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Which ion in each of the following pairs would you expect to be more strongly hydrated? Why? a. \(\mathrm{Na}^{+}\) or \(\mathrm{Mg}^{2+}\) b. \(\mathrm{Mg}^{2+}\) or \(\mathrm{Be}^{2+}\) c. \(\mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\) d. \(\mathrm{F}^{-}\) or \(\mathrm{Br}^{-}\) e. \(\mathrm{Cl}^{-}\) or \(\mathrm{ClO}_{4}^{-}\) f. \(\mathrm{ClO}_{4}^{-}\) or \(\mathrm{SO}_{4}^{2-}\)

Short Answer

Expert verified
a. \(\mathrm{Mg}^{2+}\) is more strongly hydrated because it has a higher charge than \(\mathrm{Na}^{+}\). b. \(\mathrm{Be}^{2+}\) is more strongly hydrated because it has a smaller atomic radius than \(\mathrm{Mg}^{2+}\). c. \(\mathrm{Fe}^{3+}\) is more strongly hydrated because it has a higher charge than \(\mathrm{Fe}^{2+}\). d. \(\mathrm{F}^{-}\) is more strongly hydrated because it has a smaller atomic radius than \(\mathrm{Br}^{-}\). e. \(\mathrm{Cl}^{-}\) is more strongly hydrated because it has a higher effective charge density than \(\mathrm{ClO}_{4}^{-}\). f. \(\mathrm{SO}_{4}^{2-}\) is more strongly hydrated because it has a higher negative charge than \(\mathrm{ClO}_{4}^{-}\).

Step by step solution

01

Compare the charges of each ion.

Since \(\mathrm{Mg}^{2+}\) has a higher charge than \(\mathrm{Na}^{+}\), it is more strongly hydrated. b. \(\mathrm{Mg}^{2+}\) or \(\mathrm{Be}^{2+}\)
02

Compare the sizes of each ion.

Since both ions, \(\mathrm{Mg}^{2+}\) and \(\mathrm{Be}^{2+}\), have the same charges, we need to compare their sizes. Be has a smaller atomic radius than Mg so, \(\mathrm{Be}^{2+}\) will be more strongly hydrated. c. \(\mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\)
03

Compare the charges of each ion.

Since \(\mathrm{Fe}^{3+}\) has a higher charge than \(\mathrm{Fe}^{2+}\), it is more strongly hydrated. d. \(\mathrm{F}^{-}\) or \(\mathrm{Br}^{-}\)
04

Compare the sizes of each ion.

Since both ions, \(\mathrm{F}^{-}\) and \(\mathrm{Br}^{-}\) have equal charges, we need to compare their sizes. F has a smaller atomic radius than Br, so \(\mathrm{F}^{-}\) will be more strongly hydrated. e. \(\mathrm{Cl}^{-}\) or \(\mathrm{ClO}_{4}^{-}\)
05

Compare the effective charge densities of each ion.

Though both ions have equal charges, the \(\mathrm{ClO}_{4}^{-}\) ion is much larger compared to the \(\mathrm{Cl}^{-}\) ion due to the presence of the oxygen atoms. Therefore, \(\mathrm{Cl}^{-}\) has a higher effective charge density than \(\mathrm{ClO}_{4}^{-}\) and will be more strongly hydrated. f. \(\mathrm{ClO}_{4}^{-}\) or \(\mathrm{SO}_{4}^{2-}\)
06

Compare the charges of each ion.

