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Chapter 13: Problem 70

Find the value of \(K_{\mathrm{a}}\) for the conjugate acid of the following bases. (a) pyridine, a pesticide; \(K_{\mathrm{b}}=1.5 \times 10^{-9}\) (b) aniline, an important dye intermediate; \(K_{\mathrm{b}}=3.8 \times 10^{-10}\)

Short Answer

Expert verified
The value of \(K_{\mathrm{a}}\) for pyridine's conjugate acid is \(6.67 \times 10^{-6}\), and the value of \(K_{\mathrm{a}}\) for aniline's conjugate acid is \(2.63 \times 10^{-5}\).

Step by step solution

01

Identify the given values

For each base, we are given their respective values of \(K_{\mathrm{b}}\). Write down the values: (a) Pyridine: \(K_{\mathrm{b}} = 1.5 \times 10^{-9}\) (b) Aniline: \(K_{\mathrm{b}} = 3.8 \times 10^{-10}\) ##Step 2: Recall the relationship between \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\)##
02

Recall the relationship between \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\)

The relationship between the \(K_{\mathrm{a}}\) for a conjugate acid and the \(K_{\mathrm{b}}\) for a weak base is given by: $$ K_{\mathrm{a}} \cdot K_{\mathrm{b}} = K_{\mathrm{w}} $$ ##Step 3: Calculate \(K_{\mathrm{a}}\) for pyridine's conjugate acid##
03

Calculate \(K_{\mathrm{a}}\) for pyridine's conjugate acid

Using the relationship between \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\), we can calculate \(K_{\mathrm{a}}\) for the conjugate acid of pyridine. Since \(K_{\mathrm{b}}=1.5 \times 10^{-9}\) and \(K_{\mathrm{w}} = 1 \times 10^{-14}\), we have: $$ K_{\mathrm{a}} = \frac{K_{\mathrm{w}}}{K_{\mathrm{b}}} = \frac{1 \times 10^{-14}}{1.5 \times 10^{-9}} $$ $$ K_{\mathrm{a}} = 6.67 \times 10^{-6} $$ The value of \(K_{\mathrm{a}}\) for pyridine's conjugate acid is \(6.67 \times 10^{-6}\). ##Step 4: Calculate \(K_{\mathrm{a}}\) for aniline's conjugate acid##
04

Calculate \(K_{\mathrm{a}}\) for aniline's conjugate acid

Using the relationship between \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\), we can calculate \(K_{\mathrm{a}}\) for the conjugate acid of aniline. Since \(K_{\mathrm{b}}=3.8 \times 10^{-10}\) and \(K_{\mathrm{w}} = 1 \times 10^{-14}\), we have: $$ K_{\mathrm{a}} = \frac{K_{\mathrm{w}}}{K_{\mathrm{b}}} = \frac{1 \times 10^{-14}}{3.8 \times 10^{-10}} $$ $$ K_{\mathrm{a}} = 2.63 \times 10^{-5} $$ The value of \(K_{\mathrm{a}}\) for aniline's conjugate acid is \(2.63 \times 10^{-5}\). The conjugate acid of pyridine has a \(K_{\mathrm{a}}\) value of \(6.67 \times 10^{-6}\), and the conjugate acid of aniline has a \(K_{\mathrm{a}}\) value of \(2.63 \times 10^{-5}\).

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Most popular questions from this chapter

WEB The pH of a \(2.642 M\) solution of a weak acid, \(\mathrm{HB}\), is \(5.32\). What is \(K_{\mathrm{a}}\) for the weak acid?

Using the Brønsted-Lowry model, write an equation to show why each of the following species produces a basic aqueous solution. (a) \(\mathrm{NH}_{3}\) (b) \(\mathrm{NO}_{2}^{-}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (d) \(\mathrm{CO}_{3}{ }^{2-}\) (e) \(\mathrm{F}^{-}\) (f) \(\mathrm{HCO}_{3}^{-}\)

Using the Brønsted-Lowry model, write equations to show why the following species behave as weak acids in water. (a) \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{+}\) (b) \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) (c) \(\mathrm{H}_{2} \mathrm{~S}\) (d) \(\mathrm{HPO}_{4}^{2-}\) (e) \(\mathrm{HClO}_{2}\) (f) \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{+}\)

Write a balanced equation showing how the \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ion can be either a Bronsted-Lowry acid or a Bronsted-Lowry base.

Solution \(\mathrm{X}\) has \(\mathrm{pH}\) 11.7. Solution \(\mathrm{Y}\) has \(\left[\mathrm{OH}^{-}\right]=4.5 \times 10^{-2}\). Which solution is more basic? Which has the higher \(\mathrm{pOH}\) ?
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