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For each of the following pairs, predict which substance is more soluble in water. a. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) or \(\mathrm{NH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{CN}\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) d. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) f. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) or \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Short Answer

Expert verified
a. \(\mathrm{NH}_{3}\) is more soluble. b. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) is more soluble. c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) is more soluble. d. \(\mathrm{CH}_{3} \mathrm{OH}\) is more soluble. e. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) is more soluble. f. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) is more soluble.

Step by step solution

01

a. Comparing \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) and \(\mathrm{NH}_{3}\)

Both of these substances are polar and will engage in hydrogen bonding with water. While both will be soluble, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) has an additional methyl group which is nonpolar and may cause a slight decrease in solubility. Therefore, \(\mathrm{NH}_{3}\) is more soluble in water.
02

b. Comparing \(\mathrm{CH}_{3} \mathrm{CN}\) and \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\)

The polarity of \(\mathrm{CH}_{3} \mathrm{CN}\) is due to the cyano group, while the polarity of \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) is due to hydrogen bonding with the oxygen. In this case, \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) is more soluble in water due to stronger hydrogen bonding.
03

c. Comparing \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) is polar due to the hydroxyl group, which can engage in hydrogen bonding with water. On the other hand, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) is nonpolar and has only weak van der Waals forces. Therefore, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) is more soluble in water.
04

d. Comparing \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

Both of these substances are polar due to the hydroxyl group. However, \(\mathrm{CH}_{3} \mathrm{OH}\) has a smaller nonpolar region than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), making it slightly more soluble in water.
05

e. Comparing \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) and \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\)

Both of these substances have a polar hydroxyl group capable of hydrogen bonding. However, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) has a larger nonpolar region than \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\), which decreases its solubility in water. Therefore, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) is more soluble in water.
06

f. Comparing \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Both of these substances are polar. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) engages in hydrogen bonding with water, while \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) can form hydrogen bonds and also dissociate into ions. In this case, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) is more soluble in water due to its ability to dissociate into ions, which further increases the solubility.

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Most popular questions from this chapter

You make \(20.0 \mathrm{g}\) of a sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) and NaCl mixture and dissolve it in \(1.00 \mathrm{kg}\) water. The freezing point of this solution is found to be \(-0.426^{\circ} \mathrm{C}\). Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

The freezing-point depression of a 0.091-m solution of CsCl is \(0.320^{\circ} \mathrm{C} .\) The freezing-point depression of a \(0.091-m\) solution of \(\mathrm{CaCl}_{2}\) is \(0.440^{\circ} \mathrm{C}\). In which solution does ion association appear to be greater? Explain.

What is ion pairing?

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For an acid or a base, when is the normality of a solution equal to the molarity of the solution and when are the two concentration units different?

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