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Chapter 14: Problem 102

Calculate the percentage of pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) that forms pyridinium ion, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), in a \(0.10-M\) aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\).

Short Answer

Expert verified
The percentage of pyridine that forms pyridinium ion in a \(0.10-M\) aqueous solution is approximately \(0.0013\%\).

Step by step solution

01

Write the chemical equation for the reaction

The equation for the reaction of pyridine with water is as follows: \[C_5H_5N_{(aq)} + H_2O_{(l)} \rightleftharpoons C_5H_5NH^+_{(aq)} + OH^{-}_{(aq)}\]
02

Write the equilibrium expression for the reaction

The equilibrium expression for the reaction is given by the \(K_b\) expression, which can be written as: \[K_b = \frac{[C_5H_5NH^+][OH^-]}{[C_5H_5N]}\]
03

Create the ICE table

We can create an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations of all species in the reaction: | | C5H5N | C5H5NH+ | OH- | |------|-----|-------|----| | I | 0.10 M | 0 M | 0 M | | C | -x | +x | +x | | E | 0.10-x | x | x | Here, "I" stands for the initial concentration of the species, "C" represents the change in concentration due to the reaction, and "E" denotes the equilibrium concentration.
04

Substitute the equilibrium concentrations into the \(K_b\) expression

Using the ICE table, we can substitute the equilibrium concentrations into the \(K_b\) expression: \[K_b = 1.7 \times 10^{-9} = \frac{x(x)}{0.10 - x}\]
05

Solve for x

Since \(K_b\) is very small, we can assume that x is small compared to the initial concentration of pyridine, i.e., \(x << 0.10\). Therefore, we can approximate \(0.10 - x \approx 0.10\). This simplifies the equation: \[1.7 \times 10^{-9} = \frac{x^2}{0.10}\] Solve for x: \[x = \sqrt{(1.7 \times 10^{-9})(0.10)} = 1.3 \times 10^{-5}\]
06

Calculate the percentage of pyridine that forms pyridinium ion

Now that we have the value of x, which represents the concentration of the pyridinium ion at equilibrium, we can calculate the percentage of pyridine that forms the pyridinium ion: \[\text{Percentage} = \frac{[C_5H_5NH^+]}{[C_5H_5N]_{initial}} \times 100\%\] \[\text{Percentage} = \frac{1.3 \times 10^{-5}}{0.10} \times 100\% = 1.3 \times 10^{-4} \times 100\% = 0.0013\%\] Thus, the percentage of pyridine that forms pyridinium ion in a \(0.10-M\) aqueous solution is approximately \(0.0013\%\).

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Most popular questions from this chapter

Consider a \(0.60-M\) solution of \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\), lactic acid \(\left(K_{\mathrm{a}}=\right.\) \(\left.1.4 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) ii. \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}\) iii. \(\mathrm{H}^{+}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{OH}^{-}\) b. Complete the following ICE table in terms of \(x\), the amount \((\mathrm{mol} / \mathrm{L})\) of lactic acid that dissociates to reach equilibrium. c. What is the equilibrium concentration for \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{3}^{-}\) ? d. Calculate the \(\mathrm{pH}\) of the solution.

Classify each of the following as a strong acid, weak acid, strong base, or weak base in aqueous solution. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{HNO}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) d. \(\mathrm{NaOH}\) e. \(\mathrm{NH}_{3}\) f. \(\mathrm{HF}\) g. h. \(\mathrm{Ca}(\mathrm{OH})_{2}\) i. \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. \(\mathrm{Li}_{2} \mathrm{O}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{SrO}\)

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.

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{~g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).
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