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Chapter 16: Problem 50

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), (B) \(\mathrm{HCO}_{3}^{-}\)

Short Answer

Expert verified
(a) For \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\): 1. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} (aq) \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}(aq) + \mathrm{H}^{+}(aq)\) and \(K_{a} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}]}\) 2. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} (aq) + \mathrm{H}_{2}\mathrm{O} (l) \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}(aq) + \mathrm{H}_{3}\mathrm{O}^{+}(aq)\) and \(K_{a} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}][\mathrm{H}_{3}\mathrm{O}^{+}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}]}\) (b) For \(\mathrm{HCO}_{3}^{-}\): 1. \(\mathrm{HCO}_{3}^{-}(aq) \rightleftharpoons \mathrm{CO}_{3}^{2-}(aq) + \mathrm{H}^{+}(aq)\) and \(K_{a} = \frac{[\mathrm{CO}_{3}^{2-}][\mathrm{H}^{+}]}{[\mathrm{HCO}_{3}^{-}]}\) 2. \(\mathrm{HCO}_{3}^{-}(aq) + \mathrm{H}_{2}\mathrm{O} (l) \rightleftharpoons \mathrm{CO}_{3}^{2-}(aq) + \mathrm{H}_{3}\mathrm{O}^{+}(aq)\) and \(K_{a} = \frac{[\mathrm{CO}_{3}^{2-}][\mathrm{H}_{3}\mathrm{O}^{+}]}{[\mathrm{HCO}_{3}^{-}]}\)

Step by step solution

01

(a) Chemical Equation with \(\mathrm{H}^{+}(a q)\)

For \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), the acid dissociation reaction will be: \[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} (aq) \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}(aq) + \mathrm{H}^{+}(aq)\]
02

(a) \(K_{a}\) Expression with \(\mathrm{H}^{+}(a q)\)

The \(K_{a}\) expression for the reaction above is: \[K_{a} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}][\mathrm{H}^{+}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}]}\]
03

(a) Chemical Equation with Hydronium Ion

For \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), the acid dissociation reaction with hydronium ion will be: \[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH} (aq) + \mathrm{H}_{2}\mathrm{O} (l) \rightleftharpoons \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}(aq) + \mathrm{H}_{3}\mathrm{O}^{+}(aq)\]
04

(a) \(K_{a}\) Expression with Hydronium Ion

The \(K_{a}\) expression for the reaction above is: \[K_{a} = \frac{[\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{COO}^{-}][\mathrm{H}_{3}\mathrm{O}^{+}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}]}\]
05

(b) Chemical Equation with \(\mathrm{H}^{+}(a q)\)

For \(\mathrm{HCO}_{3}^{-}\), the acid dissociation reaction will be: \[\mathrm{HCO}_{3}^{-}(aq) \rightleftharpoons \mathrm{CO}_{3}^{2-}(aq) + \mathrm{H}^{+}(aq)\]
06

(b) \(K_{a}\) Expression with \(\mathrm{H}^{+}(a q)\)

The \(K_{a}\) Expression for the reaction above is: \[K_{a} = \frac{[\mathrm{CO}_{3}^{2-}][\mathrm{H}^{+}]}{[\mathrm{HCO}_{3}^{-}]}\]
07

(b) Chemical Equation with Hydronium Ion

For \(\mathrm{HCO}_{3}^{-}\), the acid dissociation reaction with hydronium ion will be: \[\mathrm{HCO}_{3}^{-}(aq) + \mathrm{H}_{2}\mathrm{O} (l) \rightleftharpoons \mathrm{CO}_{3}^{2-}(aq) + \mathrm{H}_{3}\mathrm{O}^{+}(aq)\]
08

(b) \(K_{a}\) Expression with Hydronium Ion

The \(K_{a}\) Expression for the reaction above is: \[K_{a} = \frac{[\mathrm{CO}_{3}^{2-}][\mathrm{H}_{3}\mathrm{O}^{+}]}{[\mathrm{HCO}_{3}^{-}]}\]

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium represents a state in an ongoing chemical process where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no observable change in the concentrations of reactants and products over time. It occurs in reversible reactions, such as acid dissociation in water, and is dynamic, which means that the reactions continue to occur, but since they are at identical rates in both directions, the system appears to be at rest.

