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Chapter 16: Problem 75

Phenol, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH},\) has a \(K_{a}\) of $1.3 \times 10^{-10}$ (a) Write out the \(K_{a}\) reaction for phenol. (b) Calculate \(K_{b}\) for phenol's conjugate base. (c) Is phenol a stronger or weaker acid than water?

Short Answer

Expert verified
(a) The acid dissociation reaction of phenol is: \(C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+\), and its equilibrium constant expression is: \(K_a = \frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}\). (b) The \(K_b\) value for the conjugate base of phenol is \(7.69 \times 10^{-5}\), calculated using the relationship between \(K_a\) and \(K_b\), which is \(K_a \times K_b = K_w\). (c) Phenol is a stronger acid than water because its \(K_a\) value (1.3 x 10^{-10}) is greater than that of water (1.0 x 10^{-14}).

Step by step solution

01

Acid dissociation of phenol

The acid dissociation reaction of phenol can be represented as follows: \[C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+\]
02

Equilibrium constant (\(K_a\)) expression

The equilibrium constant expression for the above reaction is given by: \[K_a = \frac{[C_6H_5O^-][H^+]}{[C_6H_5OH]}\] Now we have the \(K_a\) reaction for phenol with its equilibrium constant expression. #b) Calculating \(K_b\) for phenol's conjugate base#
03

Understanding the relationship between \(K_a\) and \(K_b\)

For a conjugate acid-base pair, the relationship between their respective equilibrium constants (\(K_a\) and \(K_b\)) is given by: \[K_a \times K_b = K_w\] \(K_w\) is the ion product of water, which is equal to \(1.0 \times 10^{-14} \) at \(25^{\circ}C\). We are given \(K_a = 1.3 \times 10^{-10}\) for the phenol, so we can calculate \(K_b\) for its conjugate base (\(C_6H_5O^-\)) using this relationship.
04

Calculating \(K_b\)

Using the given \(K_a\) value, we can calculate the \(K_b\) value for phenol's conjugate base as follows: \[K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-10}}\] \[K_b = 7.69 \times 10^{-5}\] The \(K_b\) value for the conjugate base of phenol is \(7.69 \times 10^{-5}\). #c) Comparing the acidity of phenol with water#
05

Comparing \(K_a\) values

To determine if phenol is a stronger or weaker acid than water, we need to compare their respective \(K_a\) values. The \(K_a\) value of water is \(K_w = 1.0 \times 10^{-14}\) (since water is amphoteric, its \(K_a\) value is equal to its \(K_b\) value). Given that the \(K_a\) value of phenol is \(1.3 \times 10^{-10}\), we can compare the two values to determine the acidity.
06

Determining the stronger acid

Based on the \(K_a\) values, we can see that: \[K_a(C_6H_5OH) = 1.3 \times 10^{-10} > 1.0 \times 10^{-14} = K_a(H_2O)\] Since the \(K_a\) value for phenol is greater than that of water, phenol is a stronger acid than water.

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Most popular questions from this chapter

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 \(\mathrm{AlCl}_{3}\) ? (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 \(\mathrm{H}^{+}\) ?

The amino acid glycine $\left(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)$ can participate in the following equilibria in water: $\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons$ $$ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \quad K_{\mathrm{a}}=4.3 \times 10^{-3} $$ $$ \begin{aligned} \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} & \rightleftharpoons \\ &{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_{\mathrm{b}}=6.0 \times 10^{-5} \end{aligned} $$ (a) Use the values of \(K_{a}\) and \(K_{b}\) to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: $$ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-} $$ (b) What is the pH of a 0.050 Maqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with \(\mathrm{pH} 13\) ? With \(\mathrm{pH}\) ?

(a) Write a chemical equation that illustrates the autoionization of water. (b) Write the expression for the ionproduct constant for water, $K_{w} .(\mathbf{c})$ If a solution is described as basic, which of the following is true: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right]\), (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right],\) or (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)

Identify the Brønsted-Lowry acid and the Brønsted-Lowry base on the left side of each of the following equations, and also identify the conjugate acid and conjugate base of each on the right side: (a) $\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)$ (b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons$ $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) $$ \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q) $$

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{NO}_{3}^{-}\) or \(\mathrm{NO}_{2}^{-},\) (b) \(\mathrm{PO}_{4}^{3-}\) or \(\mathrm{AsO}_{4}^{3-}\), (c) \(\mathrm{HCO}_{3}^{-}\) or \(\mathrm{CO}_{3}^{2-}\).
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