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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
When added to water, polar solutes dissolve well due to interaction with water's polar molecules, while nonpolar solutes have poor solubility. KF dissociates into ions, and glucose dissolves without dissociating. RbCl is highly soluble, while AgCl has poor solubility due to strong ionic bonds. HNO₃ dissociates completely, making the solution acidic, while CO does not dissolve well or alter the pH.

Step by step solution

01

(a. Polar Solute vs Nonpolar Solute)

When a polar solute is added to water, it will dissolve well because both the solute and water are polar in nature. The polar solute's molecules have areas of positive and negative charges, and these charges interact with water's polar molecules through intermolecular forces. This interaction is the main reason behind the solute's solubility in water. On the other hand, when a nonpolar solute is added to water, it won't dissolve well because the nonpolar solute's molecules lack areas of positive and negative charges. As a result, interactions with water's polar molecules are minimum, resulting in poor solubility.
02

(b. KF vs C6H12O6)

When potassium fluoride (KF) is added to water, it dissolves well and dissociates into potassium (K⁺) and fluoride (F⁻) ions. This is because KF is an ionic compound and water can break the ionic bonds due to its high polarity. On the other hand,\_C\_{6}H\_{12}O\_{6}\_ (glucose) is a polar covalent compound, so it dissolves well in water due to its polar nature. But unlike KF, glucose doesn't dissociate into ions when dissolved.
03

(c. RbCl vs AgCl)

Rubidium chloride (RbCl) is a highly soluble ionic compound in water. When RbCl is added to water, it dissociates into rubidium (Rb⁺) and chloride (Cl⁻) ions, which are stabilized by hydration. On the other hand, silver chloride (AgCl) has poor solubility in water due to its low dissociation into silver (Ag⁺) and chloride (Cl⁻) ions. The reason for AgCl's poor solubility lies in its strong ionic bond, difficult to be broken by water molecules.
04

(d. HNO3 vs CO)

Nitric acid (HNO₃) is a strong acid, which means that when it's added to water, it will dissociate completely into hydrogen (H⁺) and nitrate (NO₃⁻) ions. The solution will be highly acidic. On the contrary, carbon monoxide (CO) is a nonpolar covalent gas, and it does not dissolve well in water. Also, it does not react with water to form ions and does not alter the pH of the solution.

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Most popular questions from this chapter

If \(10 . \mathrm{g}\) of \(\mathrm{AgNO}_{3}\) is available, what volume of \(0.25 \mathrm{M} \mathrm{AgNO}_{3}\) solution can be prepared?

A solution was prepared by mixing \(50.00 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) and \(100.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the molarity of the final solution of nitric acid.

Which of the following solutions of strong electrolytes contains the largest number of ions: \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}, 50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{BaCl}_{2}\), or \(75.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{Na}_{4} \mathrm{PO}_{4}\) ?

Many plants are poisonous because their stems and leaves contain oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) or sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\); when ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\) in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) in aqueous solution.

A \(30.0\) -mL sample of an unknown strong base is neutralized after the addition of \(12.0 \mathrm{~mL}\) of a \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) solution. If the unknown base concentration is \(0.0300 M\), give some possible identities for the unknown base.
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