At equilibrium, the concentration of all reactants and products remains constant, but it's important to note that 'constant' does not necessarily mean 'equal'. The proportions are defined by the equilibrium constant, which for acid-base reactions is expressed as the acid dissociation constant, Ka. Understanding this concept is crucial for predicting the behavior of acids and bases in solution and for calculating the pH of a given solution.
Ka Expression
The acid dissociation constant, Ka, is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for a chemical reaction known as the dissociation of acid molecules in aqueous solution. The stronger the acid, the larger the value of Ka, indicating a greater tendency of the acid molecules to lose a proton, forming hydronium ions (H3O+).

Mathematically, the Ka expression is derived from the equilibrium concentrations of the products and reactants involved in the acid dissociation reaction. For instance, for a generic acid dissociation represented as HA(aq) 鈫� H+(aq) + A-(aq), the Ka expression would be: \[Ka = \frac{[H^+][A^-]}{[HA]}\].This expression helps in calculating the degree of dissociation of an acid, and hence, predicting the pH of the solution.
Hydronium Ion
The hydronium ion, H3O+, plays a pivotal role in acid-base chemistry. It is formed when an acid donates a proton (H+) to water, a process that is fundamental to the nature of aqueous acid solutions. Although it's common to represent the proton in acid reactions simply as H+, in reality, protons are always associated with water molecules in an aqueous environment.

Representing the proton as the hydronium ion in chemical equations is more accurate and provides a clearer picture of the actual chemical processes. For example, when benzoic acid (C6H5COOH) dissociates in water, it produces the hydronium ion rather than free protons: \[C6H5COOH(aq) + H2O(l) \rightleftharpoons C6H5COO^-(aq) + H3O^+(aq)\].The hydronium ion concentration is a key factor in determining the pH of a solution, with an increase in H3O+ concentration resulting in a more acidic pH.
Acid-Base Reaction
Acid-base reactions are a subset of chemical reactions that involve the transfer of hydrogen ions between reactants. In the context of aqueous solutions, acids increase the hydronium ion concentration by donating a proton to water, while bases reduce the hydronium ion concentration by accepting a proton. The Br酶nsted-Lowry theory defines acids as proton donors and bases as proton acceptors.

In the provided exercise solutions, we see examples of acid dissociation, which are acid-base reactions where an acid donates a proton to water: for instance, with benzoic acid (C6H5COOH) or bicarbonate ion (HCO3-). These reactions result in the formation of their conjugate bases (C6H5COO- and CO32-, respectively) and the hydronium ion (H3O+). The expression of these reactions through chemical equations illustrates the reversible nature and equilibria of these acid-base processes, critical for understanding their impact on the solution's acidity.

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Most popular questions from this chapter

Atmospheric CO \(_{2}\) levels have risen by nearly 20\(\%\) over the past 40 years from 320 ppm to 400 ppm. (a) Given that the average \(\mathrm{pH}\) of clean, unpolluted rain today is \(5.4,\) determine the pH of unpolluted rain 40 years ago. Assume that carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) formed by the reaction of \(\mathrm{CO}_{2}\) and water is the only factor influencing pH. $$\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)$$ (b) What volume of \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) and 1.0 \(\mathrm{atm}\) is dissolved in a 20.0 -L bucket of today's rainwater?

Consider two solutions, solution \(\mathrm{A}\) and solution \(\mathrm{B} .\left[\mathrm{H}^{+}\right]\) in solution \(\mathrm{A}\) is 250 times greater than that in solution B. What is the difference in the pH values of the two solutions?

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{MSr}(\mathrm{OH})_{2}\) (b) 2.250 \(\mathrm{g}\) of LiOH in 250.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 1.00\) mL of 0.175 M NaOH diluted to \(2.00 \mathrm{L},\) (d) a solution formed by adding 5.00 \(\mathrm{mL}\) of 0.105 \(\mathrm{M} \mathrm{KOH}\) to 15.0 \(\mathrm{mL}\) of \(9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}.\)

What is the \(\mathrm{pH}\) of a solution that is \(2.5 \times 10^{-9} \mathrm{M}\) in \(\mathrm{NaOH}\) ? Does your answer make sense? What assumption do we normally make that is not valid in this case?

Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but they are frequently much more soluble as their acid salts. Assuming that pH in the stomach is \(2.5,\) indicate whether each of the following compounds would be present in the stomach as the neutral base or in the protonated form: nicotine, \(K_{b}=7 \times 10^{-7} ;\) caffeine, \(K_{b}=4 \times 10^{-14}\) ; strychnine, \(K_{b}=1 \times 10^{-6} ;\) quinine, \(K_{b}=1.1 \times 10^{-6} .\)
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