Since \(\mathrm{SO}_{4}^{2-}\) has a higher negative charge than \(\mathrm{ClO}_{4}^{-}\), it is more strongly hydrated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ion Charge
Ion charge is a crucial factor in understanding the hydration of ions in solution. Simply put, an ion with a higher charge will attract water molecules more strongly compared to an ion with a lower charge. This is because the electric field generated by a charged ion interacts with the slight positive charges on the hydrogen atoms in water.
In scenarios where you are comparing two ions like \[ \mathrm{Na}^{+} \text{ and } \mathrm{Mg}^{2+} \] or \[ \mathrm{Fe}^{2+} \text{ and } \mathrm{Fe}^{3+} \], you can expect \[ \mathrm{Mg}^{2+} \text{ or } \mathrm{Fe}^{3+} \] to form stronger hydration shells due to their higher charges.
In essence, the higher the charge of the ion, the stronger the attraction to water molecules, leading to more significant hydration.
Atomic Radius
The atomic radius of an ion plays a vital role in its hydration. Smaller ions have a greater charge concentration in a smaller space, which means that they attract water molecules more intensely. This results in stronger hydration because water molecules can get closer to the ion's charge center.
Compare \[ \mathrm{Be}^{2+} \] and \[ \mathrm{Mg}^{2+} \]. Despite having the same charge, \[ \mathrm{Be}^{2+} \] is smaller due to its smaller atomic radius, leading to it having strong hydration.
Similarly, between fluoride \[ \mathrm{F}^{-} \] and bromide \[ \mathrm{Br}^{-} \], the smaller fluoride ion will exhibit stronger hydration because of its smaller size. Therefore, a smaller atomic radius increases the hydration strength of an ion.
Effective Charge Density
Effective charge density refers to the charge of an ion in relation to its size. It's significant because it helps determine how strongly an ion interacts with water. The concept combines both ion charge and atomic radius.
Imagine two equally charged ions like \[ \mathrm{Cl}^{-} \text{ and } \mathrm{ClO}_{4}^{-} \]. Although they hold equivalent charges, \[ \mathrm{Cl}^{-} \] is smaller and thus, its effective charge density is greater, leading it to be more strongly hydrated.
Effective charge density ultimately enhances understanding of how ions behave in solutions, particularly which ions will attract water more strongly. The higher the effective charge density, the more substantial the ion's hydration.
Strong Hydration
Strong hydration is the term used to describe ions that are particularly good at attracting water molecules. The factors affecting this are the ion's charge, atomic radius, and effective charge density.
For instance, when comparing ions, such as \[ \mathrm{SO}_{4}^{2-} \text{ and } \mathrm{ClO}_{4}^{-} \], charge plays a key role. The higher negative charge of sulfate ensures more robust hydration compared to perchlorate.
Additionally, smaller ions or those with high effective charge density such as \[ \mathrm{Be}^{2+} \text{ and } \mathrm{F}^{-} \] also manifest strong hydration. Understanding strong hydration is essential because it affects how ions behave and move in solutions, which is key in many chemical and biological systems.

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Most popular questions from this chapter

Consider an aqueous solution containing sodium chloride that has a density of 1.01 \(\mathrm{g} / \mathrm{mL}\) . Assume the solution behaves ideally. The freezing point of this solution at 1.0 \(\mathrm{atm}\) is \(-1.28^{\circ} \mathrm{C}\) . Calculate the percent composition of this solution (by mass).

A solution of phosphoric acid was made by dissolving 10.0 \(\mathrm{g}\) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in 100.0 \(\mathrm{mL}\) water. The resulting volume was 104 \(\mathrm{mL}\) . Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 \(\mathrm{g} / \mathrm{cm}^{3}\) .

A 0.500 -g sample of a compound is dissolved in enough water to form 100.0 mL of solution. This solution has an osmotic pressure of 2.50 atm at \(25^{\circ} \mathrm{C}\) . If each molecule of the solute dissociates into two particles (in this solvent), what is the molar mass of this solute?

An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57\(\% \mathrm{C}\) and 5.30\(\%\) H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of \(-5.20^{\circ} \mathrm{C}\) is recorded for a solution made by dissolving 10.56 \(\mathrm{g}\) of the compound in 25.0 \(\mathrm{g}\) water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

How does \(\Delta H_{\mathrm{soln}}\) relate to deviations from Raoult’s law? Explain.

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