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Chapter 15: Problem 90

Describe the intermolecular forces and changes in bonding that lead to the formation of a basic solution when methylamine (CH \(_{3} \mathrm{NH}_{2}\) ) dissolves in water.

Short Answer

Expert verified
Answer: The primary intermolecular force involved in the dissolution of methylamine in water is dipole-dipole interaction due to the polar nature of both molecules. Methylamine behaves as a Bronsted-Lowry base by accepting a proton from water, forming ammonium and hydroxide ions, which leads to a basic solution.

Step by step solution

01

Identify intermolecular forces involved in the dissolution of methylamine in water

To start, we need to identify the type of intermolecular forces involved between the molecules of methylamine and water. Methylamine (CH\(_{3}\)NH\(_{2}\)) is a polar molecule because the nitrogen atom has a lone pair of electrons and forms polar bonds with hydrogen atoms. Water (H\(_{2}\)O) is also polar due to the difference in electronegativity between oxygen and hydrogen atoms. When a polar solute dissolves in a polar solvent, the primary intermolecular force involved is dipole-dipole interaction.
02

Discuss changes in bonding that lead to the formation of a basic solution

When methylamine dissolves in water, the lone pair of electrons on the nitrogen atom in methylamine can accept a proton (H\(^+\)) from water to form ammonium ion (CH\(_{3}\)NH\(_{3}^+\)) and hydroxide ion (OH\(^-\)) according to the following reaction: CH\(_{3}\)NH\(_{2}\) + H\(_{2}\)O \(\rightleftharpoons\) CH\(_{3}\)NH\(_{3}^+\) + OH\(^-\) Since OH\(^-\) ions are produced in the above reaction, the solution becomes basic.
03

Summarize the intermolecular forces and changes in bonding

In summary, the primary intermolecular force involved in the dissolution of methylamine in water is dipole-dipole interaction due to the polar nature of both methylamine and water molecules. Upon dissolving in water, methylamine behaves as a Bronsted-Lowry base by accepting a proton from water and forming ammonium and hydroxide ions, which leads to a basic solution.

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Most popular questions from this chapter

Which of the following salts produces an acidic solution in water: ammonium acetate, ammonium nitrate, or sodium formate?

Are all Arrhenius bases also Bronsted-Lowry bases? Are \(15.9=\) all Bronsted-Lowry bases also Arrhenius bases? If yes, explain why. If not, give a specific example to demonstrate the difference.

At equilibrium, the value of \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in \(0.125 \mathrm{M}\) of an unknown acid is \(4.07 \times 10^{-3} M .\) Determine the degree of ionization and the \(K_{\mathrm{a}}\) of this acid.

Predict which solution in each pair below will have the lower pH. a. \(2.56 \times 10^{-2} \mathrm{M} \mathrm{HCl}\) or \(4.09 \times 10^{-2} \mathrm{MHBr}\) b. \(1.00 \times 10^{-5} M\) acctic acid \(\left(K_{\mathrm{a}}=1.76 \times 10^{-5}\right)\) or \(1.00 \times 10^{-5} M\) formic acid \(\left(K_{a}=1.77 \times 10^{-4}\right)\) c. \(22 \mathrm{mM} \mathrm{CH}_{3} \mathrm{NH}_{2}\left(\mathrm{p} K_{\mathrm{b}}=3.36\right)\) or \(22 \mathrm{mM}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\left(K_{\mathrm{b}}=5.9 \times 10^{-4}\right)\) d. \(158 \mathrm{mM} \mathrm{NH}_{3}\left(\mathrm{p} K_{\mathrm{b}}=4.75\right)\) or \(158 \mathrm{m} M\) acetic acid \(\left(p K_{a}=4.75\right)\) e. \(0.00395 M \mathrm{HNO}_{3}\) or \(0.00145 M \mathrm{HClO}_{4}\) f. \(2.05 \times 10^{-1} M\) propionic acid \(\left(K_{2}=1.4 \times 10^{-5}\right)\) or \(2.05 \times 10^{-1} M\) fluoroacetic acid \(\left(K_{2}=2.6 \times 10^{-3}\right)\) g. 375 m \(M\) pyridine \(\left(p K_{b}=8.77\right)\) or \(375 \mathrm{mM}\) aniline \(\left(\mathrm{p} K_{\mathrm{b}}=9.40\right)\) h. \(0.555 M \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\left(K_{2}=3 \times 10^{-3}\right)\) or \(0.355 M \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\left(K_{2}=1 \times 10^{-4}\right)\)

In an aqueous solution of HNO \(_{3},\) which compound acts as a Bronsted-Lowry acid and which is the Bronsted-Lowry base?
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