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Chapter 16: Problem 35

A 275 mL sample of vapor in equilibrium with 1-propylamine at \(25.0^{\circ} \mathrm{C}\) is removed and dissolved in \(0.500 \mathrm{L} \mathrm{H}_{2} \mathrm{O} .\) For 1 -propylamine, \(\mathrm{p} K_{\mathrm{b}}=3.43\) and v.p. \(=316\) Torr. (a) What should be the pH of the aqueous solution? (b) How many \(\mathrm{mg}\) of \(\mathrm{NaOH}\) dissolved in \(0.500 \mathrm{L}\) of water give the same pH?

Short Answer

Expert verified
(a) The pH of the aqueous solution should be 11.79. (b) 122.6 mg of NaOH dissolved in 0.500 L of water will give the same pH.

Step by step solution

01

Calculate the concentration of 1-propylamine

\[n = PV/RT \]\[n = (316 Torr)(275 mL) / (62.364 L Torr /mol K)(298.15 K) = 0.00462 mol\]\[C = n/V = 0.00462 mol/0.500 L = 0.00923 M\]
02

Calculate the hydroxide ion concentration

For base ionization of 1-propylamine: \n\[NH3 + H2O <=> NH4+ + OH-\]Kb = [NH4+][OH-] / [NH3]. Hence, \[OH-] = √(Kb * [NH3])\]First, calculate Kb from pKb: \n\[Kb = 10^-(pKb) = 10^-3.43 = 3.72 \times 10^-4\]Next, calculate [OH-]:\n\[[OH-] = √((3.72 \times 10^-4) * 0.00923) = 6.13 \times 10^-3 M\]
03

Calculate the pH of the solution

First, calculate pOH from [OH-]: \n\[pOH = -log[OH-] = -log(6.13 \times 10^-3) = 2.21\]Then, calculate pH: \n\[pH = 14 - pOH = 14 - 2.21 = 11.79\]
04

Calculate the concentration of NaOH

Since NaOH is a strong base and fully dissociates in water, the [OH-] is the same as the concentration of the NaOH, which is [OH-] = 6.13 \times 10^-3 M.
05

Calculate the mass of NaOH

\[n = C \times V = (6.13 \times 10^-3 M) \times (0.500 L) = 3.065 \times 10^-3 mol\]Then, calculate the mass: \n\[m = n \times MW = 3.065 \times 10^-3 mol \times 40 g/mol = 0.1226 g\],or convert to mg: \n\[m = 0.1226 g \times 1000 mg/g = 122.6 mg\]

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Most popular questions from this chapter

Propionic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH},\) is \(0.42 \%\) ionized in \(0.80 \mathrm{M}\) solution. The \(K_{\mathrm{a}}\) for this acid is (a) \(1.42 \times 10^{-5}\) (b) \(1.42 \times 10^{-7} ;\) (c) \(1.77 \times 10^{-5} ;\) (d) \(6.15 \times 10^{4}\) (e) none of these.

Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\left(\mathrm{p} K_{\mathrm{b}}=8.82\right),\) forms a salt, pyridinium chloride, as a result of a reaction with HCl. Write an ionic equation to represent the hydrolysis of the pyridinium ion, and calculate the \(\mathrm{pH}\) of \(0.0482 \mathrm{M} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \mathrm{Cl}^{-}(\mathrm{aq})\).

With which of the following bases will the ionization of acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) proceed furthest toward completion (to the right): (a) \(\mathrm{H}_{2} \mathrm{O} ;\) (b) \(\mathrm{NH}_{3} ;\) (c) \(\mathrm{Cl}^{-}\) (d) \(\mathrm{NO}_{3}^{-} ?\) Explain your answer.

Write the formula of the conjugate base in the reaction of each acid with water. (a) \(\mathrm{HIO}_{3} ;\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (c) \(\mathrm{HPO}_{4}^{2-} ;\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\)

Explain why trichloroacetic acid, \(\mathrm{CCl}_{3} \mathrm{COOH},\) is a stronger acid than acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\).
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