/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Derive an expression for the rel... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 26

Derive an expression for the relationship between \(\mathrm{p} K_{\mathrm{a}}\) and \(\mathrm{p} K_{\mathrm{b}}\) for a conjugate acid-base pair. \((\mathrm{p} K=-\log K .)\)

Short Answer

Expert verified
The relationship between the acidity constant \(pK_a\) and the basicity constant \(pK_b\) for a conjugate acid-base pair is: \(pK_a + pK_b = pK_w\).

Step by step solution

01

Write expression for \(K_a\) and \(K_b\)

For the acid dissociation reaction: \[HA + H_2O \rightleftharpoons H_3O^+ + A^-\] The acidity constant, \(K_a\), is defined as: \[K_a = \frac{[H_3O^+][A^-]}{[HA]}\] Similarly, for the base dissociation reaction: \[A^- + H_2O \rightleftharpoons HA + OH^-\] The basicity constant, \(K_b\), is defined as: \[K_b = \frac{[HA][OH^-]}{[A^-]}\] Step 2: Write the expression for \(pK_a\) and \(pK_b\)
02

Write expression for \(pK_a\) and \(pK_b\)

The \(pK_a\) and \(pK_b\) are the negative logarithm of \(K_a\) and \(K_b\) respectively. We have: \[pK_a = -\log{K_a}\] \[pK_b = -\log{K_b}\] Step 3: Use the relationship between \(K_a\), \(K_b\), and \(K_w\)
03

Use the relationship between \(K_a\), \(K_b\), and \(K_w\)

The relationship between \(K_a\), \(K_b\), and the auto-ionization constant of water (\(K_w = [H_3O^+][OH^-]\)) is given by: \[K_a \cdot K_b = K_w\] Step 4: Write the relationship between \(pK_a\) and \(pK_b\)
04

Write the relationship between \(pK_a\) and \(pK_b\)

Taking the logarithm of the relation \(K_a \cdot K_b = K_w\), we get: \[\log{K_a} + \log{K_b} = \log{K_w}\] Now, multiply the whole equation by -1: \[-\log{K_a} - \log{K_b} = -\log{K_w}\] Finally, substituting the expressions for \(pK_a\) and \(pK_b\): \[pK_a + pK_b = pK_w\] Therefore, the relationship between the acidity constant \(pK_a\) and the basicity constant \(pK_b\) for a conjugate acid-base pair is: \[pK_a + pK_b = pK_w\]

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Most popular questions from this chapter

Place the species in each of the following groups in order of increasing acid strength. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) (bond energies: \(\mathrm{H}-\mathrm{O}, 467 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{S}\), \(363 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{~kJ} / \mathrm{mol})\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}\) c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}{ }^{+}\) d. \(\mathrm{NH}_{4}{ }^{+}, \mathrm{PH}_{4}{ }^{+}\) (bond energies: \(\mathrm{N}-\mathrm{H}, 391 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H}, 322\) \(\mathrm{kJ} / \mathrm{mol}\) ) Give reasons for the orders you chose.

Place the species in each of the following groups in order of increasing base strength. Give your reasoning in each case. a. \(\mathrm{IO}_{3}^{-}, \mathrm{Br} \mathrm{O}_{3}^{-}\) b. \(\mathrm{NO}_{2}^{-}, \mathrm{NO}_{3}^{-}\) c. \(\mathrm{OCl}^{-}, \mathrm{OI}^{-}\)

Trichloroacetic acid \(\left(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\right)\) is a corrosive acid that is used to precipitate proteins. The \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of trichloroacetic acid is the same as the \(\mathrm{pH}\) of a \(0.040 \mathrm{M} \mathrm{HClO}_{4}\) solution. Calculate \(K_{\mathrm{a}}\) for trichloroacetic acid.

For propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\), determine the concentration of all species present, the \(\mathrm{pH}\), and the percent dissociation of a \(0.100 M\) solution.

For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated \(\mathrm{pH}\). Calculate the \(\mathrm{pH}\) of each solution. Group I: \(0.20 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}, 0.20 \mathrm{MHCl}, 0.20 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}, 0.20\) \(M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NHCl}\) Group II: \(0.20 \mathrm{M} \mathrm{KOI}, 0.20 \mathrm{M} \mathrm{NaCN}, 0.20 \mathrm{M} \mathrm{KOCl}, 0.20 \mathrm{M}\) \(\mathrm{NaNO}_{2}\) a. the solution with the lowest \(\mathrm{pH}\) b. the solution with the highest \(\mathrm{pH}\) c. the solution with the \(\mathrm{pH}\) closest to \(7.00\)
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