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Chapter 4: Problem 13

Differentiate between what happens when the following are added to water. a. polar solute versus nonpolar solute b. KF versus \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. \(\mathrm{RbCl}\) versus \(\mathrm{AgCl}\) d. \(\mathrm{HNO}_{3}\) versus \(\mathrm{CO}\)

Short Answer

Expert verified
In summary, polar solutes dissolve well in water due to attractive interactions with polar water molecules, while nonpolar solutes do not dissolve well and form separate phases. KF and \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) both dissolve in water, but KF dissolves more readily due to its ionic nature. \(\mathrm{RbCl}\) dissolves in water, while \(\mathrm{AgCl}\) forms a precipitate due to its low solubility. \(\mathrm{HNO}_{3}\) ionizes and dissolves in water, making the solution acidic, while \(\mathrm{CO}\) has limited solubility and remains mostly in the gas phase.

Step by step solution

01

1. Polar solutes versus nonpolar solutes

When added to water, polar solutes typically dissolve well because they are attracted to the polar water molecules. The positive and negative charges of the polar solute interact with the negative oxygen and positive hydrogen atoms of the water molecules, allowing the solute to dissolve. On the other hand, nonpolar solutes do not dissolve well in water because they lack the charge needed to interact with the polar water molecules. Nonpolar solutes will minimally interact with water, clustering together and forming a separate phase.
02

2. KF versus \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

When KF (potassium fluoride) is added to water, it will readily dissolve due to its ionic nature. The ionic bonds between potassium (K+) and fluoride (F-) ions break, resulting in hydrated ions surrounded by water molecules. As for \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose), it is a polar covalent compound with multiple hydroxyl groups (-OH) that can form hydrogen bonds with water. Therefore, glucose can also dissolve in water, though to a lesser extent than KF.
03

3. \(\mathrm{RbCl}\) versus \(\mathrm{AgCl}\)

\(\mathrm{RbCl}\) (rubidium chloride) is an ionic compound with a soluble nature. When added to water, the ionic bond between rubidium (Rb+) and chloride (Cl-) ions breaks, resulting in hydrated ions surrounded by water molecules. However, \(\mathrm{AgCl}\) (silver chloride) is an ionic compound with low solubility in water. The ionic bond between silver (Ag+) and chloride (Cl-) ions is strong, which makes the compound's dissolution process limited. Instead of fully dissolving, \(\mathrm{AgCl}\) will form a precipitate when added to water.
04

4. \(\mathrm{HNO}_{3}\) versus \(\mathrm{CO}\)

\(\mathrm{HNO}_{3}\) (nitric acid) is a polar covalent compound that ionizes in water, releasing hydrogen (H+) ions and nitrate (NO3-) ions, making the solution acidic. Due to its polarity and ability to form hydrogen bonds with water, \(\mathrm{HNO}_{3}\) is highly soluble in water. On the other hand, \(\mathrm{CO}\) (carbon monoxide) is a nonpolar covalent compound with a very limited solubility in water. Since its molecular structure lacks charges or functional groups capable of interacting with water molecules, \(\mathrm{CO}\) doesn't readily dissolve in water and remains mostly in the gas phase.

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Most popular questions from this chapter

Douglasite is a mineral with the formula \(2 \mathrm{KCl} \cdot \mathrm{FeCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\). Calculate the mass percent of douglasite in a \(455.0-\mathrm{mg}\) sample if it took \(37.20 \mathrm{~mL}\) of a \(0.1000-M \mathrm{AgNO}_{3}\) solution to precipitate all the \(\mathrm{Cl}^{-}\) as \(\mathrm{AgCl}\). Assume the douglasite is the only source of chloride ion.

Consider the reaction of \(19.0 \mathrm{~g}\) of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and \(29.0 \mathrm{~g}\) of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture.

Assign oxidation numbers to all the atoms in each of the following. a. \(\mathrm{SrCr}_{2} \mathrm{O}_{7}\) b. \(\mathrm{CuCl}_{2}\) c. \(\mathrm{O}_{2}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) e. \(\mathrm{MgCO}_{3}\) f. \(\mathrm{Ag}\) g. \(\mathrm{PbSO}_{3}\) h. \(\mathrm{PbO}_{2}\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) j. \(\mathrm{CO}_{2}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) l. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

In the spectroscopic analysis of many substances, a series of standard solutions of known concentration are measured to generate a calibration curve. How would you prepare standard solutions containing \(10.0,25.0,50.0,75.0\), and \(100 . \mathrm{ppm}\) of copper from a commercially produced \(1000.0\) -ppm solution? Assume each solution has a final volume of \(100.0 \mathrm{~mL}\). (See Exercise 121 for definitions.)

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{UO}_{2}^{2+}\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}\) c. \(\mathrm{NaBiO}_{3}\) d. As \(_{4}\) e. \(\mathrm{HAsO}_{2}\) f. \(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\) g. \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) h. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) i. \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)
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