Figure 17.23, what two precipitating agents could be used? Using these agents, indicate which of the cations is A, which is B, and which is C.[Section 17.7]

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Textbook Question

Which of these statements about the common-ion effect is most correct? (a) The solubility of a salt MA is decreased in a solution that already contains either M+ or A-. (b) Common ions alter the equilibrium constant for the reaction of an ionic solid with water. (c) The common-ion effect does not apply to unusual ions like SO32 - . (d) The solubility of a salt MA is affected equally by the addition of either A- or a noncommon ion.

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Textbook Question

Consider the equilibrium B1aq2 + H2O1l2 Ξ” HB+1aq2 + OH-1aq2. Suppose that a salt of HB+1aq2 is added to a solution of B1aq2 at equilibrium. (c) Will the pH of the solution increase, decrease, or stay the same?

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Textbook Question

(b) Calculate the percent ionization of 0.0075 M butanoic acid in a solution containing 0.085 M sodium butanoate.

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Textbook Question

(a) Calculate the percent ionization of 0.125 M lactic acid 1Ka = 1.4 * 10-42.

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Textbook Question

Which of the following solutions is a buffer? (a) A solution made by mixing 100 mL of 0.100 M CH3COOH and 50 mL of 0.100 M NaOH, (b) a solution made by mixing 100 mL of 0.100 M CH3COOH and 500 mL of 0.100 M NaOH, (c) A solution made by mixing 100 mL of 0.100 M CH3COOH and 50 mL of 0.100 M HCl, (d) A solution made by mixing 100 mL of 0.100 M CH3COOK and 50 mL of 0.100 M KCl.

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